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[{"Name":"Lectures And Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro to Thermodynamics","Duration":"2m 36s","ChapterTopicVideoID":11919,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/11919.jpeg","UploadDate":"2018-03-29T10:50:43.4000000","DurationForVideoObject":"PT2M36S","Description":null,"MetaTitle":"Intro to Thermodynamics: Video + Workbook | Proprep","MetaDescription":"Thermodynamics - Lectures And Exercises. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/thermodynamics/lectures-and-exercises/vid12347","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.370","Text":"Hello. In this lesson,"},{"Start":"00:02.370 ","End":"00:05.985","Text":"we\u0027re going to be learning an introduction to thermodynamics."},{"Start":"00:05.985 ","End":"00:10.680","Text":"Thermo means heat and dynamics means movement."},{"Start":"00:10.680 ","End":"00:14.595","Text":"Why is the movement of heat important?"},{"Start":"00:14.595 ","End":"00:22.035","Text":"Thermodynamics started with the invention of the steam engine."},{"Start":"00:22.035 ","End":"00:27.930","Text":"That is actually what started or what kicked off the industrial revolution."},{"Start":"00:27.930 ","End":"00:33.825","Text":"The science of thermodynamics can be traced back to the industrial revolution,"},{"Start":"00:33.825 ","End":"00:37.120","Text":"approximately 200 years ago."},{"Start":"00:37.120 ","End":"00:39.350","Text":"There are lots of engines,"},{"Start":"00:39.350 ","End":"00:41.975","Text":"there is a steam engine, diesel engine,"},{"Start":"00:41.975 ","End":"00:48.530","Text":"engines that use different types of petrol and all of these come from thermodynamics."},{"Start":"00:48.530 ","End":"00:55.950","Text":"All of these engines work by utilizing a difference in temperature."},{"Start":"00:56.080 ","End":"01:00.155","Text":"Up until today, these engines,"},{"Start":"01:00.155 ","End":"01:03.740","Text":"which use these heat differences,"},{"Start":"01:03.740 ","End":"01:05.990","Text":"and you use the laws of thermodynamics,"},{"Start":"01:05.990 ","End":"01:15.105","Text":"haven\u0027t been able to become 100 percent or even very close to 100 percent efficiency."},{"Start":"01:15.105 ","End":"01:20.635","Text":"A lot of the energy is lost to heat rather than to producing work,"},{"Start":"01:20.635 ","End":"01:24.250","Text":"which is what we actually want the engines to be doing."},{"Start":"01:24.250 ","End":"01:26.665","Text":"Later on in this chapter,"},{"Start":"01:26.665 ","End":"01:32.155","Text":"we\u0027re going to be learning about a theoretical engine called the Carnot engine."},{"Start":"01:32.155 ","End":"01:40.090","Text":"This theoretical engine is considered the most efficient engine and what scientists"},{"Start":"01:40.090 ","End":"01:43.840","Text":"today that work in the field of thermodynamics are trying to do is"},{"Start":"01:43.840 ","End":"01:48.070","Text":"to create a real-life replica of the Carnot engine."},{"Start":"01:48.070 ","End":"01:57.290","Text":"That means to create an engine that works in reality at close to 100 percent efficiency."},{"Start":"01:57.290 ","End":"02:01.150","Text":"To conclude, thermodynamics is"},{"Start":"02:01.150 ","End":"02:05.934","Text":"the study of the relationship between heat and temperature,"},{"Start":"02:05.934 ","End":"02:08.550","Text":"with work and energy."},{"Start":"02:08.550 ","End":"02:15.005","Text":"Thermodynamics can be applied to lots and lots of different studies and researches."},{"Start":"02:15.005 ","End":"02:17.315","Text":"Thermodynamics is used extensively in"},{"Start":"02:17.315 ","End":"02:22.445","Text":"aerospace engineering and cell biology in the study of black holes,"},{"Start":"02:22.445 ","End":"02:26.480","Text":"in chemistry, biomedical engineering, and even economics."},{"Start":"02:26.480 ","End":"02:30.455","Text":"What are we going to be learning in this chapter is very useful,"},{"Start":"02:30.455 ","End":"02:33.250","Text":"not just for physicists."},{"Start":"02:33.250 ","End":"02:36.160","Text":"That\u0027s the end of this lesson."}],"ID":12347},{"Watched":false,"Name":"Heat, Temperature and Heat Capacity","Duration":"13m 39s","ChapterTopicVideoID":11920,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.610","Text":"Hello. In this lesson,"},{"Start":"00:02.610 ","End":"00:04.905","Text":"we\u0027re going to be learning about the terms heat,"},{"Start":"00:04.905 ","End":"00:07.665","Text":"temperature, and heat capacity."},{"Start":"00:07.665 ","End":"00:10.469","Text":"Heat is a type of energy."},{"Start":"00:10.469 ","End":"00:17.340","Text":"Heat is the thermal energy transferred from a hotter system to a cooler system."},{"Start":"00:17.340 ","End":"00:22.005","Text":"Now, heat is generally denoted by the letter Q,"},{"Start":"00:22.005 ","End":"00:25.230","Text":"sometimes also by a q,"},{"Start":"00:25.230 ","End":"00:33.390","Text":"and its units are joules because heat is a type of energy."},{"Start":"00:33.390 ","End":"00:36.730","Text":"The units are in joules."},{"Start":"00:36.730 ","End":"00:42.965","Text":"Now, heat is also called a process quantity"},{"Start":"00:42.965 ","End":"00:50.545","Text":"because heat is defined by the process with which energy can be transferred."},{"Start":"00:50.545 ","End":"00:53.765","Text":"Let\u0027s say you\u0027ve baked some cookies and"},{"Start":"00:53.765 ","End":"00:57.380","Text":"you reach into the oven to take the cookie tray out."},{"Start":"00:57.380 ","End":"01:00.830","Text":"We know that the tray is going to be very, very hot."},{"Start":"01:00.830 ","End":"01:03.215","Text":"In regards to heat,"},{"Start":"01:03.215 ","End":"01:07.700","Text":"we don\u0027t speak about the heat of the tray,"},{"Start":"01:07.700 ","End":"01:15.400","Text":"but rather we speak about the heat transferred from the cookie tray onto your hands."},{"Start":"01:15.400 ","End":"01:22.005","Text":"Now a final note on heat is that it\u0027s an extensive property."},{"Start":"01:22.005 ","End":"01:27.470","Text":"That means that heat is dependent on how many molecules are"},{"Start":"01:27.470 ","End":"01:33.125","Text":"in the object or the system that we\u0027re dealing with."},{"Start":"01:33.125 ","End":"01:39.320","Text":"If we have a system with 1,000 molecules or a million molecules,"},{"Start":"01:39.320 ","End":"01:43.100","Text":"the heat of that system is going to be different."},{"Start":"01:43.100 ","End":"01:45.260","Text":"Then we have temperature."},{"Start":"01:45.260 ","End":"01:48.365","Text":"Now, temperature is different to heat."},{"Start":"01:48.365 ","End":"01:52.805","Text":"Temperature is the measure of the average kinetic energy of a system."},{"Start":"01:52.805 ","End":"01:57.605","Text":"We take the kinetic energy of every single atom or molecule in a system,"},{"Start":"01:57.605 ","End":"02:02.615","Text":"and then we take the average kinetic energy of all of the molecules or atoms,"},{"Start":"02:02.615 ","End":"02:05.799","Text":"and then that will give us the temperature."},{"Start":"02:05.799 ","End":"02:11.840","Text":"The temperature is typically measured in units of degrees,"},{"Start":"02:11.840 ","End":"02:19.555","Text":"so degrees Celsius or degrees Fahrenheit or Kelvin."},{"Start":"02:19.555 ","End":"02:26.930","Text":"These are examples of units that can be used for temperature."},{"Start":"02:26.930 ","End":"02:33.815","Text":"Temperature is denoted by the letter T. Now,"},{"Start":"02:33.815 ","End":"02:37.670","Text":"temperature is an intensive property."},{"Start":"02:37.670 ","End":"02:39.140","Text":"What does that mean?"},{"Start":"02:39.140 ","End":"02:46.749","Text":"Temperature is independent on the mass or how many molecules you have in the system."},{"Start":"02:46.749 ","End":"02:48.545","Text":"That\u0027s different to heat."},{"Start":"02:48.545 ","End":"02:54.515","Text":"Heat is dependent on how many molecules are in the system or the mass of the system,"},{"Start":"02:54.515 ","End":"02:57.985","Text":"temperature is independent of it."},{"Start":"02:57.985 ","End":"03:03.215","Text":"This property that the fact that temperature is an intensive property"},{"Start":"03:03.215 ","End":"03:08.945","Text":"means that chemists or scientists can use the melting point to"},{"Start":"03:08.945 ","End":"03:14.330","Text":"identify pure substances because the temperature at which"},{"Start":"03:14.330 ","End":"03:21.085","Text":"a substance melts is independent of the amount of the sample that we have."},{"Start":"03:21.085 ","End":"03:25.520","Text":"As we said, temperature is the average kinetic energy of a system,"},{"Start":"03:25.520 ","End":"03:32.600","Text":"so if we take our mug of hot tea again or a nice cold glass of iced tea,"},{"Start":"03:32.600 ","End":"03:37.970","Text":"we know that the kinetic energy in the mug of hot tea is"},{"Start":"03:37.970 ","End":"03:44.245","Text":"going to be higher than the average kinetic energy of a glass of iced tea."},{"Start":"03:44.245 ","End":"03:47.870","Text":"Now we\u0027re going to speak about heat capacity."},{"Start":"03:47.870 ","End":"03:50.990","Text":"Heat capacity is how much energy is"},{"Start":"03:50.990 ","End":"03:55.610","Text":"required in order to change the temperature of a substance."},{"Start":"03:55.610 ","End":"04:00.800","Text":"First of all, we can see that heat capacity connects these 2 terms,"},{"Start":"04:00.800 ","End":"04:03.310","Text":"heat and temperature, together."},{"Start":"04:03.310 ","End":"04:08.915","Text":"Now, heat capacity is also an extensive property."},{"Start":"04:08.915 ","End":"04:12.215","Text":"It depends on the amount of substance that we have,"},{"Start":"04:12.215 ","End":"04:14.030","Text":"just like with heat."},{"Start":"04:14.030 ","End":"04:16.530","Text":"The clue was in the name, both have heat inside,"},{"Start":"04:16.530 ","End":"04:18.875","Text":"so they\u0027re both extensive properties."},{"Start":"04:18.875 ","End":"04:24.840","Text":"Heat capacity is unique to every different type of substance."},{"Start":"04:25.220 ","End":"04:30.305","Text":"If we\u0027re speaking about water molecules, H_2O,"},{"Start":"04:30.305 ","End":"04:37.040","Text":"they will have a heat capacity which is different to the molecules in,"},{"Start":"04:37.040 ","End":"04:40.805","Text":"for instance, some liquid metal."},{"Start":"04:40.805 ","End":"04:45.045","Text":"Now, heat capacity has 2 subcategories."},{"Start":"04:45.045 ","End":"04:47.885","Text":"There\u0027s either the specific heat capacity"},{"Start":"04:47.885 ","End":"04:51.220","Text":"or it\u0027s also sometimes called the specific heat,"},{"Start":"04:51.220 ","End":"04:55.580","Text":"and there is also the molar heat capacity."},{"Start":"04:55.580 ","End":"04:59.840","Text":"The specific heat capacity is the energy required to raise the temperature"},{"Start":"04:59.840 ","End":"05:04.349","Text":"of 1 gram of substance by 1 degree Kelvin."},{"Start":"05:04.349 ","End":"05:07.490","Text":"The molar heat capacity is the energy required to raise"},{"Start":"05:07.490 ","End":"05:12.840","Text":"the temperature of 1 mole of substance by 1 Kelvin."},{"Start":"05:12.840 ","End":"05:15.785","Text":"That\u0027s the difference."},{"Start":"05:15.785 ","End":"05:18.695","Text":"The units are as follows."},{"Start":"05:18.695 ","End":"05:23.245","Text":"The specific heat capacity is dependent on grams,"},{"Start":"05:23.245 ","End":"05:28.640","Text":"so the units are going to be joules per"},{"Start":"05:28.640 ","End":"05:36.635","Text":"grams Kelvin where gram and Kelvin are in the denominator."},{"Start":"05:36.635 ","End":"05:41.060","Text":"The molar heat capacity is going to be very similar,"},{"Start":"05:41.060 ","End":"05:46.250","Text":"it\u0027s going to be joules per mole Kelvin,"},{"Start":"05:46.250 ","End":"05:51.180","Text":"where again, both mole and Kelvin are in the denominator."},{"Start":"05:51.950 ","End":"05:56.630","Text":"The specific heat capacity or the molar heat capacity, but in general,"},{"Start":"05:56.630 ","End":"06:01.940","Text":"heat capacity, can be calculated for different materials."},{"Start":"06:01.940 ","End":"06:05.645","Text":"That\u0027s no problem. The temperature we can measure"},{"Start":"06:05.645 ","End":"06:09.970","Text":"a change in temperature simply by using a thermometer,"},{"Start":"06:09.970 ","End":"06:17.085","Text":"but how can we measure heat where heat is this energy transfer?"},{"Start":"06:17.085 ","End":"06:24.025","Text":"How can we see how much energy has been transferred to or from a specific system?"},{"Start":"06:24.025 ","End":"06:26.780","Text":"This is where a very,"},{"Start":"06:26.780 ","End":"06:29.665","Text":"very useful equation comes into play."},{"Start":"06:29.665 ","End":"06:38.870","Text":"Here\u0027s our very useful equation for measuring this heat transfer for thermal energy."},{"Start":"06:38.870 ","End":"06:42.155","Text":"Q is, of course,"},{"Start":"06:42.155 ","End":"06:45.140","Text":"the heat transfer that we\u0027re trying to calculate."},{"Start":"06:45.140 ","End":"06:52.360","Text":"This m is the mass of the system measured in grams."},{"Start":"06:52.360 ","End":"06:56.840","Text":"This C, because we\u0027re measuring the mass of the system,"},{"Start":"06:56.840 ","End":"07:03.780","Text":"this C is of course going to be the specific heat capacity."},{"Start":"07:04.160 ","End":"07:08.675","Text":"I forgot to say that heat capacity is denoted by"},{"Start":"07:08.675 ","End":"07:16.970","Text":"either a C or a c. These are the units depending on if we\u0027re using the specific heat,"},{"Start":"07:16.970 ","End":"07:22.745","Text":"such as here because we\u0027re dealing with mass or the molar heat capacity."},{"Start":"07:22.745 ","End":"07:27.560","Text":"But then we\u0027ll have the number of moles instead of mass."},{"Start":"07:27.560 ","End":"07:32.460","Text":"Then our Delta T is our change in temperature."},{"Start":"07:33.100 ","End":"07:40.010","Text":"Our change in temperature is usually going to be measured by"},{"Start":"07:40.010 ","End":"07:47.795","Text":"using the change in its degrees Celsius or in the degrees Kelvin."},{"Start":"07:47.795 ","End":"07:51.650","Text":"For this equation, you can also use a change in degrees Fahrenheit,"},{"Start":"07:51.650 ","End":"07:55.219","Text":"but then you have to change the equation slightly."},{"Start":"07:55.219 ","End":"07:59.450","Text":"What\u0027s important to remember is that the jump between"},{"Start":"07:59.450 ","End":"08:03.290","Text":"1 degree centigrade is the same as the jump between"},{"Start":"08:03.290 ","End":"08:07.190","Text":"1 degree Kelvin because centigrade and Kelvin are"},{"Start":"08:07.190 ","End":"08:11.660","Text":"pretty much the same things just shifted along the number line slightly,"},{"Start":"08:11.660 ","End":"08:19.860","Text":"so 0 degrees Celsius is just negative 273 degrees Kelvin,"},{"Start":"08:19.860 ","End":"08:22.830","Text":"so it\u0027s just a shift."},{"Start":"08:22.830 ","End":"08:26.885","Text":"What\u0027s important to remember is if you\u0027re taking the change in temperature,"},{"Start":"08:26.885 ","End":"08:30.320","Text":"you have to find the change in"},{"Start":"08:30.320 ","End":"08:35.100","Text":"temperature either in degrees Celsius or in degrees Kelvin."},{"Start":"08:36.380 ","End":"08:43.999","Text":"Mass and our specific heat capacity can only have positive values,"},{"Start":"08:43.999 ","End":"08:46.850","Text":"which means that the value of Q,"},{"Start":"08:46.850 ","End":"08:49.325","Text":"whether it\u0027s a positive or a negative value,"},{"Start":"08:49.325 ","End":"08:54.510","Text":"is dependent on our Delta T over here."},{"Start":"08:54.510 ","End":"08:59.225","Text":"Now, Delta T or change in temperature is defined as"},{"Start":"08:59.225 ","End":"09:04.660","Text":"the temperature at the end minus the initial temperature."},{"Start":"09:04.660 ","End":"09:08.220","Text":"Let\u0027s see what we can learn from this."},{"Start":"09:08.220 ","End":"09:12.470","Text":"Let\u0027s scroll down to have a little bit more space."},{"Start":"09:12.470 ","End":"09:15.575","Text":"If our change in temperature,"},{"Start":"09:15.575 ","End":"09:20.130","Text":"Delta T is bigger than 0,"},{"Start":"09:20.130 ","End":"09:25.035","Text":"so Delta T is a positive value,"},{"Start":"09:25.035 ","End":"09:28.550","Text":"so then that means that our heat,"},{"Start":"09:28.550 ","End":"09:34.410","Text":"Q, is also going to be a positive value."},{"Start":"09:34.420 ","End":"09:36.470","Text":"What does this mean?"},{"Start":"09:36.470 ","End":"09:38.975","Text":"If our Q is a positive value,"},{"Start":"09:38.975 ","End":"09:45.090","Text":"then that means that energy in the system has increased."},{"Start":"09:45.770 ","End":"09:49.714","Text":"There\u0027s an energy increase in the system."},{"Start":"09:49.714 ","End":"09:53.060","Text":"What does an energy increase in the system mean?"},{"Start":"09:53.060 ","End":"09:55.955","Text":"Also, we can see that from this equation over here."},{"Start":"09:55.955 ","End":"10:00.045","Text":"That means that the temperature of the system increases."},{"Start":"10:00.045 ","End":"10:06.970","Text":"That means that our final temperature is larger than our initial temperature."},{"Start":"10:06.970 ","End":"10:09.690","Text":"This can go both ways."},{"Start":"10:09.690 ","End":"10:12.095","Text":"If our change in temperature is positive,"},{"Start":"10:12.095 ","End":"10:15.500","Text":"then that means our heat is positive."},{"Start":"10:15.500 ","End":"10:18.915","Text":"What does our heat if it\u0027s positive mean?"},{"Start":"10:18.915 ","End":"10:22.060","Text":"That means that there\u0027s an energy increase in the system."},{"Start":"10:22.060 ","End":"10:24.860","Text":"What does an energy increase in the system mean?"},{"Start":"10:24.860 ","End":"10:30.950","Text":"It means that the temperature of the system has increased and that is denoted"},{"Start":"10:30.950 ","End":"10:38.035","Text":"by saying that the final temperature is greater than the initial temperature."},{"Start":"10:38.035 ","End":"10:43.250","Text":"Conversely, if our temperature change is negative,"},{"Start":"10:43.250 ","End":"10:45.500","Text":"so that means smaller than 0,"},{"Start":"10:45.500 ","End":"10:48.980","Text":"so that also defines the sign of our heat,"},{"Start":"10:48.980 ","End":"10:53.360","Text":"so that means that our heat is negative as well,"},{"Start":"10:53.360 ","End":"10:59.985","Text":"it\u0027s less than 0, which means that there\u0027s an energy decrease in the system."},{"Start":"10:59.985 ","End":"11:03.545","Text":"What does an energy decrease in the system mean?"},{"Start":"11:03.545 ","End":"11:05.915","Text":"Of course, that again works both ways."},{"Start":"11:05.915 ","End":"11:13.730","Text":"That means that the final temperature is less than the initial temperature."},{"Start":"11:13.730 ","End":"11:18.865","Text":"That means that the temperature of the system has decreased."},{"Start":"11:18.865 ","End":"11:23.825","Text":"That means if we started off with a temperature of 100 degrees Celsius,"},{"Start":"11:23.825 ","End":"11:28.175","Text":"then the temperature now is 80 degrees Celsius."},{"Start":"11:28.175 ","End":"11:33.335","Text":"Of course, these inequalities we can also get from the beginning of the statement,"},{"Start":"11:33.335 ","End":"11:35.430","Text":"it means the same thing."},{"Start":"11:35.630 ","End":"11:45.580","Text":"Now let\u0027s take a look at the specific heat capacity of water and gold."},{"Start":"11:45.580 ","End":"11:50.570","Text":"The specific heat capacity of water is given as approximately"},{"Start":"11:50.570 ","End":"11:56.650","Text":"4,186 joules per kilogram Kelvin."},{"Start":"11:56.650 ","End":"11:58.490","Text":"Here we\u0027re dealing with kilograms."},{"Start":"11:58.490 ","End":"12:00.020","Text":"Usually, we deal just with grams,"},{"Start":"12:00.020 ","End":"12:01.925","Text":"but it doesn\u0027t really matter."},{"Start":"12:01.925 ","End":"12:05.225","Text":"That means that in order to raise"},{"Start":"12:05.225 ","End":"12:11.255","Text":"the temperature of 1 kilogram of water by 1 degree Kelvin,"},{"Start":"12:11.255 ","End":"12:14.960","Text":"which as we know is the same jump as by 1 degree Celsius,"},{"Start":"12:14.960 ","End":"12:24.550","Text":"we have to put into our system 4,186 joules of energy."},{"Start":"12:24.550 ","End":"12:26.705","Text":"On the other hand,"},{"Start":"12:26.705 ","End":"12:28.040","Text":"let\u0027s look at gold."},{"Start":"12:28.040 ","End":"12:34.610","Text":"The specific heat capacity of gold is equal to 100 joules per kilogram Kelvin."},{"Start":"12:34.610 ","End":"12:44.505","Text":"That means that if we have 1 kilogram of gold and we want to raise it by 1 degree Kelvin,"},{"Start":"12:44.505 ","End":"12:45.960","Text":"raise the temperature,"},{"Start":"12:45.960 ","End":"12:53.500","Text":"then that means that we have to put into the system only 100 joules of energy."},{"Start":"12:53.500 ","End":"12:57.304","Text":"We can see that in order to heat water,"},{"Start":"12:57.304 ","End":"13:02.135","Text":"it\u0027s going to take significantly a much longer time"},{"Start":"13:02.135 ","End":"13:07.480","Text":"than it will take us to heat up the same weight in gold."},{"Start":"13:07.480 ","End":"13:10.740","Text":"Of course, this can go both ways."},{"Start":"13:10.740 ","End":"13:12.950","Text":"Yes, we need to exert"},{"Start":"13:12.950 ","End":"13:18.140","Text":"a lot more energy in order to heat up water than we do in order to heat up gold."},{"Start":"13:18.140 ","End":"13:20.315","Text":"However, in the same way,"},{"Start":"13:20.315 ","End":"13:23.060","Text":"because of this specific heat capacity,"},{"Start":"13:23.060 ","End":"13:28.985","Text":"once we have heated up our water and its temperature has risen by 1 degree,"},{"Start":"13:28.985 ","End":"13:34.505","Text":"it will maintain that heat for a lot longer than gold will."},{"Start":"13:34.505 ","End":"13:37.280","Text":"This is the end of the lesson on temperature,"},{"Start":"13:37.280 ","End":"13:39.960","Text":"heat, and heat capacity."}],"ID":12348},{"Watched":false,"Name":"Thermal Conductivity","Duration":"12m 54s","ChapterTopicVideoID":11921,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.025","Text":"Hello. In this lesson,"},{"Start":"00:02.025 ","End":"00:06.495","Text":"we\u0027re going to learn about the next term in our thermodynamics chapter."},{"Start":"00:06.495 ","End":"00:09.600","Text":"That term is thermal conductivity."},{"Start":"00:09.600 ","End":"00:15.570","Text":"Thermal conductivity is speaking about the heat transfer between 2 bodies."},{"Start":"00:15.570 ","End":"00:19.680","Text":"If we have 2 bodies which are in contact,"},{"Start":"00:19.680 ","End":"00:25.665","Text":"let\u0027s imagine that this body is our hot body and this body is our cold body."},{"Start":"00:25.665 ","End":"00:31.875","Text":"We know that heat is going to be transferred from our hot body to our cold body."},{"Start":"00:31.875 ","End":"00:35.270","Text":"In other words, heat is going to be"},{"Start":"00:35.270 ","End":"00:39.620","Text":"transferred from the hot body in order to heat up the cold body."},{"Start":"00:39.620 ","End":"00:45.060","Text":"The cold body is going to cool down the hot body."},{"Start":"00:45.060 ","End":"00:49.710","Text":"Thermal conductivity is talking about this heat transfer."},{"Start":"00:49.710 ","End":"00:59.480","Text":"We know that a low thermal conductivity will mean less heat transfer between 2 objects."},{"Start":"00:59.480 ","End":"01:06.810","Text":"A high thermal conductivity will mean a high heat transfer between 2 bodies."},{"Start":"01:07.340 ","End":"01:11.825","Text":"Let\u0027s see how we can calculate thermal conductivity."},{"Start":"01:11.825 ","End":"01:13.505","Text":"Here we have our hot body,"},{"Start":"01:13.505 ","End":"01:15.367","Text":"and here we have our cold body,"},{"Start":"01:15.367 ","End":"01:17.870","Text":"and we can see that they\u0027re not in contact."},{"Start":"01:17.870 ","End":"01:23.183","Text":"Both of these bodies are isolated away from each other,"},{"Start":"01:23.183 ","End":"01:27.665","Text":"so we have an insulated system."},{"Start":"01:27.665 ","End":"01:32.345","Text":"But now, what happens if we join between these 2 bodies?"},{"Start":"01:32.345 ","End":"01:35.660","Text":"Let\u0027s imagine that this black line is"},{"Start":"01:35.660 ","End":"01:40.288","Text":"some rod where 1 of the ends of the rod is touching the hot body,"},{"Start":"01:40.288 ","End":"01:44.280","Text":"and the other end of the rod is touching the cold body."},{"Start":"01:44.840 ","End":"01:49.639","Text":"Now, heat from the hot body is going to travel"},{"Start":"01:49.639 ","End":"01:54.200","Text":"from the hot body through this rod until it reaches the cold body."},{"Start":"01:54.200 ","End":"01:58.205","Text":"This heat transfer is going to continue until"},{"Start":"01:58.205 ","End":"02:01.325","Text":"the temperature in this body over here"},{"Start":"02:01.325 ","End":"02:05.210","Text":"is equal to the temperature in this body over here."},{"Start":"02:05.210 ","End":"02:12.070","Text":"Heat is going to be transferred until we reach equilibrium between the 2 bodies."},{"Start":"02:12.070 ","End":"02:18.045","Text":"Now we want to know at what rate we have this heat transfer."},{"Start":"02:18.045 ","End":"02:23.490","Text":"Let\u0027s call that H. H is the rate of heat transfer."},{"Start":"02:23.490 ","End":"02:26.055","Text":"How are we going to calculate this?"},{"Start":"02:26.055 ","End":"02:31.150","Text":"That\u0027s how much heat is transferred in a given amount of time."},{"Start":"02:31.150 ","End":"02:35.070","Text":"Delta Q is the change in heat,"},{"Start":"02:35.070 ","End":"02:37.030","Text":"so in 1 body,"},{"Start":"02:37.030 ","End":"02:42.365","Text":"divided by the timeframe that we\u0027re measuring,"},{"Start":"02:42.365 ","End":"02:51.220","Text":"so divided by Delta t. This H is the rate of heat transfer."},{"Start":"02:51.220 ","End":"02:55.143","Text":"Then our Delta Q is how much heat was transferred,"},{"Start":"02:55.143 ","End":"02:57.760","Text":"so the amount of thermal energy transferred,"},{"Start":"02:57.760 ","End":"03:01.755","Text":"divided by the time that we\u0027ve measured."},{"Start":"03:01.755 ","End":"03:05.100","Text":"Now let\u0027s see what this is equal to."},{"Start":"03:05.100 ","End":"03:10.810","Text":"Our rate of heat transfer is dependent on a few factors."},{"Start":"03:10.810 ","End":"03:18.905","Text":"The first factor is K. K is the conductivity of our material,"},{"Start":"03:18.905 ","End":"03:21.660","Text":"of our rod over here."},{"Start":"03:22.580 ","End":"03:24.980","Text":"Conductivity, and of course,"},{"Start":"03:24.980 ","End":"03:28.055","Text":"we\u0027re talking here about thermal conductivity."},{"Start":"03:28.055 ","End":"03:33.050","Text":"As we know, metals have a high thermal conductivity."},{"Start":"03:33.050 ","End":"03:35.690","Text":"If we try and heat up metal,"},{"Start":"03:35.690 ","End":"03:38.705","Text":"we know that it heats up relatively quickly."},{"Start":"03:38.705 ","End":"03:42.470","Text":"However, wood has a low thermal conductivity,"},{"Start":"03:42.470 ","End":"03:48.440","Text":"so low that we can even say that wood is maybe a thermal insulator."},{"Start":"03:48.440 ","End":"03:50.795","Text":"That\u0027s what our K is."},{"Start":"03:50.795 ","End":"03:58.820","Text":"Then we multiply this by A divided by X. What is A?"},{"Start":"03:58.820 ","End":"04:03.840","Text":"A is the cross-sectional area of the rod."},{"Start":"04:04.460 ","End":"04:09.860","Text":"A thicker rod, or a rod with a larger cross-sectional area,"},{"Start":"04:09.860 ","End":"04:13.595","Text":"will mean a higher rate of heat transfer."},{"Start":"04:13.595 ","End":"04:16.220","Text":"This is being divided by X,"},{"Start":"04:16.220 ","End":"04:21.040","Text":"where X is the length of the rod."},{"Start":"04:21.170 ","End":"04:28.655","Text":"Because we see that the length of the rod is placed in the denominator of this equation,"},{"Start":"04:28.655 ","End":"04:33.830","Text":"we can see that the longer the rod is that connects between the hot and cold bodies,"},{"Start":"04:33.830 ","End":"04:38.420","Text":"the less or the lower the rate of heat transfer."},{"Start":"04:38.480 ","End":"04:43.085","Text":"In order to have a higher rate of heat transfer,"},{"Start":"04:43.085 ","End":"04:44.900","Text":"we want a thick rod,"},{"Start":"04:44.900 ","End":"04:47.750","Text":"so a large cross-sectional area,"},{"Start":"04:47.750 ","End":"04:52.055","Text":"but we also want the rod to be as short as possible."},{"Start":"04:52.055 ","End":"04:57.305","Text":"Then the next thing that we have to multiply this by is"},{"Start":"04:57.305 ","End":"05:03.050","Text":"our Delta T. Our change in temperature,"},{"Start":"05:03.050 ","End":"05:06.380","Text":"or sorry, rather instead of changing temperature,"},{"Start":"05:06.380 ","End":"05:10.530","Text":"the difference in temperature between the 2 bodies."},{"Start":"05:10.730 ","End":"05:16.090","Text":"Delta T is the temperature difference between the 2 bodies."},{"Start":"05:16.090 ","End":"05:23.155","Text":"Let\u0027s say that our hot body has an initial temperature of 100 degrees centigrade,"},{"Start":"05:23.155 ","End":"05:28.030","Text":"and our cold body has a temperature of 0 degrees centigrade."},{"Start":"05:28.030 ","End":"05:32.950","Text":"We can see that the temperature difference between the 2 bodies is 100 minus 0,"},{"Start":"05:32.950 ","End":"05:35.700","Text":"so 100 degrees centigrade."},{"Start":"05:35.700 ","End":"05:39.910","Text":"We can see that our rate of heat transfer is going to be high."},{"Start":"05:39.910 ","End":"05:44.335","Text":"Then, heat is going to be transferred from the hot body to the cold body."},{"Start":"05:44.335 ","End":"05:49.600","Text":"Eventually, our hot body from being 100 degrees"},{"Start":"05:49.600 ","End":"05:54.675","Text":"centigrade is going to become 90 degrees centigrade,"},{"Start":"05:54.675 ","End":"06:00.148","Text":"and our cold body from being 0 degrees centigrade is going to therefore,"},{"Start":"06:00.148 ","End":"06:01.943","Text":"from this heat transfer,"},{"Start":"06:01.943 ","End":"06:05.000","Text":"become 10 degrees centigrade."},{"Start":"06:05.300 ","End":"06:14.065","Text":"Originally we saw that our Delta T at the beginning was equal to 100 degrees centigrade."},{"Start":"06:14.065 ","End":"06:16.405","Text":"But now in the second step,"},{"Start":"06:16.405 ","End":"06:18.625","Text":"we can see that our change in temperature,"},{"Start":"06:18.625 ","End":"06:23.680","Text":"or our difference in temperature between the 2 bodies is going to be 90 minus 10,"},{"Start":"06:23.680 ","End":"06:26.910","Text":"which is 80 degrees centigrade."},{"Start":"06:26.910 ","End":"06:30.040","Text":"Now we can see that as our hot body"},{"Start":"06:30.040 ","End":"06:34.365","Text":"cools and our cold body heats up due to this heat transfer,"},{"Start":"06:34.365 ","End":"06:39.020","Text":"our Delta T, so the difference in temperature between the 2 bodies,"},{"Start":"06:39.020 ","End":"06:41.660","Text":"is going to decrease."},{"Start":"06:41.660 ","End":"06:51.930","Text":"Then that means that slowly our rate of heat transfer is going to also decrease."},{"Start":"06:52.190 ","End":"06:58.865","Text":"Eventually, we\u0027ll get to a temperature of the hot body being, let\u0027s say,"},{"Start":"06:58.865 ","End":"07:03.707","Text":"60 degrees Celsius,"},{"Start":"07:03.707 ","End":"07:08.985","Text":"and our cold body will therefore be 40 degrees Celsius."},{"Start":"07:08.985 ","End":"07:14.190","Text":"Here we can see that our Delta T is equal to 60 minus 40,"},{"Start":"07:14.190 ","End":"07:16.790","Text":"is 20 degrees Celsius."},{"Start":"07:16.790 ","End":"07:21.500","Text":"You can see that our rate of heat transfer has dramatically reduced."},{"Start":"07:21.500 ","End":"07:27.035","Text":"At the beginning, we were multiplying this equation by Delta T of 100."},{"Start":"07:27.035 ","End":"07:30.710","Text":"Then as our heat transfer progresses,"},{"Start":"07:30.710 ","End":"07:35.120","Text":"so as the temperatures become closer to one another,"},{"Start":"07:35.120 ","End":"07:40.710","Text":"so the difference in temperature between the hot body and the cold body decreases,"},{"Start":"07:40.710 ","End":"07:44.940","Text":"we can see that our Delta T also decreases."},{"Start":"07:44.940 ","End":"07:51.335","Text":"That means that multiplying this whole expression by now a smaller number, this time 20."},{"Start":"07:51.335 ","End":"07:55.835","Text":"We can see that our rate of heat transfer is much less."},{"Start":"07:55.835 ","End":"07:59.530","Text":"Eventually, we\u0027ll get to thermal equilibrium."},{"Start":"07:59.530 ","End":"08:01.175","Text":"What does that mean?"},{"Start":"08:01.175 ","End":"08:07.910","Text":"Eventually, our hot body is going to have a temperature of 50 degrees Celsius."},{"Start":"08:07.910 ","End":"08:14.910","Text":"Similarly, our cold body will have a temperature of 50 degrees Celsius."},{"Start":"08:14.990 ","End":"08:21.785","Text":"Now we can see that our Delta T is equal to 50 minus 50,"},{"Start":"08:21.785 ","End":"08:23.905","Text":"which is equal to 0."},{"Start":"08:23.905 ","End":"08:31.400","Text":"Now we\u0027re multiplying this expression by 0 because our Delta T is now 0."},{"Start":"08:31.400 ","End":"08:35.990","Text":"Now we can see that there\u0027s going to be no heat transferred"},{"Start":"08:35.990 ","End":"08:41.300","Text":"between these 2 bodies because both of them are at the same temperature."},{"Start":"08:41.300 ","End":"08:45.625","Text":"This means that we have reached thermal equilibrium."},{"Start":"08:45.625 ","End":"08:48.515","Text":"If we were to draw that graphically,"},{"Start":"08:48.515 ","End":"08:52.340","Text":"let\u0027s say that this is our temperature axis,"},{"Start":"08:52.340 ","End":"08:56.770","Text":"and let\u0027s say that this is our time axis."},{"Start":"08:56.770 ","End":"08:59.505","Text":"The red represents our hot body."},{"Start":"08:59.505 ","End":"09:03.770","Text":"It starts at an initial temperature of 100, and then slowly,"},{"Start":"09:03.770 ","End":"09:10.195","Text":"it\u0027s going to get smaller until it reaches some minimum value."},{"Start":"09:10.195 ","End":"09:16.025","Text":"Conversely, our cold body in blue is going to be starting from some cold temperature."},{"Start":"09:16.025 ","End":"09:22.730","Text":"It will be heating up slowly until it reaches some point of equilibrium,"},{"Start":"09:22.730 ","End":"09:24.676","Text":"where both the red,"},{"Start":"09:24.676 ","End":"09:26.630","Text":"so the hot body and the blue,"},{"Start":"09:26.630 ","End":"09:29.929","Text":"the cold body, are at the same temperature."},{"Start":"09:29.929 ","End":"09:33.855","Text":"This will happen at some value for time."},{"Start":"09:33.855 ","End":"09:39.440","Text":"Now, what we can see is that the difference between"},{"Start":"09:39.440 ","End":"09:45.825","Text":"these 2 graphs is our rate of heat transfer."},{"Start":"09:45.825 ","End":"09:51.315","Text":"We can see that the rate of heat transfer is very fast at the beginning,"},{"Start":"09:51.315 ","End":"09:55.880","Text":"and slowly as the 2 bodies approach the same temperature,"},{"Start":"09:55.880 ","End":"09:59.600","Text":"we can see that the rate of heat transfer gradually"},{"Start":"09:59.600 ","End":"10:04.710","Text":"reduces until it gets to approximately 0."},{"Start":"10:04.930 ","End":"10:07.450","Text":"Now, as a quick note,"},{"Start":"10:07.450 ","End":"10:09.730","Text":"but we\u0027ll speak about it later in this chapter,"},{"Start":"10:09.730 ","End":"10:18.025","Text":"in order to have a more efficient engine,"},{"Start":"10:18.025 ","End":"10:20.815","Text":"we want a high rate of heat transfer."},{"Start":"10:20.815 ","End":"10:23.605","Text":"As we see, that happens when there\u0027s"},{"Start":"10:23.605 ","End":"10:29.640","Text":"a large temperature drop or a large temperature difference between the 2 bodies."},{"Start":"10:29.640 ","End":"10:38.890","Text":"Slowly, as the temperature drop between the 2 bodies reaches a lower value,"},{"Start":"10:38.890 ","End":"10:41.245","Text":"or 0 or 20 degrees,"},{"Start":"10:41.245 ","End":"10:47.960","Text":"we can see that the efficiency of our engine is going to be reduced."},{"Start":"10:47.960 ","End":"10:49.385","Text":"For an efficient engine,"},{"Start":"10:49.385 ","End":"10:55.150","Text":"we want our hot body to be very hot and our cold body to be very cold,"},{"Start":"10:55.150 ","End":"10:57.035","Text":"such that our Delta T,"},{"Start":"10:57.035 ","End":"11:00.020","Text":"our difference in temperature between the 2 bodies,"},{"Start":"11:00.020 ","End":"11:02.315","Text":"is going to be very large."},{"Start":"11:02.315 ","End":"11:07.480","Text":"That means a large rate of heat transfer and a more efficient engine."},{"Start":"11:07.480 ","End":"11:12.525","Text":"Now, let\u0027s just give 2 number examples."},{"Start":"11:12.525 ","End":"11:15.880","Text":"Let\u0027s take the metal, aluminum."},{"Start":"11:16.100 ","End":"11:24.305","Text":"It has thermal conductivity of 220."},{"Start":"11:24.305 ","End":"11:27.630","Text":"But if we look at wood,"},{"Start":"11:28.100 ","End":"11:37.835","Text":"it has a roundabout thermal conductivity of between 0.4 to 0.04,"},{"Start":"11:37.835 ","End":"11:39.775","Text":"depending on the type of wood."},{"Start":"11:39.775 ","End":"11:44.165","Text":"As we can see, wood is a thermal insulator,"},{"Start":"11:44.165 ","End":"11:47.960","Text":"so heat doesn\u0027t pass very well through wooden things."},{"Start":"11:47.960 ","End":"11:50.329","Text":"However, we can see that through aluminum,"},{"Start":"11:50.329 ","End":"11:52.190","Text":"heat passes very well."},{"Start":"11:52.190 ","End":"11:56.945","Text":"That\u0027s why when you\u0027re cooking in the kitchen and you have your pots,"},{"Start":"11:56.945 ","End":"11:59.450","Text":"most of the pots are made out of some metal,"},{"Start":"11:59.450 ","End":"12:02.720","Text":"be it aluminum or something else because we can"},{"Start":"12:02.720 ","End":"12:07.010","Text":"see that it has a high heat or thermal conductivity,"},{"Start":"12:07.010 ","End":"12:12.290","Text":"which means that heat from the fire can pass on to our pot,"},{"Start":"12:12.290 ","End":"12:13.505","Text":"and through our pot,"},{"Start":"12:13.505 ","End":"12:16.075","Text":"can go and heat up our food."},{"Start":"12:16.075 ","End":"12:21.689","Text":"Conversely, wood being a very poor thermal conductor,"},{"Start":"12:21.689 ","End":"12:24.529","Text":"and we can even say that it\u0027s a thermal insulator,"},{"Start":"12:24.529 ","End":"12:30.415","Text":"so wood is a common material to be used for the handles of the pot."},{"Start":"12:30.415 ","End":"12:33.280","Text":"Why is that? Because when we lift the pot,"},{"Start":"12:33.280 ","End":"12:36.763","Text":"we don\u0027t want the handles to be hot so that we don\u0027t burn our hand."},{"Start":"12:36.763 ","End":"12:42.170","Text":"That\u0027s why it\u0027s a good idea to use a thermal insulator for the handles,"},{"Start":"12:42.170 ","End":"12:46.580","Text":"but a thermal conductor for the section of the pot that we want to"},{"Start":"12:46.580 ","End":"12:51.162","Text":"actually pass the heat along to heat up our food."},{"Start":"12:51.162 ","End":"12:54.420","Text":"That\u0027s the end of this lesson."}],"ID":12349},{"Watched":false,"Name":"Latent Heat","Duration":"12m 50s","ChapterTopicVideoID":11922,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:04.595","Text":"we\u0027re going to be learning about latent heat."},{"Start":"00:04.595 ","End":"00:06.210","Text":"What is latent heat?"},{"Start":"00:06.210 ","End":"00:10.860","Text":"Latent heat is the amount of energy absorbed or released by"},{"Start":"00:10.860 ","End":"00:17.160","Text":"a material during its phase change that occurs at a constant temperature."},{"Start":"00:17.160 ","End":"00:19.590","Text":"What is a phase change?"},{"Start":"00:19.590 ","End":"00:26.100","Text":"That means a change in the substance\u0027s or in the material\u0027s physical state."},{"Start":"00:26.100 ","End":"00:31.650","Text":"That means that we\u0027re talking about either a change from a solid to a liquid or a gas,"},{"Start":"00:31.650 ","End":"00:33.615","Text":"or from a liquid to a gas,"},{"Start":"00:33.615 ","End":"00:35.385","Text":"or in the other way round,"},{"Start":"00:35.385 ","End":"00:39.100","Text":"from a liquid or a gas to a solid."},{"Start":"00:39.730 ","End":"00:43.700","Text":"Now, the important thing is that there\u0027s"},{"Start":"00:43.700 ","End":"00:47.915","Text":"a constant temperature during this phase change."},{"Start":"00:47.915 ","End":"00:52.370","Text":"If our material changes from a solid to a liquid,"},{"Start":"00:52.370 ","End":"00:55.445","Text":"this change is happening at a constant temperature."},{"Start":"00:55.445 ","End":"00:57.260","Text":"This is very important,"},{"Start":"00:57.260 ","End":"01:00.480","Text":"and that is latent heat."},{"Start":"01:01.070 ","End":"01:05.600","Text":"Now let\u0027s look at this graph over here."},{"Start":"01:05.600 ","End":"01:12.420","Text":"On our x-axis, we have the unit joules because we\u0027re adding energy,"},{"Start":"01:12.420 ","End":"01:14.330","Text":"or as we know, heat."},{"Start":"01:14.330 ","End":"01:16.525","Text":"Heat is thermal energy."},{"Start":"01:16.525 ","End":"01:18.245","Text":"On the y-axis,"},{"Start":"01:18.245 ","End":"01:22.565","Text":"we have our capital T, which represents temperature."},{"Start":"01:22.565 ","End":"01:26.450","Text":"If we look at this section of the graph,"},{"Start":"01:26.450 ","End":"01:30.945","Text":"we can see that we have some slant,"},{"Start":"01:30.945 ","End":"01:33.330","Text":"some gradient to the graph."},{"Start":"01:33.330 ","End":"01:36.455","Text":"We can see that this gradient is constant,"},{"Start":"01:36.455 ","End":"01:43.995","Text":"which means that the relationship between temperature and thermal energy is linear."},{"Start":"01:43.995 ","End":"01:52.520","Text":"We can see that if we take this amount of energy and we add on this amount of energy,"},{"Start":"01:52.520 ","End":"01:58.469","Text":"the change in temperature is going to be something fixed."},{"Start":"01:58.469 ","End":"02:03.345","Text":"Let\u0027s say here we\u0027re adding 3 joules,"},{"Start":"02:03.345 ","End":"02:06.745","Text":"this is our addition of energy."},{"Start":"02:06.745 ","End":"02:08.780","Text":"Then we can say that here,"},{"Start":"02:08.780 ","End":"02:11.315","Text":"that from this addition of 3 joules,"},{"Start":"02:11.315 ","End":"02:20.475","Text":"we\u0027ve raised our temperature by 1 degree Kelvin, let\u0027s say."},{"Start":"02:20.475 ","End":"02:23.270","Text":"Throughout this section of the graph,"},{"Start":"02:23.270 ","End":"02:25.564","Text":"where we can see a constant gradient,"},{"Start":"02:25.564 ","End":"02:30.540","Text":"we can see that if we add another 3 joules over here,"},{"Start":"02:30.540 ","End":"02:35.180","Text":"another 3 joules, then the height change in this graph is going to"},{"Start":"02:35.180 ","End":"02:39.340","Text":"be the exact same as this height change at this section over here,"},{"Start":"02:39.340 ","End":"02:44.765","Text":"which means that we\u0027re adding another 1 degree Kelvin of temperature."},{"Start":"02:44.765 ","End":"02:50.589","Text":"Our temperature has risen by 1 degree Kelvin over here."},{"Start":"02:50.589 ","End":"02:53.170","Text":"So another Kelvin."},{"Start":"02:53.420 ","End":"02:58.490","Text":"What we can see is that the relationship between the change in"},{"Start":"02:58.490 ","End":"03:04.470","Text":"energy is linear with our change in temperature."},{"Start":"03:05.330 ","End":"03:10.790","Text":"That\u0027s this section of the graph which has some linear gradient,"},{"Start":"03:10.790 ","End":"03:12.860","Text":"which is a positive value."},{"Start":"03:12.860 ","End":"03:15.620","Text":"But what happens when we get between this section"},{"Start":"03:15.620 ","End":"03:18.440","Text":"of the graph to this section of the graph?"},{"Start":"03:18.440 ","End":"03:22.340","Text":"Now, we can see that we have a gradient of 0."},{"Start":"03:22.340 ","End":"03:25.205","Text":"Suddenly, our graph flatlines,"},{"Start":"03:25.205 ","End":"03:30.325","Text":"so we can see that we\u0027re increasing the amount of thermal energy,"},{"Start":"03:30.325 ","End":"03:34.820","Text":"we\u0027re going along in the positive direction on this x-axis,"},{"Start":"03:34.820 ","End":"03:38.495","Text":"so increasing the thermal energy in the system."},{"Start":"03:38.495 ","End":"03:43.184","Text":"But we can see that the temperature doesn\u0027t increase or decrease,"},{"Start":"03:43.184 ","End":"03:45.680","Text":"the temperature remains constant."},{"Start":"03:45.680 ","End":"03:49.290","Text":"What on earth is going on over here?"},{"Start":"03:50.750 ","End":"03:53.095","Text":"In order to explain this,"},{"Start":"03:53.095 ","End":"03:56.740","Text":"let\u0027s imagine that this is a graph of water."},{"Start":"03:56.740 ","End":"03:58.645","Text":"Here we can see,"},{"Start":"03:58.645 ","End":"04:01.000","Text":"in which case this isn\u0027t Kelvin;"},{"Start":"04:01.000 ","End":"04:04.580","Text":"this is 1 degree centigrade that we\u0027ve added."},{"Start":"04:04.580 ","End":"04:08.715","Text":"Let\u0027s say that this is 0 degrees Celsius;"},{"Start":"04:08.715 ","End":"04:14.200","Text":"we know that water turns into ice at 0 degrees Celsius."},{"Start":"04:14.200 ","End":"04:19.980","Text":"Here, this is the graph of where we\u0027re located in the ice region."},{"Start":"04:19.980 ","End":"04:23.710","Text":"What is ice? It\u0027s just solid water."},{"Start":"04:23.710 ","End":"04:26.085","Text":"Let\u0027s imagine we\u0027re here."},{"Start":"04:26.085 ","End":"04:30.730","Text":"We\u0027re located over here at minus 80 degrees Celsius,"},{"Start":"04:30.730 ","End":"04:34.850","Text":"and then we add in 3 joules of energy."},{"Start":"04:35.570 ","End":"04:40.730","Text":"Our temperature keeps on rising and we keep adding another 3 joules,"},{"Start":"04:40.730 ","End":"04:44.060","Text":"another 3 joules until we get to minus 20 degrees,"},{"Start":"04:44.060 ","End":"04:45.830","Text":"minus 15 degrees,"},{"Start":"04:45.830 ","End":"04:47.860","Text":"and so on and so forth."},{"Start":"04:47.860 ","End":"04:51.435","Text":"We\u0027re adding a constant amount of energy,"},{"Start":"04:51.435 ","End":"04:53.225","Text":"3 joules each time,"},{"Start":"04:53.225 ","End":"04:58.715","Text":"and each time we go up 1 degree Celsius until eventually,"},{"Start":"04:58.715 ","End":"05:02.195","Text":"we get to 0 degrees Celsius,"},{"Start":"05:02.195 ","End":"05:06.545","Text":"which, as we know, is the maximum temperature that ice can be."},{"Start":"05:06.545 ","End":"05:11.705","Text":"Anything above 0 degrees and our ice will turn into water,"},{"Start":"05:11.705 ","End":"05:17.065","Text":"and anything below 0 degrees our water will turn to ice."},{"Start":"05:17.065 ","End":"05:22.000","Text":"What we can see is between these two red marks,"},{"Start":"05:22.000 ","End":"05:24.070","Text":"our gradient is equal to 0."},{"Start":"05:24.070 ","End":"05:31.390","Text":"We can see that even when I reach this temperature over here, 0 degrees Celsius,"},{"Start":"05:31.390 ","End":"05:33.900","Text":"and I add 3 joules of energy,"},{"Start":"05:33.900 ","End":"05:36.865","Text":"the temperature will not go up,"},{"Start":"05:36.865 ","End":"05:40.070","Text":"and I add another 3 joules of energy, again,"},{"Start":"05:40.070 ","End":"05:41.920","Text":"my temperature is constant;"},{"Start":"05:41.920 ","End":"05:44.230","Text":"it\u0027s still 0 degrees Celsius."},{"Start":"05:44.230 ","End":"05:51.960","Text":"What we can see is I can keep adding this thermal energy to my system,"},{"Start":"05:51.960 ","End":"05:55.540","Text":"but my temperature will remain constant,"},{"Start":"05:55.540 ","End":"05:57.314","Text":"0 degrees Celsius,"},{"Start":"05:57.314 ","End":"06:00.070","Text":"between this point and this point."},{"Start":"06:00.070 ","End":"06:05.740","Text":"Then only once I add thermal energy that passes this point over here,"},{"Start":"06:05.740 ","End":"06:09.575","Text":"only then the more thermal energy I add,"},{"Start":"06:09.575 ","End":"06:12.955","Text":"the higher the temperature will go."},{"Start":"06:12.955 ","End":"06:19.490","Text":"What we can see is between this point and this point where we started,"},{"Start":"06:19.490 ","End":"06:21.365","Text":"but obviously, it can be much colder."},{"Start":"06:21.365 ","End":"06:26.605","Text":"This whole section of the graph is representing ice."},{"Start":"06:26.605 ","End":"06:29.750","Text":"Here, we have solid water, we have ice."},{"Start":"06:29.750 ","End":"06:36.420","Text":"But now, between this point and this point,"},{"Start":"06:36.420 ","End":"06:38.115","Text":"these two green points,"},{"Start":"06:38.115 ","End":"06:43.925","Text":"we can see that here we have ice at 0 degrees,"},{"Start":"06:43.925 ","End":"06:45.770","Text":"and here at this end,"},{"Start":"06:45.770 ","End":"06:49.165","Text":"we have water at 0 degrees Celsius."},{"Start":"06:49.165 ","End":"06:55.565","Text":"Because as we know, 0 degrees Celsius is exactly the junction from"},{"Start":"06:55.565 ","End":"07:03.570","Text":"H_2_O between it being a liquid over here and a solid over here."},{"Start":"07:04.580 ","End":"07:11.750","Text":"As we can see, ice at 0 degrees Celsius has less energy because it"},{"Start":"07:11.750 ","End":"07:18.600","Text":"corresponds to the energy over here in gray on the graph,"},{"Start":"07:18.600 ","End":"07:22.895","Text":"whereas water at 0 degrees Celsius has more energy,"},{"Start":"07:22.895 ","End":"07:27.060","Text":"as it corresponds to this point over here on the graph."},{"Start":"07:27.620 ","End":"07:31.925","Text":"We can see that between these two points in green,"},{"Start":"07:31.925 ","End":"07:35.900","Text":"there is an energy change, all of this."},{"Start":"07:35.900 ","End":"07:40.205","Text":"However, we can see that the temperature has remained constant."},{"Start":"07:40.205 ","End":"07:43.390","Text":"We\u0027re still at 0 degrees centigrade."},{"Start":"07:43.390 ","End":"07:46.595","Text":"This energy change over here,"},{"Start":"07:46.595 ","End":"07:49.580","Text":"where we\u0027re increasing the energy or"},{"Start":"07:49.580 ","End":"07:54.325","Text":"decreasing the energy depending on which way we\u0027re looking,"},{"Start":"07:54.325 ","End":"07:57.585","Text":"but the temperature is constant,"},{"Start":"07:57.585 ","End":"08:01.810","Text":"all of this is latent heat."},{"Start":"08:02.210 ","End":"08:05.680","Text":"Only ones I\u0027ve put into my system,"},{"Start":"08:05.680 ","End":"08:07.750","Text":"all of this energy,"},{"Start":"08:07.750 ","End":"08:10.270","Text":"all of this thermal energy."},{"Start":"08:10.270 ","End":"08:15.340","Text":"This amount of thermal energy is called the latent heat."},{"Start":"08:15.340 ","End":"08:19.090","Text":"That means once I\u0027ve melted all of the ice,"},{"Start":"08:19.090 ","End":"08:22.474","Text":"so we\u0027re still at the same temperature of 0 degrees Celsius."},{"Start":"08:22.474 ","End":"08:24.460","Text":"However, my ice,"},{"Start":"08:24.460 ","End":"08:26.815","Text":"from being at 0 degrees Celsius,"},{"Start":"08:26.815 ","End":"08:31.660","Text":"has become water or liquid at 0 degrees Celsius."},{"Start":"08:31.660 ","End":"08:36.946","Text":"Only then, when I start adding in more energy to the system,"},{"Start":"08:36.946 ","End":"08:40.450","Text":"will my temperature begins to rise until I eventually,"},{"Start":"08:40.450 ","End":"08:42.505","Text":"from having liquid water,"},{"Start":"08:42.505 ","End":"08:47.515","Text":"get water vapor, which is just water gas."},{"Start":"08:47.515 ","End":"08:50.210","Text":"That is latent heat."},{"Start":"08:50.210 ","End":"08:52.520","Text":"The energy I need to provide into"},{"Start":"08:52.520 ","End":"08:58.670","Text":"a system for going from a solid to a liquid to melt all of the solid,"},{"Start":"08:58.670 ","End":"09:00.754","Text":"such that it\u0027s at the same temperature,"},{"Start":"09:00.754 ","End":"09:05.850","Text":"except it has experienced a phase change from solid to liquid."},{"Start":"09:05.960 ","End":"09:09.430","Text":"That is latent heat."},{"Start":"09:09.970 ","End":"09:16.490","Text":"Once we\u0027ve turned our ice into water over here,"},{"Start":"09:16.490 ","End":"09:18.875","Text":"here\u0027s our phase change,"},{"Start":"09:18.875 ","End":"09:24.750","Text":"and here we\u0027re dealing with up until this point over here, water."},{"Start":"09:24.750 ","End":"09:31.040","Text":"Then we\u0027ll get to another area with a 0 gradient."},{"Start":"09:31.040 ","End":"09:35.705","Text":"This is our phase change from water into gas."},{"Start":"09:35.705 ","End":"09:41.945","Text":"Here, again, we\u0027ll get another positive gradient where we\u0027re dealing with vapor."},{"Start":"09:41.945 ","End":"09:47.150","Text":"Now, something that is important to remember is"},{"Start":"09:47.150 ","End":"09:54.140","Text":"that when we have our phase change from ice to water to vapor,"},{"Start":"09:54.140 ","End":"09:58.880","Text":"our heat capacity, if we remember,"},{"Start":"09:58.880 ","End":"10:03.980","Text":"our heat capacity was denoted by the letter C, changes."},{"Start":"10:03.980 ","End":"10:07.780","Text":"Even though we\u0027re still dealing with H_2_O,"},{"Start":"10:07.780 ","End":"10:13.450","Text":"the heat capacity of H_2_O in solid form has"},{"Start":"10:13.450 ","End":"10:18.775","Text":"a different value to the heat capacity of H_2_O in liquid form,"},{"Start":"10:18.775 ","End":"10:24.680","Text":"which has a different heat capacity to H_2_O in gaseous form."},{"Start":"10:26.330 ","End":"10:35.010","Text":"Latent heat is represented by the letter capital L and its units are joules per gram."},{"Start":"10:35.010 ","End":"10:42.800","Text":"How much energy is needed in order for a phase change of 1 gram?"},{"Start":"10:44.270 ","End":"10:47.535","Text":"This is latent heat."},{"Start":"10:47.535 ","End":"10:51.525","Text":"What\u0027s important to take from this lesson is one,"},{"Start":"10:51.525 ","End":"10:53.915","Text":"that latent heat is joules per gram,"},{"Start":"10:53.915 ","End":"10:56.405","Text":"the definition for latent heat,"},{"Start":"10:56.405 ","End":"11:00.975","Text":"that it\u0027s the energy absorbed at a constant temperature."},{"Start":"11:00.975 ","End":"11:05.840","Text":"It\u0027s just the phase change and we have to remember that the phase is changing,"},{"Start":"11:05.840 ","End":"11:08.570","Text":"but the temperature is not."},{"Start":"11:08.570 ","End":"11:14.285","Text":"Now another important thing to remember is that the energy,"},{"Start":"11:14.285 ","End":"11:16.745","Text":"even though we\u0027re at the same temperature,"},{"Start":"11:16.745 ","End":"11:25.220","Text":"the energy of ice at 0 degrees is going to be less than the energy of water at 0 degrees."},{"Start":"11:25.220 ","End":"11:29.680","Text":"Of course, this is the same for every material."},{"Start":"11:29.680 ","End":"11:33.300","Text":"If we have some material or substance,"},{"Start":"11:33.300 ","End":"11:35.300","Text":"in its solid form,"},{"Start":"11:35.300 ","End":"11:41.810","Text":"it\u0027s going to have lower amount of energy than if it\u0027s at the exact same temperature,"},{"Start":"11:41.810 ","End":"11:43.975","Text":"but in liquid form."},{"Start":"11:43.975 ","End":"11:47.890","Text":"That\u0027s another very important thing to remember."},{"Start":"11:47.890 ","End":"11:50.370","Text":"The final thing to remember,"},{"Start":"11:50.370 ","End":"11:51.935","Text":"I\u0027ll draw it out."},{"Start":"11:51.935 ","End":"11:58.390","Text":"If we have a bathtub with water and we have some block of ice inside."},{"Start":"11:58.390 ","End":"12:04.510","Text":"As long as this block of ice is going through a phase change,"},{"Start":"12:04.510 ","End":"12:09.320","Text":"going from ice to water or from solid to liquid,"},{"Start":"12:09.320 ","End":"12:15.753","Text":"the temperature of the entire bath is going to remain constant,."},{"Start":"12:15.753 ","End":"12:20.400","Text":"Once this block has gone through the phase change,"},{"Start":"12:20.400 ","End":"12:23.540","Text":"gone from a solid to a liquid,"},{"Start":"12:23.540 ","End":"12:27.140","Text":"only then once the phase change is complete,"},{"Start":"12:27.140 ","End":"12:30.935","Text":"so that means this entire block has now become a liquid,"},{"Start":"12:30.935 ","End":"12:38.135","Text":"only then can the temperature of the bath of water or of liquid start to increase."},{"Start":"12:38.135 ","End":"12:39.680","Text":"That\u0027s very important."},{"Start":"12:39.680 ","End":"12:48.055","Text":"The temperature can only begin to increase once the phase change has completely finished."},{"Start":"12:48.055 ","End":"12:51.250","Text":"That\u0027s the end of this lesson."}],"ID":12350},{"Watched":false,"Name":"Exercise 1","Duration":"23m 14s","ChapterTopicVideoID":11923,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"Hello. In this lesson,"},{"Start":"00:02.205 ","End":"00:05.040","Text":"we\u0027re going to be answering the following questions."},{"Start":"00:05.040 ","End":"00:09.615","Text":"An urn of cross sectional area A is filled with water."},{"Start":"00:09.615 ","End":"00:13.980","Text":"The temperature of the water is 0 degrees Celsius and we\u0027re"},{"Start":"00:13.980 ","End":"00:18.240","Text":"being told that the size of the urn insulated whilst the button is not"},{"Start":"00:18.240 ","End":"00:23.580","Text":"insulated and it\u0027s resting on a heated surface of power H."},{"Start":"00:23.580 ","End":"00:26.700","Text":"The heat from the heated surface travels through the bottom of"},{"Start":"00:26.700 ","End":"00:30.615","Text":"the urn and no heat is lost to the surroundings."},{"Start":"00:30.615 ","End":"00:34.650","Text":"The bottom of the urn has thickness D and is made of copper."},{"Start":"00:34.650 ","End":"00:38.840","Text":"The heated surface is switched on and at that exact moment"},{"Start":"00:38.840 ","End":"00:43.580","Text":"an ice cube of mass m_ice is added to the urn."},{"Start":"00:43.580 ","End":"00:46.505","Text":"After a time of t_1 has passed,"},{"Start":"00:46.505 ","End":"00:49.240","Text":"the ice cube is removed from the urn."},{"Start":"00:49.240 ","End":"00:54.515","Text":"The mass of the ice cube is now half its original mass."},{"Start":"00:54.515 ","End":"00:57.545","Text":"Assume that the thermal conductivity of the copper,"},{"Start":"00:57.545 ","End":"00:59.645","Text":"the heat capacity of the ice,"},{"Start":"00:59.645 ","End":"01:03.055","Text":"and the latent heat for melting ice is given."},{"Start":"01:03.055 ","End":"01:04.880","Text":"Question number 1 is,"},{"Start":"01:04.880 ","End":"01:09.470","Text":"what was the temperature of the ice cube when it was removed from urn?"},{"Start":"01:09.470 ","End":"01:13.805","Text":"From our previous lesson dealing with latent heat,"},{"Start":"01:13.805 ","End":"01:16.415","Text":"we\u0027re able to answer this question."},{"Start":"01:16.415 ","End":"01:22.475","Text":"Now the whole idea with latent heat was that if we\u0027re going through a phase change,"},{"Start":"01:22.475 ","End":"01:24.335","Text":"if we\u0027re going from, let\u0027s say,"},{"Start":"01:24.335 ","End":"01:27.695","Text":"an ice cube to liquid water,"},{"Start":"01:27.695 ","End":"01:32.930","Text":"that means that the temperature must remain constant."},{"Start":"01:32.930 ","End":"01:40.790","Text":"We know that the ice cube when it was taken out of the urn has been halved in mass,"},{"Start":"01:40.790 ","End":"01:45.710","Text":"which means that half of the mass of the ice cube has become water,"},{"Start":"01:45.710 ","End":"01:47.810","Text":"but half of it is still ice,"},{"Start":"01:47.810 ","End":"01:53.570","Text":"which means that not all of the ice cube has melted into water,"},{"Start":"01:53.570 ","End":"01:58.655","Text":"which means that we\u0027re still in the process, a phase change."},{"Start":"01:58.655 ","End":"02:00.665","Text":"That is very important,"},{"Start":"02:00.665 ","End":"02:04.459","Text":"which means that the temperature has not changed."},{"Start":"02:04.459 ","End":"02:10.415","Text":"Now, we know that the phase change for water between it being a liquid"},{"Start":"02:10.415 ","End":"02:16.655","Text":"and a solid between it being liquid water or ice is 0 degrees Celsius."},{"Start":"02:16.655 ","End":"02:20.630","Text":"That\u0027s how we can know that the temperature of"},{"Start":"02:20.630 ","End":"02:24.980","Text":"the ice cube when it is taken out from the urn,"},{"Start":"02:24.980 ","End":"02:30.160","Text":"is going to be equal to 0 degrees Celsius."},{"Start":"02:30.160 ","End":"02:33.225","Text":"Now, question number 2,"},{"Start":"02:33.225 ","End":"02:37.685","Text":"is asking us what was ice cubes temperature at t is"},{"Start":"02:37.685 ","End":"02:43.255","Text":"equal to 0 right at the beginning of the experiment."},{"Start":"02:43.255 ","End":"02:49.609","Text":"All were given in the question is the initial mass of the ice cube."},{"Start":"02:49.609 ","End":"02:58.820","Text":"But that isn\u0027t giving us the temperature of the ice at t is equal to 0."},{"Start":"02:58.820 ","End":"03:01.040","Text":"This is what we\u0027re trying to find."},{"Start":"03:01.040 ","End":"03:04.820","Text":"Now, what we are being told is that the temperature of the water is at"},{"Start":"03:04.820 ","End":"03:08.600","Text":"0 degrees Celsius and that we have"},{"Start":"03:08.600 ","End":"03:11.930","Text":"a heated surface of power age"},{"Start":"03:11.930 ","End":"03:16.955","Text":"and the heat from the heated surface travels through the bottom of the urn."},{"Start":"03:16.955 ","End":"03:23.660","Text":"We can assume in this question that no energy or power is lost to the environment."},{"Start":"03:23.660 ","End":"03:28.190","Text":"All the power or all the energy is transferred to the water"},{"Start":"03:28.190 ","End":"03:33.990","Text":"inside the urn and subsequently also to the ice cube."},{"Start":"03:35.540 ","End":"03:37.715","Text":"Just before we carry on,"},{"Start":"03:37.715 ","End":"03:40.880","Text":"we know that the temperature of the ice cube at t is equal to"},{"Start":"03:40.880 ","End":"03:44.960","Text":"0 can be a maximum of 0 degrees Celsius."},{"Start":"03:44.960 ","End":"03:47.650","Text":"But it can be, of course, a lot less,"},{"Start":"03:47.650 ","End":"03:52.350","Text":"it could be minus 30 degrees Celsius, for example."},{"Start":"03:52.350 ","End":"03:54.800","Text":"What we\u0027re going to do is we\u0027re going to use"},{"Start":"03:54.800 ","End":"03:58.370","Text":"this equation where we see that the power is equal to"},{"Start":"03:58.370 ","End":"04:05.645","Text":"the change in heat or the change in thermal energy divided by the change in time."},{"Start":"04:05.645 ","End":"04:07.475","Text":"Or in other words,"},{"Start":"04:07.475 ","End":"04:13.145","Text":"the total heat is transferred and the total amount of time is going to be equal to"},{"Start":"04:13.145 ","End":"04:20.700","Text":"the power of the heated surface multiplied by the time."},{"Start":"04:21.700 ","End":"04:27.680","Text":"This is the amount of heat that has been put in to the system."},{"Start":"04:27.680 ","End":"04:30.530","Text":"But now what we want to say that this is equal"},{"Start":"04:30.530 ","End":"04:34.890","Text":"to what has been happening inside the system."},{"Start":"04:34.970 ","End":"04:37.755","Text":"What has been happening to our system?"},{"Start":"04:37.755 ","End":"04:41.309","Text":"So 2 things have been happening."},{"Start":"04:41.309 ","End":"04:45.530","Text":"Our first thing was when the ice cube was dropped into the water,"},{"Start":"04:45.530 ","End":"04:50.170","Text":"the ice cube itself was heated or warmed."},{"Start":"04:50.170 ","End":"04:53.870","Text":"Case for the first thing that happened was that the temperature of"},{"Start":"04:53.870 ","End":"04:57.830","Text":"the ice cube was increased from its original temperature,"},{"Start":"04:57.830 ","End":"05:00.395","Text":"G02, the heated surface,"},{"Start":"05:00.395 ","End":"05:02.345","Text":"and then after that,"},{"Start":"05:02.345 ","End":"05:06.950","Text":"the other set of heat that was in or thermal energy that was put in,"},{"Start":"05:06.950 ","End":"05:10.850","Text":"was the thermal energy required to melt the ice."},{"Start":"05:10.850 ","End":"05:18.240","Text":"Because we know that at a certain time our ice cube starts to melt."},{"Start":"05:18.530 ","End":"05:22.400","Text":"All we\u0027ve done here is we\u0027ve used this equation to say"},{"Start":"05:22.400 ","End":"05:25.505","Text":"that the total heat that was put into a system,"},{"Start":"05:25.505 ","End":"05:30.050","Text":"or the total amount of thermal energy put into a system is equal to"},{"Start":"05:30.050 ","End":"05:35.405","Text":"the power of the heated surface multiplied by the time that it is running."},{"Start":"05:35.405 ","End":"05:37.700","Text":"Then we said that obviously"},{"Start":"05:37.700 ","End":"05:42.530","Text":"the total thermal energy or heat put into a system is going to be equal"},{"Start":"05:42.530 ","End":"05:50.900","Text":"to the heat needed to get to the place where we are now."},{"Start":"05:50.900 ","End":"05:56.030","Text":"In our system that included ice cube being heated until it got"},{"Start":"05:56.030 ","End":"06:01.385","Text":"to 0 degrees Celsius plus the energy needed in order to melt the ice,"},{"Start":"06:01.385 ","End":"06:04.890","Text":"because we\u0027re told that the ice cube is melting."},{"Start":"06:06.130 ","End":"06:11.735","Text":"What is the heat required to heat the ice?"},{"Start":"06:11.735 ","End":"06:14.530","Text":"We have this equation over here."},{"Start":"06:14.530 ","End":"06:20.755","Text":"First of all, we know that the heat is going to be equal to the mass of our ice,"},{"Start":"06:20.755 ","End":"06:25.815","Text":"which we\u0027re given is equal to m_ice,"},{"Start":"06:25.815 ","End":"06:28.590","Text":"then we multiply that by c,"},{"Start":"06:28.590 ","End":"06:30.360","Text":"which is the heat capacity."},{"Start":"06:30.360 ","End":"06:33.760","Text":"We\u0027re given the heat capacity of the ice,"},{"Start":"06:33.760 ","End":"06:36.625","Text":"so we\u0027re multiplying this by c_ice,"},{"Start":"06:36.625 ","End":"06:42.505","Text":"and then we have to multiply it by the change in temperature."},{"Start":"06:42.505 ","End":"06:49.009","Text":"We\u0027re multiplying by Delta T. What is this change in temperature?"},{"Start":"06:49.220 ","End":"06:51.995","Text":"What is this change in temperature?"},{"Start":"06:51.995 ","End":"06:58.490","Text":"We know that the change in temperature is equal to the final temperature,"},{"Start":"06:58.490 ","End":"07:00.380","Text":"and we\u0027re speaking specifically about the ice."},{"Start":"07:00.380 ","End":"07:07.135","Text":"The final temperature of the ice minus the initial temperature of the ice."},{"Start":"07:07.135 ","End":"07:10.160","Text":"In our case, we know that the final temperature of"},{"Start":"07:10.160 ","End":"07:12.830","Text":"the ice from question number 1 is equal to"},{"Start":"07:12.830 ","End":"07:18.365","Text":"0 degrees Celsius minus our initial temperature of the ice,"},{"Start":"07:18.365 ","End":"07:20.435","Text":"which is what we\u0027re trying to find."},{"Start":"07:20.435 ","End":"07:27.570","Text":"Let\u0027s say that T_i is equal to T_ice,"},{"Start":"07:27.570 ","End":"07:29.982","Text":"a T is equal to 0."},{"Start":"07:29.982 ","End":"07:35.625","Text":"Now let\u0027s talk about the energy needed in order to melt the ice."},{"Start":"07:35.625 ","End":"07:38.445","Text":"We know that to melt the ice,"},{"Start":"07:38.445 ","End":"07:42.750","Text":"we multiply the latent heat for melting ice,"},{"Start":"07:42.750 ","End":"07:51.535","Text":"which he has given as Land we multiply it by the amount of ice that was melted."},{"Start":"07:51.535 ","End":"07:55.900","Text":"Here we have our latent heat for melting ice, so let\u0027s call, it,"},{"Start":"07:55.900 ","End":"07:58.435","Text":"L ice even though in the question it\u0027s given as L,"},{"Start":"07:58.435 ","End":"08:00.580","Text":"just to remind you, and then,"},{"Start":"08:00.580 ","End":"08:02.740","Text":"how much ice was melted."},{"Start":"08:02.740 ","End":"08:07.360","Text":"We\u0027re being told that at a certain time,"},{"Start":"08:07.360 ","End":"08:11.845","Text":"the mass of the ice cube is now half its original mass."},{"Start":"08:11.845 ","End":"08:19.360","Text":"What\u0027s left is just half the mass of the ice cube at the start."},{"Start":"08:19.360 ","End":"08:26.485","Text":"This is how the total heat that we put into the system is divided."},{"Start":"08:26.485 ","End":"08:30.580","Text":"We have the heat needed in order to raise the temperature of the ice,"},{"Start":"08:30.580 ","End":"08:36.040","Text":"and the heat needed in order to turn the ice from a solid into a liquid."},{"Start":"08:36.040 ","End":"08:38.720","Text":"From ice into water."},{"Start":"08:39.390 ","End":"08:42.085","Text":"What is this time t?"},{"Start":"08:42.085 ","End":"08:46.645","Text":"This is our time t_1 because we know that 1/2 of the ice,"},{"Start":"08:46.645 ","End":"08:49.615","Text":"we\u0027re speaking at the final time,"},{"Start":"08:49.615 ","End":"08:52.930","Text":"half of the ice has already melted."},{"Start":"08:52.930 ","End":"08:56.980","Text":"Then the question we\u0027re being told that that happens at t_1."},{"Start":"08:56.980 ","End":"09:02.260","Text":"Now we can see that we have an equation with just 1 unknown,"},{"Start":"09:02.260 ","End":"09:03.940","Text":"which is this t_i,"},{"Start":"09:03.940 ","End":"09:05.680","Text":"that we don\u0027t know what it is,"},{"Start":"09:05.680 ","End":"09:09.505","Text":"and this is exactly what we\u0027re trying to solve for this question."},{"Start":"09:09.505 ","End":"09:16.450","Text":"Now, all that is left to do is to rearrange this equation in order to isolate out t_i."},{"Start":"09:16.450 ","End":"09:19.090","Text":"That\u0027s the initial temperature of the ice."},{"Start":"09:19.090 ","End":"09:22.820","Text":"That\u0027s how we\u0027ve answered this question."},{"Start":"09:23.490 ","End":"09:28.270","Text":"Once you\u0027ve rearranged this equation over here, okay,"},{"Start":"09:28.270 ","End":"09:30.340","Text":"substituting in for Delta T,"},{"Start":"09:30.340 ","End":"09:35.200","Text":"negative t_i, which is our variable that we\u0027re trying to find."},{"Start":"09:35.200 ","End":"09:37.990","Text":"You should get this answer."},{"Start":"09:37.990 ","End":"09:41.005","Text":"This is the answer for question Number 2."},{"Start":"09:41.005 ","End":"09:44.395","Text":"Now let\u0027s take a look at question Number 3."},{"Start":"09:44.395 ","End":"09:46.615","Text":"Question Number 3 is asking us,"},{"Start":"09:46.615 ","End":"09:50.995","Text":"had the ice cube not been removed from the urn,"},{"Start":"09:50.995 ","End":"09:55.690","Text":"how long would it have taken for the ice cube to fully melt?"},{"Start":"09:55.690 ","End":"10:01.150","Text":"Here, the ice cube isn\u0027t taken out after this time, t_1."},{"Start":"10:01.150 ","End":"10:05.980","Text":"We\u0027re wondering the whole mass of the ice cube to fully melt,"},{"Start":"10:05.980 ","End":"10:08.620","Text":"and how long will that take?"},{"Start":"10:08.620 ","End":"10:12.100","Text":"What we\u0027re trying to find is t_2."},{"Start":"10:12.100 ","End":"10:17.995","Text":"So this question we\u0027re going to solve exactly like we did with question Number 2."},{"Start":"10:17.995 ","End":"10:23.080","Text":"Again, we\u0027re going to say that the total heat in the system"},{"Start":"10:23.080 ","End":"10:27.760","Text":"is equal to the power of the surface multiplied by our time frame,"},{"Start":"10:27.760 ","End":"10:29.725","Text":"which here is t_2."},{"Start":"10:29.725 ","End":"10:36.879","Text":"Now I\u0027m going to have a tilde on top of this t_2 to show that in this question,"},{"Start":"10:36.879 ","End":"10:39.190","Text":"this is our unknown."},{"Start":"10:39.190 ","End":"10:43.720","Text":"The total heat in the system is given by this equation."},{"Start":"10:43.720 ","End":"10:46.750","Text":"But what does it mean in our system?"},{"Start":"10:46.750 ","End":"10:53.860","Text":"It\u0027s equal to the amount of heat needed in order to raise the temperature of the ice."},{"Start":"10:53.860 ","End":"10:59.035","Text":"As we know, that is equal to this equation over here."},{"Start":"10:59.035 ","End":"11:06.310","Text":"The mass of the ice multiplied by the heat capacity of the ice,"},{"Start":"11:06.310 ","End":"11:09.354","Text":"which is given to us in the question,"},{"Start":"11:09.354 ","End":"11:14.230","Text":"multiplied by our Delta T."},{"Start":"11:14.230 ","End":"11:21.325","Text":"Plus we have the heat required in order to melt the ice."},{"Start":"11:21.325 ","End":"11:25.180","Text":"That is going to be our latent heat for melting ice,"},{"Start":"11:25.180 ","End":"11:31.045","Text":"which is L ice multiplied by the amount of ice that we want to have melted."},{"Start":"11:31.045 ","End":"11:32.710","Text":"Because that\u0027s what we want."},{"Start":"11:32.710 ","End":"11:37.225","Text":"That all of the ice cube is melted."},{"Start":"11:37.225 ","End":"11:41.590","Text":"We\u0027re going to multiply that by the total mass of the ice cube."},{"Start":"11:41.590 ","End":"11:44.920","Text":"Before we multiplied by 1/2 the mass of the ice cube"},{"Start":"11:44.920 ","End":"11:49.090","Text":"because the question was about half the ice cube being melted."},{"Start":"11:49.090 ","End":"11:54.985","Text":"But here we\u0027re multiplying by the full mass because all of the ice cube is being melted."},{"Start":"11:54.985 ","End":"11:56.860","Text":"Then for this question,"},{"Start":"11:56.860 ","End":"12:02.105","Text":"we just have to isolate out our t_2 to get her out and say,"},{"Start":"12:02.105 ","End":"12:05.085","Text":"this is our answer for question Number 3."},{"Start":"12:05.085 ","End":"12:08.070","Text":"A quick note, this Delta T over here is"},{"Start":"12:08.070 ","End":"12:12.765","Text":"the exact same Delta T that we had in question Number 2."},{"Start":"12:12.765 ","End":"12:19.090","Text":"Our Delta T is our final temperature minus our initial temperature where we know"},{"Start":"12:19.090 ","End":"12:22.990","Text":"our final temperature is 0 degrees Celsius minus"},{"Start":"12:22.990 ","End":"12:28.195","Text":"our initial temperature which is this what we found in Question 2."},{"Start":"12:28.195 ","End":"12:32.470","Text":"So if you want, you can also sub that into this onset."},{"Start":"12:32.470 ","End":"12:35.930","Text":"Now let\u0027s answer question Number 4."},{"Start":"12:37.380 ","End":"12:42.925","Text":"Now the heated surface is removed from"},{"Start":"12:42.925 ","End":"12:47.770","Text":"underneath on the bottom of the urn is now also insulated."},{"Start":"12:47.770 ","End":"12:49.270","Text":"Aside from the walls of the urn,"},{"Start":"12:49.270 ","End":"12:52.430","Text":"the bottom is now also insulated."},{"Start":"12:52.530 ","End":"12:58.990","Text":"Then the ice cube of the same mass M ice is added to the urn."},{"Start":"12:58.990 ","End":"13:02.875","Text":"Our question is, will the ice cube\u0027s mass increase,"},{"Start":"13:02.875 ","End":"13:05.680","Text":"decrease or remain the same?"},{"Start":"13:05.680 ","End":"13:09.280","Text":"First of all, we need to remember that the temperature of"},{"Start":"13:09.280 ","End":"13:13.105","Text":"the water at this tide is 0 degrees Celsius."},{"Start":"13:13.105 ","End":"13:16.990","Text":"It\u0027s still water, but it\u0027s water at 0 degrees Celsius."},{"Start":"13:16.990 ","End":"13:20.830","Text":"We now know that the urn is completely insulated"},{"Start":"13:20.830 ","End":"13:25.375","Text":"so that no heat can be transferred to heating up the water,"},{"Start":"13:25.375 ","End":"13:31.520","Text":"and the heat from the water can\u0027t be transferred and lost to the environment."},{"Start":"13:32.460 ","End":"13:35.500","Text":"Let\u0027s see how we can do this."},{"Start":"13:35.500 ","End":"13:38.590","Text":"We have temperature of our ice cube,"},{"Start":"13:38.590 ","End":"13:40.510","Text":"which we got to question Number 2,"},{"Start":"13:40.510 ","End":"13:47.545","Text":"and we know that the temperature of our ice cube initially,"},{"Start":"13:47.545 ","End":"13:55.420","Text":"this is what we got in question Number 2 is less than 0 degrees Celsius."},{"Start":"13:55.420 ","End":"14:03.535","Text":"This we know. Once the ice cube is placed into the water,"},{"Start":"14:03.535 ","End":"14:06.400","Text":"we know that heat from the water is"},{"Start":"14:06.400 ","End":"14:11.545","Text":"transferred to the ice cube and it heats up the ice cube."},{"Start":"14:11.545 ","End":"14:17.590","Text":"Now we know that our system will never exceed 0 degrees Celsius because"},{"Start":"14:17.590 ","End":"14:24.400","Text":"the water is at 0 degrees Celsius and the ice is obviously less than 0 degrees Celsius."},{"Start":"14:24.400 ","End":"14:29.365","Text":"Once the 2 temperatures between the water and the ice cube balance out,"},{"Start":"14:29.365 ","End":"14:32.740","Text":"we know that we\u0027re going to be located at a temperature"},{"Start":"14:32.740 ","End":"14:36.850","Text":"somewhere between the the initial temperature of the ice cube,"},{"Start":"14:36.850 ","End":"14:39.775","Text":"which is much less than 0 degrees Celsius,"},{"Start":"14:39.775 ","End":"14:43.480","Text":"and the initial temperature of the water,"},{"Start":"14:43.480 ","End":"14:45.580","Text":"which is 0 degrees Celsius."},{"Start":"14:45.580 ","End":"14:49.510","Text":"We know that water freezes over and becomes a solid,"},{"Start":"14:49.510 ","End":"14:53.485","Text":"becomes ice at 0 degrees Celsius."},{"Start":"14:53.485 ","End":"14:57.595","Text":"If we go lower than 0 degrees Celsius,"},{"Start":"14:57.595 ","End":"15:03.680","Text":"that means that our water is going to be freezing and turning into ice."},{"Start":"15:03.680 ","End":"15:11.260","Text":"We know that even though we\u0027re going to be located at some temperature between here,"},{"Start":"15:11.260 ","End":"15:13.390","Text":"so let\u0027s call this x."},{"Start":"15:13.390 ","End":"15:16.720","Text":"We\u0027re going to be at a balance between"},{"Start":"15:16.720 ","End":"15:21.175","Text":"our initial temperature of ice and our initial temperature of the water."},{"Start":"15:21.175 ","End":"15:25.000","Text":"We know that this is in the range where water freezes."},{"Start":"15:25.000 ","End":"15:28.630","Text":"We know that the water in the urn is going to freeze,"},{"Start":"15:28.630 ","End":"15:32.410","Text":"or at least some of it is going to freeze which means that"},{"Start":"15:32.410 ","End":"15:37.390","Text":"the ice cube\u0027s mass is going to therefore increase."},{"Start":"15:37.390 ","End":"15:41.740","Text":"This is our answer to question number 4."},{"Start":"15:41.740 ","End":"15:44.770","Text":"Now, before we finish the lesson,"},{"Start":"15:44.770 ","End":"15:48.115","Text":"we can see that we haven\u0027t used lots"},{"Start":"15:48.115 ","End":"15:52.360","Text":"of the different variables or values that we were given in the question,"},{"Start":"15:52.360 ","End":"15:56.920","Text":"such as the cross-sectional area of the urn,"},{"Start":"15:56.920 ","End":"15:59.890","Text":"such as the thermal conductivity of"},{"Start":"15:59.890 ","End":"16:06.020","Text":"the copper and such as the thickness of the bottom of the urn."},{"Start":"16:06.090 ","End":"16:13.165","Text":"These values are generally used when we\u0027re trying to find the temperature at equilibrium."},{"Start":"16:13.165 ","End":"16:18.355","Text":"What does equilibrium in thermodynamics mean?"},{"Start":"16:18.355 ","End":"16:24.120","Text":"If we have a high power for the heated surface,"},{"Start":"16:24.120 ","End":"16:25.425","Text":"then as we know,"},{"Start":"16:25.425 ","End":"16:29.670","Text":"not all of the heat from the heated surface will be transferred to the water."},{"Start":"16:29.670 ","End":"16:33.750","Text":"Why is that? Because the bottom of the urn is made out of"},{"Start":"16:33.750 ","End":"16:37.900","Text":"copper and has a thickness D. Some of the energy,"},{"Start":"16:37.900 ","End":"16:44.425","Text":"thermal energy from the heated surface will be wasted and heating up the copper,"},{"Start":"16:44.425 ","End":"16:50.305","Text":"and will be wasted to the environment and won\u0027t go on to heat up the water."},{"Start":"16:50.305 ","End":"16:56.170","Text":"However, if the power of the heated surface is a lower value,"},{"Start":"16:56.170 ","End":"17:01.030","Text":"then we will have more heat transfer to the water"},{"Start":"17:01.030 ","End":"17:06.430","Text":"itself because less energy will be wasted at the bottom of the urn."},{"Start":"17:06.430 ","End":"17:10.915","Text":"When we\u0027re dealing with equilibrium in thermodynamics,"},{"Start":"17:10.915 ","End":"17:12.445","Text":"we\u0027re trying to find,"},{"Start":"17:12.445 ","End":"17:15.385","Text":"given a certain power,"},{"Start":"17:15.385 ","End":"17:20.770","Text":"what has to be the difference in temperature between the 2 bodies such"},{"Start":"17:20.770 ","End":"17:29.200","Text":"that all of the heat from the heated surface is transferred to the water."},{"Start":"17:29.670 ","End":"17:33.565","Text":"In other words, that all of the heat from"},{"Start":"17:33.565 ","End":"17:39.775","Text":"the heated surface will go through the copper and be transferred."},{"Start":"17:39.775 ","End":"17:45.520","Text":"All the heat will be transferred by the copper to the water."},{"Start":"17:45.520 ","End":"17:47.980","Text":"Now, also notice that if"},{"Start":"17:47.980 ","End":"17:53.875","Text":"the temperature jump or the difference in temperature between the 2 bodies is too big,"},{"Start":"17:53.875 ","End":"18:00.430","Text":"our power won\u0027t be enough in order to reach from the heated surface to the water."},{"Start":"18:00.430 ","End":"18:04.970","Text":"Let\u0027s see what this temperature difference has to be."},{"Start":"18:05.490 ","End":"18:09.460","Text":"We\u0027re going to be using this equation over here."},{"Start":"18:09.460 ","End":"18:15.955","Text":"First of all, H is the power going from the heated surface through"},{"Start":"18:15.955 ","End":"18:19.285","Text":"the copper base and into the water"},{"Start":"18:19.285 ","End":"18:23.905","Text":"so that is what this is and it\u0027s given to us over here."},{"Start":"18:23.905 ","End":"18:29.755","Text":"H is the power from the heated surface through the copper into the water."},{"Start":"18:29.755 ","End":"18:33.415","Text":"This is equal to K,"},{"Start":"18:33.415 ","End":"18:35.920","Text":"which we\u0027re given in the question is"},{"Start":"18:35.920 ","End":"18:40.555","Text":"the thermal conductivity specific to copper in this question."},{"Start":"18:40.555 ","End":"18:43.660","Text":"Then we\u0027re multiplying all of this by A,"},{"Start":"18:43.660 ","End":"18:46.750","Text":"the cross-sectional area of the urn,"},{"Start":"18:46.750 ","End":"18:50.230","Text":"which we know is also given to us as A,"},{"Start":"18:50.230 ","End":"18:54.700","Text":"and we\u0027re dividing it by the distance that it needs to"},{"Start":"18:54.700 ","End":"19:01.030","Text":"travel between the heated surface and until it reaches the bottom of the urn."},{"Start":"19:01.030 ","End":"19:06.550","Text":"That means that we\u0027re dealing with the thickness of the base of the urn."},{"Start":"19:06.550 ","End":"19:09.940","Text":"It has to travel from the heated surface through the thickness"},{"Start":"19:09.940 ","End":"19:13.165","Text":"of the base of the urn until it reaches the water."},{"Start":"19:13.165 ","End":"19:17.830","Text":"We know that that is D. That\u0027s the thickness of"},{"Start":"19:17.830 ","End":"19:22.510","Text":"the base of the urn and then all of this is multiplied"},{"Start":"19:22.510 ","End":"19:27.385","Text":"by delta T. Delta T is the temperature jump from"},{"Start":"19:27.385 ","End":"19:33.355","Text":"1 side of the copper to the other side of the copper."},{"Start":"19:33.355 ","End":"19:38.500","Text":"In our case, our delta T is"},{"Start":"19:38.500 ","End":"19:41.545","Text":"the difference in the temperature between the 2 sides"},{"Start":"19:41.545 ","End":"19:44.965","Text":"of the copper bottom or the copper base of the urn."},{"Start":"19:44.965 ","End":"19:47.845","Text":"We know that on 1 side of the base of the urn,"},{"Start":"19:47.845 ","End":"19:49.735","Text":"we have the temperature of the water."},{"Start":"19:49.735 ","End":"19:53.785","Text":"We have the water which is at 0 degrees,"},{"Start":"19:53.785 ","End":"19:56.920","Text":"and at the other side of the base of the urn,"},{"Start":"19:56.920 ","End":"20:00.155","Text":"we have the heated surface."},{"Start":"20:00.155 ","End":"20:02.850","Text":"That we don\u0027t know what the temperature is,"},{"Start":"20:02.850 ","End":"20:07.290","Text":"so let\u0027s just call it Ts for T surface,"},{"Start":"20:07.290 ","End":"20:09.255","Text":"the temperature of the heated surface."},{"Start":"20:09.255 ","End":"20:12.040","Text":"That\u0027s what our delta T is."},{"Start":"20:12.050 ","End":"20:15.330","Text":"As we can see, if the difference in"},{"Start":"20:15.330 ","End":"20:18.735","Text":"temperature between the water and the heated surface is small,"},{"Start":"20:18.735 ","End":"20:21.525","Text":"that means that delta T will be small."},{"Start":"20:21.525 ","End":"20:25.655","Text":"Delta T is proportional to our power"},{"Start":"20:25.655 ","End":"20:31.795","Text":"H. If we have a small jump in temperature between the water and the heated surface,"},{"Start":"20:31.795 ","End":"20:37.600","Text":"that means that our power is going to be also very, very low."},{"Start":"20:37.600 ","End":"20:42.740","Text":"That means that we\u0027re going to have a low efficiency for heat transfer."},{"Start":"20:43.110 ","End":"20:49.540","Text":"What is actually happening when we have a low rate of heat transfer?"},{"Start":"20:49.540 ","End":"20:56.395","Text":"It means that there is enough time for the heat to be lost to the environment."},{"Start":"20:56.395 ","End":"21:00.220","Text":"That\u0027s what happens when heat transfer is slow."},{"Start":"21:00.220 ","End":"21:03.085","Text":"We want our heat transfer to be faster."},{"Start":"21:03.085 ","End":"21:05.260","Text":"That means we want a larger H,"},{"Start":"21:05.260 ","End":"21:11.545","Text":"which means that we need a larger temperature jump between the temperature of the water"},{"Start":"21:11.545 ","End":"21:18.715","Text":"and the temperature of the heated plate or the heated surface."},{"Start":"21:18.715 ","End":"21:22.270","Text":"Now, notice in all the questions that we were asked,"},{"Start":"21:22.270 ","End":"21:25.720","Text":"we just assumed that there was no heat loss to"},{"Start":"21:25.720 ","End":"21:27.940","Text":"the environment and that all the heat from"},{"Start":"21:27.940 ","End":"21:31.030","Text":"the surface was transferred straight to the water."},{"Start":"21:31.030 ","End":"21:33.280","Text":"What we\u0027re doing right now,"},{"Start":"21:33.280 ","End":"21:37.220","Text":"let\u0027s call this star."},{"Start":"21:37.290 ","End":"21:39.685","Text":"Bonus question."},{"Start":"21:39.685 ","End":"21:44.440","Text":"What we\u0027re doing right now is we\u0027re seeing what we have to do."},{"Start":"21:44.440 ","End":"21:51.430","Text":"The conditions that we have to meet in order for us not to lose this heat."},{"Start":"21:51.430 ","End":"21:55.885","Text":"In order to have a heat transfer at the correct rate"},{"Start":"21:55.885 ","End":"22:02.120","Text":"such that it is the most efficient and no heat is lost to the environment."},{"Start":"22:02.280 ","End":"22:10.465","Text":"Here our only unknown is the temperature of our heated surface Ts."},{"Start":"22:10.465 ","End":"22:13.645","Text":"H, K, A, D,"},{"Start":"22:13.645 ","End":"22:18.144","Text":"and 0 is given to us in the question."},{"Start":"22:18.144 ","End":"22:23.890","Text":"That means that all we have to do is to isolate out this Ts and then we can find"},{"Start":"22:23.890 ","End":"22:31.070","Text":"the temperature that the heated surface needs to be at such that there is no energy loss."},{"Start":"22:31.350 ","End":"22:37.840","Text":"If you\u0027re ever being asked in thermodynamics to find the point at equilibrium,"},{"Start":"22:37.840 ","End":"22:41.650","Text":"that means the point at which there is no energy loss in the system"},{"Start":"22:41.650 ","End":"22:46.030","Text":"and that means that you have to use this equation over here, the H,"},{"Start":"22:46.030 ","End":"22:49.135","Text":"your power is equal to K,"},{"Start":"22:49.135 ","End":"22:53.620","Text":"which is the thermal conductivity of a certain material multiplied by A,"},{"Start":"22:53.620 ","End":"22:59.170","Text":"the cross-sectional area divided by the distance the thermal energy"},{"Start":"22:59.170 ","End":"23:04.840","Text":"needs to travel multiplied by the temperature jump between the 2 bodies."},{"Start":"23:04.840 ","End":"23:11.980","Text":"In here it would be between the heated surface and the water."},{"Start":"23:11.980 ","End":"23:14.930","Text":"That\u0027s the end of this question."}],"ID":12351},{"Watched":false,"Name":"Exercise 2","Duration":"11m 6s","ChapterTopicVideoID":11924,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"A lead, ball of mass m,"},{"Start":"00:02.235 ","End":"00:05.985","Text":"is dropped from a height h. After hitting the floor,"},{"Start":"00:05.985 ","End":"00:11.085","Text":"the ball bounces back to height of 1/10th of its original height."},{"Start":"00:11.085 ","End":"00:14.385","Text":"The heat capacity of the lead is given,"},{"Start":"00:14.385 ","End":"00:17.355","Text":"the latent heat of the lead is also given,"},{"Start":"00:17.355 ","End":"00:22.005","Text":"and the melting point of the lead is given."},{"Start":"00:22.005 ","End":"00:23.580","Text":"Question number 1 is,"},{"Start":"00:23.580 ","End":"00:28.930","Text":"how much heat is transferred to the ball due to it hitting the floor?"},{"Start":"00:29.660 ","End":"00:35.130","Text":"As we know, heat is in units of energy,"},{"Start":"00:35.130 ","End":"00:38.640","Text":"because heat is simply thermal energy."},{"Start":"00:38.640 ","End":"00:45.260","Text":"We can say that the heat transferred to the floor is equal to the change in"},{"Start":"00:45.260 ","End":"00:54.180","Text":"energy due to the ball falling from a height and then bouncing back up slightly."},{"Start":"00:54.180 ","End":"00:58.085","Text":"What we have here is potential energy."},{"Start":"00:58.085 ","End":"01:04.235","Text":"The change in energy is simply going to be the initial potential energy,"},{"Start":"01:04.235 ","End":"01:10.745","Text":"which is the mass of the ball of lead multiplied by gravity,"},{"Start":"01:10.745 ","End":"01:13.335","Text":"multiplied by its original height,"},{"Start":"01:13.335 ","End":"01:16.080","Text":"minus its final energy."},{"Start":"01:16.080 ","End":"01:18.075","Text":"What is its final energy?"},{"Start":"01:18.075 ","End":"01:20.825","Text":"It\u0027s its mass multiplied by g,"},{"Start":"01:20.825 ","End":"01:23.750","Text":"multiplied by its final height, which we\u0027re given,"},{"Start":"01:23.750 ","End":"01:28.200","Text":"is equal to h divided by 10."},{"Start":"01:28.910 ","End":"01:35.435","Text":"In other words, all the energy that I lost due to this change in height,"},{"Start":"01:35.435 ","End":"01:37.010","Text":"this reduction in heights,"},{"Start":"01:37.010 ","End":"01:40.675","Text":"I lost to heat energy."},{"Start":"01:40.675 ","End":"01:45.490","Text":"Now let\u0027s answer question number 2."},{"Start":"01:45.740 ","End":"01:47.840","Text":"Question number 2 is,"},{"Start":"01:47.840 ","End":"01:53.760","Text":"what is the change in internal energy due to the ball hitting the floor?"},{"Start":"01:54.440 ","End":"01:59.465","Text":"Question number 2 is exactly the same as question number 1."},{"Start":"01:59.465 ","End":"02:04.370","Text":"All of the energy that we lost due to this change in height,"},{"Start":"02:04.370 ","End":"02:06.800","Text":"so this change in potential energy,"},{"Start":"02:06.800 ","End":"02:10.325","Text":"was therefore lost to heat."},{"Start":"02:10.325 ","End":"02:18.635","Text":"Of course, this heat is going to raise the internal energy of the ball."},{"Start":"02:18.635 ","End":"02:25.685","Text":"We can say that the heat is simply equal to our Delta E over here,"},{"Start":"02:25.685 ","End":"02:35.025","Text":"which we got was equal to mgh minus mgh divided by 10."},{"Start":"02:35.025 ","End":"02:39.889","Text":"This change in energy resulted in heat,"},{"Start":"02:39.889 ","End":"02:46.200","Text":"which this heat therefore raises the internal energy of the ball."},{"Start":"02:46.460 ","End":"02:50.995","Text":"Now let\u0027s answer question number 3."},{"Start":"02:50.995 ","End":"02:54.815","Text":"The temperature of the ball before it\u0027s full is T_0."},{"Start":"02:54.815 ","End":"02:59.400","Text":"Calculate its temperature after it hits the ground."},{"Start":"02:59.810 ","End":"03:02.490","Text":"We know that from the fall,"},{"Start":"03:02.490 ","End":"03:09.440","Text":"the ball lost energy and this energy was transferred to heat, to thermal energy."},{"Start":"03:09.440 ","End":"03:11.990","Text":"What we want to know is how"},{"Start":"03:11.990 ","End":"03:19.400","Text":"this thermal energy has increased or decreased the temperature of the ball."},{"Start":"03:19.400 ","End":"03:23.090","Text":"We know that the relationship between heat"},{"Start":"03:23.090 ","End":"03:26.630","Text":"and temperature is given by this equation: Q,"},{"Start":"03:26.630 ","End":"03:28.340","Text":"heat or thermal energy,"},{"Start":"03:28.340 ","End":"03:33.605","Text":"is equal to mass multiplied by heat capacity,"},{"Start":"03:33.605 ","End":"03:36.205","Text":"multiplied by Delta T,"},{"Start":"03:36.205 ","End":"03:38.920","Text":"the change in temperature."},{"Start":"03:40.250 ","End":"03:43.565","Text":"Our Q we already have,"},{"Start":"03:43.565 ","End":"03:49.580","Text":"it\u0027s equal to mgh"},{"Start":"03:49.580 ","End":"03:55.275","Text":"minus mgh divided by 10."},{"Start":"03:55.275 ","End":"03:57.245","Text":"This is equal to the mass,"},{"Start":"03:57.245 ","End":"03:59.780","Text":"which is given to us in the question,"},{"Start":"03:59.780 ","End":"04:01.760","Text":"multiplied by the heat capacity,"},{"Start":"04:01.760 ","End":"04:03.950","Text":"which is also given to us in the question,"},{"Start":"04:03.950 ","End":"04:07.445","Text":"multiplied by Delta T. What is Delta T?"},{"Start":"04:07.445 ","End":"04:12.975","Text":"It\u0027s our final temperature where this is our unknown,"},{"Start":"04:12.975 ","End":"04:15.390","Text":"so I\u0027ll add a tilde on top,"},{"Start":"04:15.390 ","End":"04:18.260","Text":"minus our initial temperature,"},{"Start":"04:18.260 ","End":"04:23.370","Text":"which is given to us in the question T naught or T_0."},{"Start":"04:23.370 ","End":"04:31.680","Text":"Now, all we have to do is we have to do some algebra to isolate out this T final."},{"Start":"04:32.020 ","End":"04:34.585","Text":"Once you do some algebra,"},{"Start":"04:34.585 ","End":"04:38.710","Text":"we can see that the final temperature is going to"},{"Start":"04:38.710 ","End":"04:43.740","Text":"be the initial temperature plus 9/10ths gh"},{"Start":"04:43.740 ","End":"04:48.145","Text":"divided by C. We can see that"},{"Start":"04:48.145 ","End":"04:54.840","Text":"this increase in thermal energy is going to result in an increase in temperature."},{"Start":"04:54.840 ","End":"04:59.045","Text":"Now let\u0027s answer question number 4."},{"Start":"04:59.045 ","End":"05:00.985","Text":"Question number 4 is,"},{"Start":"05:00.985 ","End":"05:02.680","Text":"after hitting the floor,"},{"Start":"05:02.680 ","End":"05:08.005","Text":"the lead ball will always bounce to 1/10th of its original height."},{"Start":"05:08.005 ","End":"05:13.610","Text":"All the energy from the collision with the ground has transferred to the ball."},{"Start":"05:13.610 ","End":"05:20.005","Text":"From what height must the ball be dropped such that it will completely melt?"},{"Start":"05:20.005 ","End":"05:23.990","Text":"We\u0027re trying to find the ideal heights to drop the ball"},{"Start":"05:23.990 ","End":"05:28.860","Text":"from such that the entire ball is going to melt."},{"Start":"05:29.090 ","End":"05:33.900","Text":"What we want to calculate right now is how much heat energy"},{"Start":"05:33.900 ","End":"05:37.820","Text":"needs to be put in in order to increase the temperature of"},{"Start":"05:37.820 ","End":"05:40.710","Text":"the lead ball from before its fall which is"},{"Start":"05:40.710 ","End":"05:46.640","Text":"T_0 to its melting point in temperature, which is this."},{"Start":"05:46.640 ","End":"05:53.370","Text":"Then we want to add in our latent heat in order to actually melt the ball."},{"Start":"05:54.170 ","End":"05:58.460","Text":"The thermal energy that I have to put in in order to get"},{"Start":"05:58.460 ","End":"06:04.280","Text":"a temperature increase is going to be equal to the equation m,"},{"Start":"06:04.280 ","End":"06:06.665","Text":"the mass of the ball,"},{"Start":"06:06.665 ","End":"06:09.755","Text":"multiplied by its heat capacity,"},{"Start":"06:09.755 ","End":"06:16.480","Text":"multiplied by our Delta T. The temperature change that we want."},{"Start":"06:16.480 ","End":"06:18.660","Text":"Let\u0027s plug in our values."},{"Start":"06:18.660 ","End":"06:19.980","Text":"The mass of the ball,"},{"Start":"06:19.980 ","End":"06:21.985","Text":"in the question is given as m,"},{"Start":"06:21.985 ","End":"06:27.960","Text":"our heat capacity of lead is given to us as C_pb."},{"Start":"06:28.280 ","End":"06:33.600","Text":"Then we\u0027re multiplying it by the change in temperature that we want."},{"Start":"06:33.600 ","End":"06:38.780","Text":"We said that we want it to go from its original temperature before the ball fell,"},{"Start":"06:38.780 ","End":"06:40.715","Text":"which was at a temperature of T_0,"},{"Start":"06:40.715 ","End":"06:45.755","Text":"to its melting point temperature because we want the ball to completely melt,"},{"Start":"06:45.755 ","End":"06:49.160","Text":"which means that the ball has to be at its melting point."},{"Start":"06:49.160 ","End":"06:51.380","Text":"The change in temperature,"},{"Start":"06:51.380 ","End":"06:53.675","Text":"how much we want to raise the temperature,"},{"Start":"06:53.675 ","End":"06:55.460","Text":"is our final temperature,"},{"Start":"06:55.460 ","End":"06:58.550","Text":"T, melting of Pb."},{"Start":"06:58.550 ","End":"07:00.035","Text":"Pb, of course,"},{"Start":"07:00.035 ","End":"07:04.054","Text":"is the chemical symbol for lead,"},{"Start":"07:04.054 ","End":"07:08.160","Text":"minus the original temperature, which was T_0."},{"Start":"07:08.240 ","End":"07:12.650","Text":"This is the thermal energy that we have to put in to"},{"Start":"07:12.650 ","End":"07:15.530","Text":"increase the temperature of the ball from"},{"Start":"07:15.530 ","End":"07:19.890","Text":"its original temperature to its melting point temperature."},{"Start":"07:19.910 ","End":"07:25.325","Text":"Now, our ball of lead is at the correct temperature for melting,"},{"Start":"07:25.325 ","End":"07:28.265","Text":"but now we need the ball to actually melt,"},{"Start":"07:28.265 ","End":"07:32.585","Text":"which means that we need to use our latent heat."},{"Start":"07:32.585 ","End":"07:38.360","Text":"Latent heat is obviously given in units of joules per gram."},{"Start":"07:38.360 ","End":"07:43.385","Text":"Let\u0027s say that we have to add this extra energy for melting."},{"Start":"07:43.385 ","End":"07:46.920","Text":"In order to get it in the units of joules,"},{"Start":"07:46.920 ","End":"07:48.620","Text":"we\u0027re going to use latent heat,"},{"Start":"07:48.620 ","End":"07:52.850","Text":"which is joules per gram, multiplied by gram."},{"Start":"07:52.850 ","End":"07:55.670","Text":"How much it\u0027s of mass m?"},{"Start":"07:55.670 ","End":"08:01.970","Text":"We\u0027re going to multiply our L_fus,"},{"Start":"08:01.970 ","End":"08:05.520","Text":"multiplied by the mass of the ball,"},{"Start":"08:05.520 ","End":"08:11.580","Text":"and then we have units for energy."},{"Start":"08:11.580 ","End":"08:17.585","Text":"Now, what we have to do with these two is to find"},{"Start":"08:17.585 ","End":"08:23.310","Text":"the total energy needed in order to melt the entire ball,"},{"Start":"08:23.310 ","End":"08:27.555","Text":"we\u0027re going to add these two up."},{"Start":"08:27.555 ","End":"08:34.415","Text":"The total energy is going to be equal to Q plus"},{"Start":"08:34.415 ","End":"08:42.020","Text":"E. This is the total energy that I need in order to melt my lead ball."},{"Start":"08:42.020 ","End":"08:44.840","Text":"But now my question is actually from what height"},{"Start":"08:44.840 ","End":"08:48.170","Text":"must the ball be dropped such that it will melt?"},{"Start":"08:48.170 ","End":"08:52.175","Text":"I need to find out from what height I need to drop my ball"},{"Start":"08:52.175 ","End":"08:57.690","Text":"from so that it will generate this much heat energy."},{"Start":"08:57.690 ","End":"09:01.145","Text":"What we could see from the previous questions is,"},{"Start":"09:01.145 ","End":"09:03.125","Text":"where did our thermal energy come from?"},{"Start":"09:03.125 ","End":"09:07.505","Text":"It came from this change in potential energy."},{"Start":"09:07.505 ","End":"09:14.010","Text":"That\u0027s great. We can see that potential energy involves our variable h, height."},{"Start":"09:14.010 ","End":"09:17.585","Text":"What we\u0027re going to do is we\u0027re going to see by"},{"Start":"09:17.585 ","End":"09:21.925","Text":"using a similar equation using potential energy,"},{"Start":"09:21.925 ","End":"09:27.665","Text":"how much energy from height is going to be converted to this thermal energy?"},{"Start":"09:27.665 ","End":"09:33.275","Text":"What do we can see at the beginning of the question right over here,"},{"Start":"09:33.275 ","End":"09:34.985","Text":"is that after hitting the floor,"},{"Start":"09:34.985 ","End":"09:39.865","Text":"the ball bounces back to a height of h divided by 10."},{"Start":"09:39.865 ","End":"09:45.080","Text":"At the end, after the ball has been dropped and reach the ground,"},{"Start":"09:45.080 ","End":"09:49.835","Text":"it bounces back to 1/10th of its original height. What does that mean?"},{"Start":"09:49.835 ","End":"09:55.280","Text":"It\u0027s lost 9/10ths of its original height."},{"Start":"09:55.280 ","End":"09:59.135","Text":"It\u0027s lost 90 percent of its original height,"},{"Start":"09:59.135 ","End":"10:06.210","Text":"which means that it\u0027s lost 90 percent of its potential energy."},{"Start":"10:06.210 ","End":"10:08.250","Text":"Where has that energy gone to?"},{"Start":"10:08.250 ","End":"10:12.890","Text":"It\u0027s being transferred to this thermal heat energy."},{"Start":"10:13.160 ","End":"10:18.830","Text":"That means that the total energy that we need in order to do this is going to be"},{"Start":"10:18.830 ","End":"10:23.915","Text":"equal to mg multiplied by the height difference from its original height,"},{"Start":"10:23.915 ","End":"10:27.390","Text":"which is 10h divided by 10,"},{"Start":"10:27.390 ","End":"10:29.850","Text":"to its final height,"},{"Start":"10:29.850 ","End":"10:32.295","Text":"which is 1h divided by 10."},{"Start":"10:32.295 ","End":"10:35.233","Text":"We\u0027re going to multiply this by 9/10h."},{"Start":"10:35.233 ","End":"10:42.195","Text":"9/10h is the height difference."},{"Start":"10:42.195 ","End":"10:43.955","Text":"Now, all that we have to do,"},{"Start":"10:43.955 ","End":"10:49.610","Text":"is we have to isolate out this h in order to find the height that the ball must be"},{"Start":"10:49.610 ","End":"10:53.480","Text":"dropped from given this information"},{"Start":"10:53.480 ","End":"10:56.180","Text":"and these values that we\u0027ve been given in the question."},{"Start":"10:56.180 ","End":"11:00.995","Text":"I just did the algebra and this is the answer that you should get."},{"Start":"11:00.995 ","End":"11:03.920","Text":"You can check this yourself."},{"Start":"11:03.920 ","End":"11:06.780","Text":"That\u0027s the end of this lesson."}],"ID":12352},{"Watched":false,"Name":"Exercise 3","Duration":"19m 24s","ChapterTopicVideoID":11917,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:05.010","Text":"we\u0027re going to be dealing with the energy loss of an oven."},{"Start":"00:05.010 ","End":"00:09.510","Text":"This is your regular standard large kitchen oven"},{"Start":"00:09.510 ","End":"00:12.435","Text":"that you can probably find in your kitchen."},{"Start":"00:12.435 ","End":"00:16.035","Text":"Let\u0027s look at what we have over here."},{"Start":"00:16.035 ","End":"00:23.085","Text":"First of all, this K^3 is the outer layer of the oven."},{"Start":"00:23.085 ","End":"00:28.920","Text":"Now the outer layer of the oven is made out of tin."},{"Start":"00:28.920 ","End":"00:31.335","Text":"Let\u0027s write it down over here."},{"Start":"00:31.335 ","End":"00:33.045","Text":"This is made out of tin,"},{"Start":"00:33.045 ","End":"00:36.990","Text":"and tin has very good thermal conductance."},{"Start":"00:36.990 ","End":"00:41.360","Text":"Now in an oven, we can imagine that we want thermal resistivity,"},{"Start":"00:41.360 ","End":"00:43.255","Text":"because we want to insulate the oven."},{"Start":"00:43.255 ","End":"00:45.270","Text":"We don\u0027t want thermal conductance."},{"Start":"00:45.270 ","End":"00:48.020","Text":"Why is the outer layer of the oven made out of tin,"},{"Start":"00:48.020 ","End":"00:49.595","Text":"which is a good conductor?"},{"Start":"00:49.595 ","End":"00:51.630","Text":"We\u0027ll get to that later,"},{"Start":"00:51.630 ","End":"00:54.570","Text":"and its thermal conductance is 60."},{"Start":"00:54.570 ","End":"01:00.160","Text":"Then we can see that its thickness is 1 millimeter."},{"Start":"01:00.160 ","End":"01:06.710","Text":"Then the area of this whole outer layer which is enveloping"},{"Start":"01:06.710 ","End":"01:15.750","Text":"this air gap and the inner layer of the oven has a total area of 3 meters squared."},{"Start":"01:16.510 ","End":"01:21.655","Text":"Then this inner layer in white is our air."},{"Start":"01:21.655 ","End":"01:26.519","Text":"We have just some air in the space,"},{"Start":"01:26.519 ","End":"01:30.735","Text":"that\u0027s our K^2. This is air."},{"Start":"01:30.735 ","End":"01:36.890","Text":"What we can see is that its thermal conductivity is 0.265."},{"Start":"01:36.890 ","End":"01:42.160","Text":"What we can see is that air has very low thermal conductance."},{"Start":"01:42.160 ","End":"01:46.115","Text":"We can say that air is a thermal resistor."},{"Start":"01:46.115 ","End":"01:53.925","Text":"We can see its thermal conductance is 0.265 is much less than that of tin, which was 60."},{"Start":"01:53.925 ","End":"01:59.205","Text":"This thickness of this air layer is 10 millimeters."},{"Start":"01:59.205 ","End":"02:03.475","Text":"The total area of this air is also"},{"Start":"02:03.475 ","End":"02:08.075","Text":"3 meters squared because the difference between these two is low,"},{"Start":"02:08.075 ","End":"02:09.830","Text":"we can see that we\u0027re dealing here,"},{"Start":"02:09.830 ","End":"02:13.940","Text":"we go in 1 millimeters and then 10 millimeters what\u0027s"},{"Start":"02:13.940 ","End":"02:19.350","Text":"a total of 11 millimeters when we\u0027re speaking about 3 meters squared."},{"Start":"02:19.540 ","End":"02:23.820","Text":"This also has an area of 3 meters squared."},{"Start":"02:23.820 ","End":"02:27.550","Text":"Then the next layer is K^1."},{"Start":"02:27.550 ","End":"02:32.680","Text":"Now, K^1 is some kind of ceramic material."},{"Start":"02:32.870 ","End":"02:39.115","Text":"As we can see, it has a thermal conductivity of 0.4."},{"Start":"02:39.115 ","End":"02:44.215","Text":"We can also see that its thermal conductivity is very low."},{"Start":"02:44.215 ","End":"02:49.310","Text":"It\u0027s only slightly higher than the thermal conductivity of air,"},{"Start":"02:49.310 ","End":"02:52.325","Text":"but we can see that it has a very low conductance,"},{"Start":"02:52.325 ","End":"02:58.475","Text":"especially relative to tin that has a high conductance and its thickness."},{"Start":"02:58.475 ","End":"03:02.195","Text":"The thickness of the inner layer of the oven is 2 millimeters."},{"Start":"03:02.195 ","End":"03:06.575","Text":"Again, the total surface area is also 3 meters squared."},{"Start":"03:06.575 ","End":"03:10.055","Text":"Because this total of 13 millimeters"},{"Start":"03:10.055 ","End":"03:13.745","Text":"isn\u0027t going to make a big difference to this 3 meters squared area,"},{"Start":"03:13.745 ","End":"03:16.320","Text":"so we can just ignore it."},{"Start":"03:17.210 ","End":"03:20.445","Text":"Now, what is the point of this oven?"},{"Start":"03:20.445 ","End":"03:26.945","Text":"We have our soup, which we\u0027re trying to heat up at 200 degrees Celsius or centigrade."},{"Start":"03:26.945 ","End":"03:30.910","Text":"What we want is we want to design"},{"Start":"03:30.910 ","End":"03:35.410","Text":"our oven such that once our oven reaches this temperature,"},{"Start":"03:35.410 ","End":"03:41.740","Text":"there\u0027ll be very little energy lost to the surroundings such that our oven"},{"Start":"03:41.740 ","End":"03:50.500","Text":"can stay at this 200 degrees Celsius with minimal energy being put in constantly."},{"Start":"03:50.500 ","End":"03:58.490","Text":"What we want is to keep the heat in this area over here where the food is."},{"Start":"03:58.490 ","End":"04:06.570","Text":"Question Number 1 is what is the oven\u0027s thermal resistance?"},{"Start":"04:06.570 ","End":"04:09.150","Text":"How are we going to do this?"},{"Start":"04:09.150 ","End":"04:13.340","Text":"Our outer layer or middle layer and our inner layer can"},{"Start":"04:13.340 ","End":"04:17.690","Text":"simply be considered like resistors, thermal resistance."},{"Start":"04:17.690 ","End":"04:22.400","Text":"What we\u0027re going to do is we\u0027re going to calculate the thermal resistance of each layer."},{"Start":"04:22.400 ","End":"04:27.245","Text":"Then we can see that we have a resistor attached to another resistor,"},{"Start":"04:27.245 ","End":"04:28.700","Text":"attached to another resistor."},{"Start":"04:28.700 ","End":"04:31.820","Text":"How do we find the total resistance?"},{"Start":"04:31.820 ","End":"04:36.750","Text":"We\u0027re just going to add up all the resistances together."},{"Start":"04:36.820 ","End":"04:38.930","Text":"Now, as we know,"},{"Start":"04:38.930 ","End":"04:43.010","Text":"thermal resistivity is given by this equation;"},{"Start":"04:43.010 ","End":"04:50.000","Text":"R is for resistance and it\u0027s equal to the width of"},{"Start":"04:50.000 ","End":"04:57.425","Text":"the layer W divided by the total area of the layer,"},{"Start":"04:57.425 ","End":"05:01.925","Text":"multiplied by its thermal conductance"},{"Start":"05:01.925 ","End":"05:09.470","Text":"K. This is an equation for your equation sheets."},{"Start":"05:09.470 ","End":"05:15.200","Text":"Now let\u0027s do the calculation for each one of these layers."},{"Start":"05:15.200 ","End":"05:18.035","Text":"The resistance of the first layer,"},{"Start":"05:18.035 ","End":"05:21.110","Text":"which is this inner layer of ceramic."},{"Start":"05:21.110 ","End":"05:24.005","Text":"That\u0027s going to be equal to the width of the layer,"},{"Start":"05:24.005 ","End":"05:25.970","Text":"which is 2 millimeters but of course,"},{"Start":"05:25.970 ","End":"05:27.470","Text":"we want this in meters."},{"Start":"05:27.470 ","End":"05:37.685","Text":"That\u0027s going to be equal to 0.002 divided by the area which is 3 meters squared,"},{"Start":"05:37.685 ","End":"05:44.405","Text":"so 3 multiplied by the thermal conductance of this layer,"},{"Start":"05:44.405 ","End":"05:48.865","Text":"which we were given is equal to 0.4."},{"Start":"05:48.865 ","End":"05:51.815","Text":"Once we plug this into our calculator,"},{"Start":"05:51.815 ","End":"06:00.685","Text":"we\u0027ll get to the resistivity of the inner layer is approximately equal to 0.001."},{"Start":"06:00.685 ","End":"06:05.090","Text":"Now let\u0027s look at the resistance of our inner layer,"},{"Start":"06:05.090 ","End":"06:07.310","Text":"which is this area."},{"Start":"06:07.310 ","End":"06:09.755","Text":"First of all, the width is 10 millimeters,"},{"Start":"06:09.755 ","End":"06:18.665","Text":"that\u0027s equal to 0.01 meters divided by the area which is 3 meters squared,"},{"Start":"06:18.665 ","End":"06:20.990","Text":"multiplied by the thermal conductance,"},{"Start":"06:20.990 ","End":"06:25.410","Text":"which of air was equal to 0.265."},{"Start":"06:26.900 ","End":"06:30.710","Text":"Again, once we plug this into our calculator,"},{"Start":"06:30.710 ","End":"06:37.190","Text":"we\u0027ll get that this is equal to approximately 0.012."},{"Start":"06:37.190 ","End":"06:41.660","Text":"Now let\u0027s find the resistivity of the outer layer."},{"Start":"06:41.660 ","End":"06:43.820","Text":"The layer made out of tin."},{"Start":"06:43.820 ","End":"06:46.580","Text":"We have its thickness which is equal to"},{"Start":"06:46.580 ","End":"06:54.395","Text":"0.001 divided by the total area which is 3 meters squared,"},{"Start":"06:54.395 ","End":"06:58.595","Text":"multiplied by the conductivity of this tin,"},{"Start":"06:58.595 ","End":"07:02.700","Text":"which was equal to 60."},{"Start":"07:02.990 ","End":"07:05.760","Text":"Then this is approximately,"},{"Start":"07:05.760 ","End":"07:07.965","Text":"it works out a very small number."},{"Start":"07:07.965 ","End":"07:10.250","Text":"It\u0027s approximately equal to 0."},{"Start":"07:10.250 ","End":"07:15.110","Text":"Now we can see that the total resistivity is just equal to"},{"Start":"07:15.110 ","End":"07:21.050","Text":"R_1 plus R_2 plus R_3."},{"Start":"07:21.050 ","End":"07:24.815","Text":"That\u0027s equal to 0.001"},{"Start":"07:24.815 ","End":"07:31.710","Text":"plus 0.012 plus 0."},{"Start":"07:31.710 ","End":"07:34.065","Text":"The total resistivity, therefore,"},{"Start":"07:34.065 ","End":"07:41.100","Text":"is equal to 0.013."},{"Start":"07:41.100 ","End":"07:44.150","Text":"Of course, all of these numbers are approximate,"},{"Start":"07:44.150 ","End":"07:48.541","Text":"so we can do the squiggly equal sign."},{"Start":"07:48.541 ","End":"07:51.955","Text":"This is the answer to Question Number 1,"},{"Start":"07:51.955 ","End":"07:54.775","Text":"and now let\u0027s see what this means."},{"Start":"07:54.775 ","End":"07:58.480","Text":"As we can see, the outer layer, rarely,"},{"Start":"07:58.480 ","End":"08:03.520","Text":"it doesn\u0027t provide us with any thermal resistance."},{"Start":"08:03.520 ","End":"08:05.815","Text":"It\u0027s approximately equal to 0."},{"Start":"08:05.815 ","End":"08:08.230","Text":"We can also see that the inner layer,"},{"Start":"08:08.230 ","End":"08:11.890","Text":"which is also quite a good thermal resistor,"},{"Start":"08:11.890 ","End":"08:15.835","Text":"it doesn\u0027t really provide us with that much thermal resistance."},{"Start":"08:15.835 ","End":"08:18.010","Text":"That\u0027s a little bit, but not that much."},{"Start":"08:18.010 ","End":"08:23.995","Text":"What we can see that is critical to this thermal resistance,"},{"Start":"08:23.995 ","End":"08:29.050","Text":"so to keeping the heat inside this inner section of the oven,"},{"Start":"08:29.050 ","End":"08:31.240","Text":"is our layer of air."},{"Start":"08:31.240 ","End":"08:34.750","Text":"Air is a very good thermal resistor,"},{"Start":"08:34.750 ","End":"08:38.725","Text":"and this is what really keeps the heat in."},{"Start":"08:38.725 ","End":"08:43.105","Text":"In actual fact, our outer layer of tin,"},{"Start":"08:43.105 ","End":"08:45.925","Text":"which we can see is quite a good thermal conductor,"},{"Start":"08:45.925 ","End":"08:47.320","Text":"we don\u0027t really need it."},{"Start":"08:47.320 ","End":"08:50.650","Text":"We\u0027re just trying to use it to keep the air in."},{"Start":"08:50.650 ","End":"08:53.930","Text":"That\u0027s all it\u0027s really being used for."},{"Start":"08:54.150 ","End":"08:58.195","Text":"Now, let\u0027s answer Question Number 2."},{"Start":"08:58.195 ","End":"09:01.060","Text":"Question Number 2 is how much energy is required in"},{"Start":"09:01.060 ","End":"09:04.359","Text":"order to give the oven at 200 degrees Celsius?"},{"Start":"09:04.359 ","End":"09:09.249","Text":"As we know, once the oven reaches 200 degrees Celsius,"},{"Start":"09:09.249 ","End":"09:11.665","Text":"if we unplug the oven,"},{"Start":"09:11.665 ","End":"09:14.905","Text":"slowly, this thermal energy, the heat,"},{"Start":"09:14.905 ","End":"09:20.650","Text":"is going to move to the surroundings and the oven will cool down."},{"Start":"09:20.650 ","End":"09:24.265","Text":"That means that we have to plug the oven into"},{"Start":"09:24.265 ","End":"09:30.430","Text":"the electricity box and we have to constantly pump"},{"Start":"09:30.430 ","End":"09:35.350","Text":"in electrical energy in order to convert that electrical energy into"},{"Start":"09:35.350 ","End":"09:40.972","Text":"heat energy in order to maintain the temperature inside here at a constant."},{"Start":"09:40.972 ","End":"09:44.215","Text":"So that it\u0027s always at 200 degrees Celsius."},{"Start":"09:44.215 ","End":"09:46.840","Text":"How much energy do we need to pump in,"},{"Start":"09:46.840 ","End":"09:49.255","Text":"in order to maintain this temperature?"},{"Start":"09:49.255 ","End":"09:53.470","Text":"What are we going to do in order to answer this question is we\u0027re going to"},{"Start":"09:53.470 ","End":"09:58.120","Text":"redraw this oven as a simple diagram."},{"Start":"09:58.120 ","End":"10:02.935","Text":"What we know about the oven is that we can think of it as a box."},{"Start":"10:02.935 ","End":"10:05.050","Text":"We have our oven,"},{"Start":"10:05.050 ","End":"10:10.700","Text":"which is this box of a temperature of 200 degrees Celsius."},{"Start":"10:10.700 ","End":"10:18.280","Text":"Then we know that it is attached to some rod that has a resistance of R_T,"},{"Start":"10:18.280 ","End":"10:24.655","Text":"which we calculated in the previous question and it is touching this cold box,"},{"Start":"10:24.655 ","End":"10:28.450","Text":"which we\u0027re being told is just in the oven scenario,"},{"Start":"10:28.450 ","End":"10:30.235","Text":"it\u0027s just the outside of the room."},{"Start":"10:30.235 ","End":"10:34.030","Text":"Here, we can just consider the room as a whole,"},{"Start":"10:34.030 ","End":"10:39.950","Text":"as a different box with a temperature of 15 degrees Celsius."},{"Start":"10:40.020 ","End":"10:45.730","Text":"What do we know? We know that we have heat that is being"},{"Start":"10:45.730 ","End":"10:50.665","Text":"lost to the environment that\u0027s traveling through this rod to the outside room."},{"Start":"10:50.665 ","End":"10:53.215","Text":"In the oven, heat is leaving the oven,"},{"Start":"10:53.215 ","End":"10:58.765","Text":"it\u0027s seeping out to the outside. What does that mean?"},{"Start":"10:58.765 ","End":"11:01.075","Text":"That means that in this direction,"},{"Start":"11:01.075 ","End":"11:03.703","Text":"we have this Q_out."},{"Start":"11:03.703 ","End":"11:10.660","Text":"So we have heat leaving the oven to the room but in the question,"},{"Start":"11:10.660 ","End":"11:14.680","Text":"we want to keep the oven temperature constant at this 200 degrees"},{"Start":"11:14.680 ","End":"11:19.165","Text":"C. That means that through electrical energy,"},{"Start":"11:19.165 ","End":"11:21.370","Text":"we\u0027re going to be pumping in heat,"},{"Start":"11:21.370 ","End":"11:23.260","Text":"pumping in thermal energy."},{"Start":"11:23.260 ","End":"11:25.450","Text":"That\u0027s going to be Q_in."},{"Start":"11:25.450 ","End":"11:28.030","Text":"We know that because we\u0027re not heating the oven,"},{"Start":"11:28.030 ","End":"11:30.581","Text":"but we\u0027re maintaining this constant temperature."},{"Start":"11:30.581 ","End":"11:37.000","Text":"We know that Q_in has to be equal to Q_out because if Q_in is bigger than Q_out,"},{"Start":"11:37.000 ","End":"11:39.700","Text":"then the oven will increase in temperature."},{"Start":"11:39.700 ","End":"11:41.110","Text":"If it\u0027s less than Q_out,"},{"Start":"11:41.110 ","End":"11:42.640","Text":"then the oven will still cool,"},{"Start":"11:42.640 ","End":"11:45.080","Text":"but just more slowly."},{"Start":"11:45.210 ","End":"11:51.790","Text":"We\u0027re trying to calculate what this Q_in is and we know that it\u0027s equal to Q_out."},{"Start":"11:51.790 ","End":"11:58.220","Text":"That means that all we have to do is we have to calculate what our Q_out is equal to."},{"Start":"11:58.650 ","End":"12:02.620","Text":"Here is a useful equation and this,"},{"Start":"12:02.620 ","End":"12:06.310","Text":"you\u0027re also going to have to write down in your equation sheets."},{"Start":"12:06.310 ","End":"12:10.800","Text":"We know that Q is our heat or our thermal energy,"},{"Start":"12:10.800 ","End":"12:18.225","Text":"and we know that the first derivative of something is the rate of change of that thing."},{"Start":"12:18.225 ","End":"12:23.730","Text":"Here, we\u0027re looking at the rate of change of heat."},{"Start":"12:23.730 ","End":"12:26.845","Text":"That\u0027s exactly that. What is this equal to?"},{"Start":"12:26.845 ","End":"12:31.390","Text":"This is equal to the difference in temperature between the 2 bodies."},{"Start":"12:31.390 ","End":"12:36.520","Text":"Here, the difference in temperature would be 200 minus 15 divided"},{"Start":"12:36.520 ","End":"12:41.815","Text":"by the resistance between the 2 objects."},{"Start":"12:41.815 ","End":"12:45.460","Text":"Here, the resistance is this R_Total."},{"Start":"12:45.460 ","End":"12:48.745","Text":"That is the resistance of the inner layer,"},{"Start":"12:48.745 ","End":"12:51.860","Text":"the air layer, and the outer layer, together."},{"Start":"12:53.400 ","End":"12:57.130","Text":"This is an equation to write down and remember,"},{"Start":"12:57.130 ","End":"12:58.780","Text":"and now let\u0027s work it out."},{"Start":"12:58.780 ","End":"13:03.100","Text":"The rate of change of the heat is the difference in temperature."},{"Start":"13:03.100 ","End":"13:10.330","Text":"Two hundred degrees Celsius minus 15 degrees Celsius divided by the total resistance,"},{"Start":"13:10.330 ","End":"13:12.130","Text":"which is what we found in Question 1,"},{"Start":"13:12.130 ","End":"13:14.990","Text":"which is equal to 0.013."},{"Start":"13:15.960 ","End":"13:21.670","Text":"The rate of change of heat is approximately equal to"},{"Start":"13:21.670 ","End":"13:28.930","Text":"14,260 watts"},{"Start":"13:28.930 ","End":"13:33.290","Text":"or approximately 14 kilowatts."},{"Start":"13:34.170 ","End":"13:39.500","Text":"This is the answer to Question Number 2."},{"Start":"13:39.630 ","End":"13:44.890","Text":"This is how much heat is leaving the oven,"},{"Start":"13:44.890 ","End":"13:47.920","Text":"this is Q_out, and what we wanted to know was"},{"Start":"13:47.920 ","End":"13:51.145","Text":"what is Q_in and we know that they\u0027re both equal."},{"Start":"13:51.145 ","End":"13:55.030","Text":"This is Q_in."},{"Start":"13:55.030 ","End":"13:57.745","Text":"This is how much energy has to be"},{"Start":"13:57.745 ","End":"14:02.110","Text":"provided in order to maintain the temperature of the oven."},{"Start":"14:02.110 ","End":"14:05.690","Text":"Now, let\u0027s answer Question Number 3."},{"Start":"14:05.970 ","End":"14:11.875","Text":"We\u0027re now told that the oven is not insulated. What does that mean?"},{"Start":"14:11.875 ","End":"14:13.930","Text":"That means that this air layer,"},{"Start":"14:13.930 ","End":"14:15.610","Text":"which was our main insulation,"},{"Start":"14:15.610 ","End":"14:19.420","Text":"so this empty space now doesn\u0027t exist."},{"Start":"14:19.420 ","End":"14:24.220","Text":"That means that our inner layer and outer layer are touching."},{"Start":"14:24.220 ","End":"14:27.820","Text":"We don\u0027t have this layer K_2."},{"Start":"14:27.820 ","End":"14:32.695","Text":"Question 3 is, how much energy is required"},{"Start":"14:32.695 ","End":"14:38.420","Text":"now in order for the oven to stay at 200 degrees Celsius?"},{"Start":"14:38.430 ","End":"14:42.145","Text":"This is a similar question to Question Number 2,"},{"Start":"14:42.145 ","End":"14:47.380","Text":"the only difference is we\u0027re not taking into account this inner layer."},{"Start":"14:47.380 ","End":"14:53.590","Text":"We\u0027re going to use this same equation over here for the rate of heat loss."},{"Start":"14:53.590 ","End":"15:01.915","Text":"What we can see is we can do this Q̇ is equal to the difference in temperature Delta T,"},{"Start":"15:01.915 ","End":"15:03.595","Text":"which again it\u0027s the same."},{"Start":"15:03.595 ","End":"15:06.700","Text":"We\u0027re trying to keep the oven at 200 degrees Celsius and"},{"Start":"15:06.700 ","End":"15:09.981","Text":"the temperature outside is 15 degrees Celsius."},{"Start":"15:09.981 ","End":"15:13.885","Text":"We\u0027re dividing this by the resistance."},{"Start":"15:13.885 ","End":"15:19.075","Text":"We\u0027re going to go back to Question Number 1."},{"Start":"15:19.075 ","End":"15:25.540","Text":"We need to take into account the resistance of the outer layer and of the inner layer,"},{"Start":"15:25.540 ","End":"15:28.540","Text":"but not of the middle layer because it now doesn\u0027t exist."},{"Start":"15:28.540 ","End":"15:33.445","Text":"The resistance is going to be equal to our R_1,"},{"Start":"15:33.445 ","End":"15:34.810","Text":"which if we remember,"},{"Start":"15:34.810 ","End":"15:42.630","Text":"is equal to 0.001 plus the resistance of our outer layer."},{"Start":"15:42.630 ","End":"15:45.310","Text":"Here, that\u0027s our R_3,"},{"Start":"15:45.310 ","End":"15:47.000","Text":"so that was equal to,"},{"Start":"15:47.000 ","End":"15:49.460","Text":"and we did the calculation for Question Number 1,"},{"Start":"15:49.460 ","End":"15:52.125","Text":"approximately equal to 0."},{"Start":"15:52.125 ","End":"15:58.370","Text":"This means that our heat loss is going to be equal to approximately once we put"},{"Start":"15:58.370 ","End":"16:06.115","Text":"this into the calculator, 185,000 watts."},{"Start":"16:06.115 ","End":"16:13.715","Text":"We can see that the amount of energy that has to be put in to our oven,"},{"Start":"16:13.715 ","End":"16:18.935","Text":"when this middle layer of just air which is insulating it,"},{"Start":"16:18.935 ","End":"16:27.970","Text":"is going to increase our energy loss by a factor of approximately 12 or so."},{"Start":"16:27.970 ","End":"16:30.605","Text":"We can see that the cost of running this oven,"},{"Start":"16:30.605 ","End":"16:33.905","Text":"if we just take away this middle layer of air,"},{"Start":"16:33.905 ","End":"16:37.640","Text":"which costs nothing, increases the cost"},{"Start":"16:37.640 ","End":"16:42.560","Text":"exponentially to keep this oven at this constant temperature."},{"Start":"16:42.560 ","End":"16:48.350","Text":"This middle layer of insulation due to air is very important."},{"Start":"16:49.320 ","End":"16:52.390","Text":"This is the answer to Question Number 3,"},{"Start":"16:52.390 ","End":"16:55.315","Text":"and now, let\u0027s look at Question Number 4."},{"Start":"16:55.315 ","End":"16:59.250","Text":"Again, the middle layer now doesn\u0027t exist,"},{"Start":"16:59.250 ","End":"17:01.475","Text":"there\u0027s no insulation over here."},{"Start":"17:01.475 ","End":"17:03.140","Text":"Question Number 4 is,"},{"Start":"17:03.140 ","End":"17:07.685","Text":"if the same energy as Question Number 2 is put into the system,"},{"Start":"17:07.685 ","End":"17:12.105","Text":"what will be the temperature inside the oven?"},{"Start":"17:12.105 ","End":"17:16.860","Text":"All we have to do is we just have to rearrange this equation."},{"Start":"17:17.110 ","End":"17:26.125","Text":"We can write that Delta T is going to be equal to Q̇ multiplied by R,"},{"Start":"17:26.125 ","End":"17:36.245","Text":"where Delta T is the temperature in the oven minus the temperature outside of the oven."},{"Start":"17:36.245 ","End":"17:40.475","Text":"That is equal to our energy"},{"Start":"17:40.475 ","End":"17:43.655","Text":"that has to be put into the system from Question Number 2, which is here,"},{"Start":"17:43.655 ","End":"17:46.910","Text":"multiplied by the resistance,"},{"Start":"17:46.910 ","End":"17:48.980","Text":"which is this resistance over here,"},{"Start":"17:48.980 ","End":"17:54.070","Text":"the resistance of the system when this middle layer is gone."},{"Start":"17:54.070 ","End":"17:56.790","Text":"T_in is our unknown,"},{"Start":"17:56.790 ","End":"17:58.635","Text":"so that\u0027s what we\u0027re trying to find,"},{"Start":"17:58.635 ","End":"18:00.580","Text":"minus our T_out,"},{"Start":"18:00.580 ","End":"18:03.260","Text":"which is 15 degrees Celsius,"},{"Start":"18:03.260 ","End":"18:07.485","Text":"is equal to our Q̇ from Question Number 2."},{"Start":"18:07.485 ","End":"18:13.280","Text":"Which is 14,260 multiplied by"},{"Start":"18:13.280 ","End":"18:16.370","Text":"the resistance that we found in Question Number"},{"Start":"18:16.370 ","End":"18:19.784","Text":"3 because we\u0027re dealing with this lack of insulation."},{"Start":"18:19.784 ","End":"18:26.190","Text":"Which is equal to 0.001 plus 0, so 0.001."},{"Start":"18:27.030 ","End":"18:30.455","Text":"Now, if we scroll down a little bit,"},{"Start":"18:30.455 ","End":"18:32.930","Text":"so we can say that the temperature inside"},{"Start":"18:32.930 ","End":"18:36.820","Text":"the oven given the energy put in as same Question 2,"},{"Start":"18:36.820 ","End":"18:40.490","Text":"and this lack of insulation is going to be equal to"},{"Start":"18:40.490 ","End":"18:48.440","Text":"14.26 plus 15 degrees Celsius,"},{"Start":"18:48.440 ","End":"18:57.220","Text":"which is simply equal to 29.26 degrees Celsius inside the oven."},{"Start":"18:57.220 ","End":"19:02.300","Text":"When we don\u0027t have this insulating layer and we still put in the amount of energy that we"},{"Start":"19:02.300 ","End":"19:07.985","Text":"needed to maintain the temperature at 200 degrees Celsius with the insulating layer."},{"Start":"19:07.985 ","End":"19:14.270","Text":"So the temperature inside will just be less than 30 degrees Celsius."},{"Start":"19:14.270 ","End":"19:20.235","Text":"We can see that this insulating layer of air is super important."},{"Start":"19:20.235 ","End":"19:24.500","Text":"That is the end of this lesson."}],"ID":12353},{"Watched":false,"Name":"Ideal Gases","Duration":"10m 10s","ChapterTopicVideoID":11925,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:04.905","Text":"we\u0027re going to be speaking about ideal gases."},{"Start":"00:04.905 ","End":"00:07.695","Text":"In reality, no gas is ideal,"},{"Start":"00:07.695 ","End":"00:11.490","Text":"and there\u0027s lots of different very small interactions between"},{"Start":"00:11.490 ","End":"00:15.840","Text":"the gas molecules which will make calculations very complicated."},{"Start":"00:15.840 ","End":"00:19.875","Text":"The scientists developed the idea of an ideal gas,"},{"Start":"00:19.875 ","End":"00:27.685","Text":"which looks at the molecules in a gas as very small hard spheres which,"},{"Start":"00:27.685 ","End":"00:30.035","Text":"aside from colliding with one another,"},{"Start":"00:30.035 ","End":"00:34.530","Text":"don\u0027t have any other interactions between one another."},{"Start":"00:34.850 ","End":"00:41.584","Text":"So that means that the collisions between the molecules are perfectly elastic,"},{"Start":"00:41.584 ","End":"00:45.320","Text":"and there\u0027s no attractive forces between the molecules."},{"Start":"00:45.320 ","End":"00:50.210","Text":"Now, this makes calculations for different types of gases,"},{"Start":"00:50.210 ","End":"00:55.600","Text":"especially when dealing with thermodynamics, much easier."},{"Start":"00:55.600 ","End":"01:01.730","Text":"In a gas, all the internal energy is in the form of kinetic energy,"},{"Start":"01:01.730 ","End":"01:05.120","Text":"and a change in this internal energy,"},{"Start":"01:05.120 ","End":"01:06.365","Text":"or kinetic energy,"},{"Start":"01:06.365 ","End":"01:09.780","Text":"results in a change in temperature."},{"Start":"01:10.880 ","End":"01:18.680","Text":"Ideal gases are useful in thermodynamics for multiple reasons."},{"Start":"01:18.680 ","End":"01:22.805","Text":"One is that when we burn some things, such as coal,"},{"Start":"01:22.805 ","End":"01:30.335","Text":"so we get some kind of gas from that which is what we use in order to get energy from it."},{"Start":"01:30.335 ","End":"01:35.525","Text":"So we use the idea of ideal gases and the calculations relating to this"},{"Start":"01:35.525 ","End":"01:40.975","Text":"in order to see how much work is done or how much energy we can get from burning coal."},{"Start":"01:40.975 ","End":"01:47.690","Text":"Also, we use in air conditioners the idea of taking this energy from"},{"Start":"01:47.690 ","End":"01:55.625","Text":"the gases and converting it into heat or into an increase or decrease in temperature."},{"Start":"01:55.625 ","End":"01:57.320","Text":"Ideal gases are very,"},{"Start":"01:57.320 ","End":"02:00.326","Text":"very useful in industry,"},{"Start":"02:00.326 ","End":"02:01.549","Text":"in steam engines,"},{"Start":"02:01.549 ","End":"02:02.840","Text":"and so on and so forth."},{"Start":"02:02.840 ","End":"02:06.860","Text":"Now let\u0027s take a look at the relevant equations."},{"Start":"02:06.860 ","End":"02:10.340","Text":"The first equation that we need to know is this equation"},{"Start":"02:10.340 ","End":"02:13.475","Text":"over here which is for internal energy."},{"Start":"02:13.475 ","End":"02:17.670","Text":"U is equal to the internal energy,"},{"Start":"02:19.060 ","End":"02:24.080","Text":"so T is the temperature of the gas."},{"Start":"02:24.080 ","End":"02:27.650","Text":"R is the gas constant,"},{"Start":"02:27.650 ","End":"02:31.782","Text":"and also sometimes called the universal gas constant."},{"Start":"02:31.782 ","End":"02:39.005","Text":"The universal gas constant is equal to 8.31 joules per kelvin mole."},{"Start":"02:39.005 ","End":"02:42.515","Text":"Notice that your temperature has to also be given"},{"Start":"02:42.515 ","End":"02:46.500","Text":"in kelvin in order for this to work out,"},{"Start":"02:46.500 ","End":"02:55.440","Text":"and then, this small or lowercase n is how many moles of gas is in the sample."},{"Start":"02:55.820 ","End":"02:58.140","Text":"Now what is this Beta?"},{"Start":"02:58.140 ","End":"03:02.165","Text":"This Beta is simply a constant which is related"},{"Start":"03:02.165 ","End":"03:06.725","Text":"to if the sample of gas is monatomic or diatomic."},{"Start":"03:06.725 ","End":"03:11.030","Text":"For instance, if the gas is made out of nitrogen,"},{"Start":"03:11.030 ","End":"03:14.501","Text":"so we know that nitrogen is in pairs."},{"Start":"03:14.501 ","End":"03:17.105","Text":"We always have the gas nitrogen N_2,"},{"Start":"03:17.105 ","End":"03:20.135","Text":"so that\u0027s a diatomic gas."},{"Start":"03:20.135 ","End":"03:25.420","Text":"But there are other gases that can be monatomic,"},{"Start":"03:25.420 ","End":"03:28.225","Text":"such as for instance, neon,"},{"Start":"03:28.225 ","End":"03:30.860","Text":"so that\u0027s denoted by Ne, neon,"},{"Start":"03:30.860 ","End":"03:37.600","Text":"and that those atoms of this gas of neon just are alone."},{"Start":"03:37.600 ","End":"03:42.075","Text":"This Beta refers to the degrees of freedom,"},{"Start":"03:42.075 ","End":"03:45.660","Text":"so we don\u0027t really need to know anything,"},{"Start":"03:45.660 ","End":"03:48.416","Text":"what is important to know is this,"},{"Start":"03:48.416 ","End":"03:53.330","Text":"that if we\u0027re dealing with a monatomic gas,"},{"Start":"03:53.330 ","End":"03:56.555","Text":"so mono, such as neon,"},{"Start":"03:56.555 ","End":"04:01.020","Text":"Beta is equal to 3 divided by 2,"},{"Start":"04:01.130 ","End":"04:12.250","Text":"and that if we\u0027re dealing with a diatomic gas like nitrogen,"},{"Start":"04:12.250 ","End":"04:18.370","Text":"then Beta, you\u0027ll substitute for Beta in this equation, 5/2."},{"Start":"04:19.610 ","End":"04:22.430","Text":"This is the equation for internal energy,"},{"Start":"04:22.430 ","End":"04:23.870","Text":"and as we can see,"},{"Start":"04:23.870 ","End":"04:26.240","Text":"we\u0027re using in this specific equation,"},{"Start":"04:26.240 ","End":"04:27.800","Text":"when we have a lowercase n,"},{"Start":"04:27.800 ","End":"04:29.525","Text":"we\u0027re speaking about moles,"},{"Start":"04:29.525 ","End":"04:31.715","Text":"and when dealing with the temperature,"},{"Start":"04:31.715 ","End":"04:33.890","Text":"we\u0027re using degrees Kelvin."},{"Start":"04:33.890 ","End":"04:38.805","Text":"How do we convert between degrees Celsius to Kelvin?"},{"Start":"04:38.805 ","End":"04:44.955","Text":"If we have degrees Celsius, some number,"},{"Start":"04:44.955 ","End":"04:52.117","Text":"then what we\u0027re going to do is we\u0027re going to add 273 to that number,"},{"Start":"04:52.117 ","End":"04:57.910","Text":"and then that\u0027s going to give us the amount of degrees in Kelvin."},{"Start":"04:58.070 ","End":"05:05.615","Text":"Let\u0027s say we\u0027re being told that the temperature of the room is 27 degrees Celsius,"},{"Start":"05:05.615 ","End":"05:09.215","Text":"and now what we want to do is convert this into Kelvin."},{"Start":"05:09.215 ","End":"05:13.085","Text":"So what we\u0027re going to do in order to get into Kelvin,"},{"Start":"05:13.085 ","End":"05:16.535","Text":"we\u0027re going to add on 273,"},{"Start":"05:16.535 ","End":"05:23.150","Text":"and then we\u0027ll get that room temperature in degrees Kelvin is equal to 300 Kelvin."},{"Start":"05:23.150 ","End":"05:28.486","Text":"So 27 degrees Celsius is the same as 300 degrees Kelvin."},{"Start":"05:28.486 ","End":"05:29.960","Text":"We just do a shift."},{"Start":"05:29.960 ","End":"05:35.760","Text":"It\u0027s important to remember this as well."},{"Start":"05:36.410 ","End":"05:40.200","Text":"This was our first equation that is important,"},{"Start":"05:40.200 ","End":"05:44.695","Text":"and remember to also write this down so that you know what Beta is equal to,"},{"Start":"05:44.695 ","End":"05:46.210","Text":"depending on in the question,"},{"Start":"05:46.210 ","End":"05:48.835","Text":"if we\u0027re being told that the gas is monatomic or diatomic,"},{"Start":"05:48.835 ","End":"05:53.470","Text":"and remember this little conversion between degrees Celsius to Kelvin."},{"Start":"05:53.470 ","End":"05:58.850","Text":"Now the next thing that we\u0027re going to learn is the ideal gas law."},{"Start":"06:00.350 ","End":"06:06.130","Text":"An ideal gas can be characterized by 3 state variables."},{"Start":"06:06.130 ","End":"06:15.220","Text":"The 3 state variables are pressure, temperature, and volume."},{"Start":"06:16.430 ","End":"06:19.680","Text":"What is a state variable?"},{"Start":"06:19.680 ","End":"06:25.340","Text":"A state variable is a precisely measurable physical property,"},{"Start":"06:25.340 ","End":"06:27.625","Text":"such as pressure, volume or temperature,"},{"Start":"06:27.625 ","End":"06:29.755","Text":"we can measure that very easily,"},{"Start":"06:29.755 ","End":"06:32.865","Text":"that characterize a system,"},{"Start":"06:32.865 ","End":"06:40.770","Text":"and these state variables are independent of how the system was brought to that state."},{"Start":"06:41.510 ","End":"06:44.420","Text":"Let\u0027s look at this equation."},{"Start":"06:44.420 ","End":"06:53.255","Text":"In this equation, we have that pressure multiplied by volume is equal to n,"},{"Start":"06:53.255 ","End":"07:02.600","Text":"which we already saw is the amount of moles in this system, multiplied by R,"},{"Start":"07:02.600 ","End":"07:04.640","Text":"which is the gas constant,"},{"Start":"07:04.640 ","End":"07:09.455","Text":"so I\u0027m just going to write GC for gas constant which we saw over here,"},{"Start":"07:09.455 ","End":"07:13.665","Text":"multiplied by T which, as we saw,"},{"Start":"07:13.665 ","End":"07:17.210","Text":"is equal to temp, so the temperature,"},{"Start":"07:17.210 ","End":"07:22.530","Text":"and remember that this is also measured in Kelvin."},{"Start":"07:22.550 ","End":"07:27.950","Text":"This is exactly like this side of the equation."},{"Start":"07:27.950 ","End":"07:32.750","Text":"So what is this uppercase or capital N over here?"},{"Start":"07:32.750 ","End":"07:37.040","Text":"This is the number of molecules that are in the sample."},{"Start":"07:37.040 ","End":"07:41.657","Text":"Here we were taking into account the number of moles in the sample,"},{"Start":"07:41.657 ","End":"07:48.065","Text":"and here we\u0027re converting using Avogadro\u0027s constant the number of molecules,"},{"Start":"07:48.065 ","End":"07:51.215","Text":"so Capital N is the number of molecules in the sample."},{"Start":"07:51.215 ","End":"07:54.500","Text":"What about this lowercase k?"},{"Start":"07:54.500 ","End":"07:57.815","Text":"Here we saw that R was the gas constant,"},{"Start":"07:57.815 ","End":"08:01.805","Text":"but the gas constant is only when we\u0027re using moles."},{"Start":"08:01.805 ","End":"08:04.270","Text":"But here we\u0027re using the number of molecules,"},{"Start":"08:04.270 ","End":"08:05.960","Text":"so instead of the gas constant,"},{"Start":"08:05.960 ","End":"08:10.140","Text":"we\u0027re going to be using the Boltzmann constant."},{"Start":"08:10.370 ","End":"08:13.775","Text":"The Boltzmann constant, which is k,"},{"Start":"08:13.775 ","End":"08:19.040","Text":"is equal to 1.38066 times 10"},{"Start":"08:19.040 ","End":"08:26.150","Text":"to the negative 23 joules per Kelvin."},{"Start":"08:26.150 ","End":"08:28.040","Text":"So this is just a constant,"},{"Start":"08:28.040 ","End":"08:30.365","Text":"and then T is, of course,"},{"Start":"08:30.365 ","End":"08:36.929","Text":"the temperature of the gas in Kelvin."},{"Start":"08:37.790 ","End":"08:43.685","Text":"The ideal gas law is useful for us in order to calculate"},{"Start":"08:43.685 ","End":"08:51.080","Text":"the change in one of the physical or state values of the gas."},{"Start":"08:51.080 ","End":"08:53.510","Text":"If we have the gas at a certain pressure,"},{"Start":"08:53.510 ","End":"08:56.360","Text":"volume, and initial temperature,"},{"Start":"08:56.360 ","End":"08:58.265","Text":"if we change the temperature,"},{"Start":"08:58.265 ","End":"09:02.840","Text":"then either the pressure or the volume will also change."},{"Start":"09:02.840 ","End":"09:07.725","Text":"So depending on how we constrain our ideal gas,"},{"Start":"09:07.725 ","End":"09:13.498","Text":"if we have some jar with one liter of gas inside,"},{"Start":"09:13.498 ","End":"09:15.304","Text":"so if we increase the temperature,"},{"Start":"09:15.304 ","End":"09:17.930","Text":"we know that the volume is going to remain the"},{"Start":"09:17.930 ","End":"09:22.175","Text":"same because it\u0027s constrained to be within the jar,"},{"Start":"09:22.175 ","End":"09:25.190","Text":"so that means that if the temperature increases,"},{"Start":"09:25.190 ","End":"09:32.070","Text":"we know that the pressure is also going to increase."},{"Start":"09:32.390 ","End":"09:34.835","Text":"So that is what we can see."},{"Start":"09:34.835 ","End":"09:37.773","Text":"Both sides of the equation have to be equal."},{"Start":"09:37.773 ","End":"09:42.455","Text":"So if our volume is constant and our temperature goes up,"},{"Start":"09:42.455 ","End":"09:45.620","Text":"in order for this side to also go up,"},{"Start":"09:45.620 ","End":"09:48.455","Text":"we need our pressure to increase as well."},{"Start":"09:48.455 ","End":"09:53.123","Text":"This is why this equation is very useful."},{"Start":"09:53.123 ","End":"09:55.670","Text":"The equation for internal energy,"},{"Start":"09:55.670 ","End":"09:59.990","Text":"we can see that Beta n and R are all constants,"},{"Start":"09:59.990 ","End":"10:06.310","Text":"so we can see that the internal energy is only dependent on the temperature."},{"Start":"10:06.310 ","End":"10:10.570","Text":"That is the end of our lesson."}],"ID":12354},{"Watched":false,"Name":"Thermodynamic Processes","Duration":"14m 57s","ChapterTopicVideoID":11926,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.310","Text":"Hello. In this lesson,"},{"Start":"00:02.310 ","End":"00:06.510","Text":"we\u0027re going to be speaking about different thermodynamic processes."},{"Start":"00:06.510 ","End":"00:11.370","Text":"As we know, thermodynamics is the study of how heat moves."},{"Start":"00:11.370 ","End":"00:14.310","Text":"We\u0027ve seen that heat always spreads out,"},{"Start":"00:14.310 ","End":"00:20.650","Text":"so we always have heat moving from a hot place to a colder place."},{"Start":"00:21.590 ","End":"00:27.690","Text":"What we want to do is we want to know how heat moves in a system or between"},{"Start":"00:27.690 ","End":"00:34.920","Text":"systems where a system is simply a area or an object that we\u0027re looking at."},{"Start":"00:34.920 ","End":"00:44.590","Text":"System could be your fridge or your body or your computer\u0027s cooling system."},{"Start":"00:44.590 ","End":"00:47.270","Text":"What we\u0027re going to do is we\u0027re going to learn about"},{"Start":"00:47.270 ","End":"00:51.995","Text":"the thermodynamic processes which happen in these systems,"},{"Start":"00:51.995 ","End":"00:57.120","Text":"or we\u0027re going to speak about how heat moves in the different types."},{"Start":"00:57.120 ","End":"00:59.030","Text":"In this lesson, we\u0027re going to learn about"},{"Start":"00:59.030 ","End":"01:03.930","Text":"the 4 different types of thermodynamic processes."},{"Start":"01:04.940 ","End":"01:12.710","Text":"The first thermodynamic process that we\u0027re going to talk about is the isothermal process."},{"Start":"01:12.710 ","End":"01:14.990","Text":"What does isothermal mean?"},{"Start":"01:14.990 ","End":"01:20.845","Text":"It means that the temperature of the system remains constant."},{"Start":"01:20.845 ","End":"01:23.770","Text":"Here we have our equation of state."},{"Start":"01:23.770 ","End":"01:28.610","Text":"We have pressure multiplied by volume is equal to nR,"},{"Start":"01:28.610 ","End":"01:33.170","Text":"which is constant which relates to the amount"},{"Start":"01:33.170 ","End":"01:39.720","Text":"of ideal gas in our system multiplied by temperature."},{"Start":"01:40.190 ","End":"01:47.240","Text":"As long as no gas is added to our system or no gas escapes from our system,"},{"Start":"01:47.240 ","End":"01:50.320","Text":"this is meant to remain a constant."},{"Start":"01:50.320 ","End":"01:53.220","Text":"It\u0027s just the amount of gas we have."},{"Start":"01:53.220 ","End":"01:57.770","Text":"What we can see is that we have PV is equal to"},{"Start":"01:57.770 ","End":"02:03.150","Text":"some constant multiplied by T, the temperature."},{"Start":"02:03.230 ","End":"02:06.768","Text":"In an isothermal process,"},{"Start":"02:06.768 ","End":"02:09.235","Text":"our temperature remains constant,"},{"Start":"02:09.235 ","End":"02:13.830","Text":"which means that our pressure and our volume changes."},{"Start":"02:13.830 ","End":"02:15.145","Text":"Let\u0027s take a look at that."},{"Start":"02:15.145 ","End":"02:18.085","Text":"Let\u0027s say we begin with some pressure,"},{"Start":"02:18.085 ","End":"02:22.325","Text":"P_1, some volume, V_1."},{"Start":"02:22.325 ","End":"02:24.840","Text":"That is equal to a constant,"},{"Start":"02:24.840 ","End":"02:26.280","Text":"so I\u0027m just not going to write that out,"},{"Start":"02:26.280 ","End":"02:29.930","Text":"and that\u0027s equal to this temperature T_1."},{"Start":"02:29.930 ","End":"02:35.460","Text":"Then let\u0027s say we change our pressure and also our volume."},{"Start":"02:35.460 ","End":"02:39.780","Text":"Now we have P_2 multiplied by this different volume"},{"Start":"02:39.780 ","End":"02:44.160","Text":"V_2 is equal to the same temperature T_1,"},{"Start":"02:44.160 ","End":"02:47.510","Text":"because our temperature has remained the same."},{"Start":"02:47.510 ","End":"02:52.160","Text":"Then what we can do from these 2 equations is we can say that"},{"Start":"02:52.160 ","End":"02:58.590","Text":"P_1V_1 is equal to P_2V_2."},{"Start":"02:58.590 ","End":"03:02.030","Text":"Because we\u0027re dealing with an isothermal process,"},{"Start":"03:02.030 ","End":"03:07.765","Text":"iso means the same and thermal is related to the heat,"},{"Start":"03:07.765 ","End":"03:12.000","Text":"which is related to the temperature, so constant temperature."},{"Start":"03:12.000 ","End":"03:18.030","Text":"Then we can divide and what we can get is that from here we\u0027ll get"},{"Start":"03:18.030 ","End":"03:27.075","Text":"that P_1 divided by P_2 is equal to V_2 divided by V_1,"},{"Start":"03:27.075 ","End":"03:35.730","Text":"and this is the equation that must happen if we\u0027re dealing with an isothermal process."},{"Start":"03:36.440 ","End":"03:43.225","Text":"Now, I\u0027m going to rub this out and I\u0027m going to write down the important points."},{"Start":"03:43.225 ","End":"03:47.915","Text":"If you\u0027re told that you\u0027re dealing with an isothermal process,"},{"Start":"03:47.915 ","End":"03:52.850","Text":"then you have to know that the temperature is some constant value,"},{"Start":"03:52.850 ","End":"03:56.105","Text":"which means that you end up with this equation."},{"Start":"03:56.105 ","End":"03:59.630","Text":"Then if you need to calculate in the question what"},{"Start":"03:59.630 ","End":"04:03.995","Text":"the initial volume is or the final pressure,"},{"Start":"04:03.995 ","End":"04:06.904","Text":"or any one of these as unknowns,"},{"Start":"04:06.904 ","End":"04:12.840","Text":"then you can just write out this equation and work it out."},{"Start":"04:13.750 ","End":"04:19.785","Text":"The second thermodynamic process is the isochoric process."},{"Start":"04:19.785 ","End":"04:24.905","Text":"What does that mean? That means that we have a constant volume."},{"Start":"04:24.905 ","End":"04:27.935","Text":"V over here is a constant."},{"Start":"04:27.935 ","End":"04:36.120","Text":"Let\u0027s say that we have our initial pressure P_1 and our volume V_1,"},{"Start":"04:36.120 ","End":"04:38.570","Text":"and this is equal to this constant over here,"},{"Start":"04:38.570 ","End":"04:42.700","Text":"so we won\u0027t write it out, multiplied by T_1."},{"Start":"04:42.700 ","End":"04:48.050","Text":"Then let\u0027s say that we\u0027re now going to change the pressure in the system."},{"Start":"04:48.050 ","End":"04:53.505","Text":"We\u0027ll have some P_2 multiplied by this V_2."},{"Start":"04:53.505 ","End":"04:56.090","Text":"We know that our volume is constant,"},{"Start":"04:56.090 ","End":"04:58.220","Text":"so this is just V_1,"},{"Start":"04:58.220 ","End":"04:59.780","Text":"it\u0027s the same volume."},{"Start":"04:59.780 ","End":"05:05.989","Text":"Then this change in pressure results in a change in temperature."},{"Start":"05:05.989 ","End":"05:11.570","Text":"Then we can rearrange this equation by algebra so we can isolate out"},{"Start":"05:11.570 ","End":"05:19.735","Text":"this V_1 so that will be V_1 is equal to T_1 divided by P_1,"},{"Start":"05:19.735 ","End":"05:28.090","Text":"and here, V_1 is equal to T_2 divided by P_2."},{"Start":"05:28.460 ","End":"05:36.465","Text":"Then we can equate these equations, both of these."},{"Start":"05:36.465 ","End":"05:47.490","Text":"Then what we\u0027ll get is that T_1 divided by P_1 is equal to T_2 divided by P_2."},{"Start":"05:47.490 ","End":"05:53.660","Text":"Then we can rearrange this further to get our T\u0027s on the same side so we can get"},{"Start":"05:53.660 ","End":"06:02.250","Text":"the T_1 divided by T_2 is equal to P_1 divided by P_2."},{"Start":"06:02.250 ","End":"06:07.160","Text":"This is an equation for an isochoric process,"},{"Start":"06:07.160 ","End":"06:11.135","Text":"so that means that the volume is constant."},{"Start":"06:11.135 ","End":"06:15.930","Text":"Let\u0027s just give a quick example for an isothermal process."},{"Start":"06:15.930 ","End":"06:21.750","Text":"That can be when we\u0027re using latent heat to melt ice, for instance."},{"Start":"06:21.750 ","End":"06:24.935","Text":"We know that the temperature of ice,"},{"Start":"06:24.935 ","End":"06:26.765","Text":"when it\u0027s solid, solid ice,"},{"Start":"06:26.765 ","End":"06:31.685","Text":"is 0 degrees Celsius and we also know that in the process of the latent heat,"},{"Start":"06:31.685 ","End":"06:37.755","Text":"the water or the liquid ice is also at 0 degrees Celsius."},{"Start":"06:37.755 ","End":"06:41.480","Text":"We can see that the temperature is constant,"},{"Start":"06:41.480 ","End":"06:44.390","Text":"we\u0027re at 0 degrees Celsius in isothermal."},{"Start":"06:44.390 ","End":"06:49.205","Text":"However, we know that the volume of ice or water,"},{"Start":"06:49.205 ","End":"06:52.760","Text":"when it\u0027s in a liquid or a solid form is different so we can see that"},{"Start":"06:52.760 ","End":"06:57.170","Text":"our volumes have changed in the process, therefore,"},{"Start":"06:57.170 ","End":"07:02.360","Text":"in order to keep this state equation or equation of state,"},{"Start":"07:02.360 ","End":"07:06.965","Text":"our pressures are also going to change in that process."},{"Start":"07:06.965 ","End":"07:13.650","Text":"Now, let\u0027s speak about an isochoric process so that\u0027s where the volume is constant."},{"Start":"07:13.650 ","End":"07:17.510","Text":"You can imagine that we have some closed pot which"},{"Start":"07:17.510 ","End":"07:21.500","Text":"is filled with a certain amount of water, for instance,"},{"Start":"07:21.500 ","End":"07:31.675","Text":"and we raised the temperature of our pot with the water."},{"Start":"07:31.675 ","End":"07:35.720","Text":"There we can see that our temperature is changing,"},{"Start":"07:35.720 ","End":"07:39.215","Text":"and because the volume is remaining constant,"},{"Start":"07:39.215 ","End":"07:42.985","Text":"all the water is still contained in the pot."},{"Start":"07:42.985 ","End":"07:44.850","Text":"We\u0027re still not at boiling points,"},{"Start":"07:44.850 ","End":"07:47.310","Text":"so we don\u0027t have water vapor,"},{"Start":"07:47.310 ","End":"07:48.675","Text":"it\u0027s just liquid water."},{"Start":"07:48.675 ","End":"07:52.040","Text":"We can see that the pressure is going to have to"},{"Start":"07:52.040 ","End":"07:58.067","Text":"change in order to make our equation of state work."},{"Start":"07:58.067 ","End":"08:01.340","Text":"This is the equation to use if you\u0027re"},{"Start":"08:01.340 ","End":"08:06.140","Text":"dealing with an isochoric or a constant volume process."},{"Start":"08:06.140 ","End":"08:11.465","Text":"The third thermodynamic process is the isobaric process."},{"Start":"08:11.465 ","End":"08:15.635","Text":"What does that mean? That means constant pressure."},{"Start":"08:15.635 ","End":"08:18.620","Text":"Let\u0027s give an example."},{"Start":"08:18.620 ","End":"08:28.160","Text":"Imagine that we have a gas inside a movable piston and we\u0027re heating up the gas,"},{"Start":"08:28.160 ","End":"08:30.755","Text":"so we\u0027re giving the gas more energy."},{"Start":"08:30.755 ","End":"08:33.620","Text":"We know that when the gas molecules have more energy,"},{"Start":"08:33.620 ","End":"08:34.700","Text":"they move around more,"},{"Start":"08:34.700 ","End":"08:37.025","Text":"therefore, increasing the pressure."},{"Start":"08:37.025 ","End":"08:40.100","Text":"But if we\u0027re dealing with a movable piston,"},{"Start":"08:40.100 ","End":"08:44.330","Text":"then that means that the piston is expanding,"},{"Start":"08:44.330 ","End":"08:49.220","Text":"increasing the volume, and therefore giving the molecules more room to move,"},{"Start":"08:49.220 ","End":"08:54.755","Text":"therefore, maintaining a constant pressure."},{"Start":"08:54.755 ","End":"08:57.275","Text":"If we\u0027re looking at that equation,"},{"Start":"08:57.275 ","End":"09:03.245","Text":"that means that our original pressure is P_1 our original volume is V_1,"},{"Start":"09:03.245 ","End":"09:07.910","Text":"which is equal to this constant multiplied by this temperature T_1."},{"Start":"09:07.910 ","End":"09:10.790","Text":"Then let\u0027s say that we increase"},{"Start":"09:10.790 ","End":"09:16.370","Text":"the temperature which means that we can also maybe increase the pressure."},{"Start":"09:16.370 ","End":"09:20.975","Text":"However, our volume is also increasing or changing."},{"Start":"09:20.975 ","End":"09:24.665","Text":"That means that our pressure can remain the same."},{"Start":"09:24.665 ","End":"09:26.435","Text":"We have the same pressure."},{"Start":"09:26.435 ","End":"09:30.110","Text":"Now, we can isolate out the P_1 from both of these equations."},{"Start":"09:30.110 ","End":"09:35.450","Text":"We\u0027ll get that P_1 is equal to T_1 divided by V_1."},{"Start":"09:35.450 ","End":"09:41.045","Text":"Here, P_1 is equal to T_2 divided by V_2."},{"Start":"09:41.045 ","End":"09:45.739","Text":"Then we can equate these 2 equations and get therefore that T_1"},{"Start":"09:45.739 ","End":"09:51.990","Text":"divided by V_1 is equal to T_2 divided by V_2."},{"Start":"09:52.330 ","End":"09:55.730","Text":"Now we can rearrange this equation."},{"Start":"09:55.730 ","End":"10:02.727","Text":"If we want to get the ratios of the temperatures relative to the ratios of the volumes,"},{"Start":"10:02.727 ","End":"10:06.110","Text":"we can just have that T_1 divided by T_2."},{"Start":"10:06.110 ","End":"10:08.000","Text":"I\u0027ve divided both sides by T_2."},{"Start":"10:08.000 ","End":"10:10.700","Text":"Then I\u0027m multiplying both sides by V_1,"},{"Start":"10:10.700 ","End":"10:14.285","Text":"so it\u0027s equal to V_1 divided by V_2."},{"Start":"10:14.285 ","End":"10:19.070","Text":"This is the equation that you need to be using in an isobaric process,"},{"Start":"10:19.070 ","End":"10:23.400","Text":"which means a process that maintains a constant pressure."},{"Start":"10:24.280 ","End":"10:28.685","Text":"These are the 3 thermodynamic processes."},{"Start":"10:28.685 ","End":"10:31.800","Text":"Now let\u0027s speak about the last one."},{"Start":"10:32.200 ","End":"10:36.965","Text":"The fourth and final process is called the adiabatic process."},{"Start":"10:36.965 ","End":"10:40.820","Text":"It states that no heat is added."},{"Start":"10:40.820 ","End":"10:44.882","Text":"What is important to note is that nothing is constant."},{"Start":"10:44.882 ","End":"10:47.600","Text":"In isothermal we had constant temperature,"},{"Start":"10:47.600 ","End":"10:50.855","Text":"then constant volume, and then constant pressure."},{"Start":"10:50.855 ","End":"10:57.425","Text":"Here, we\u0027re not talking about any one of these PV or T values as being constant;"},{"Start":"10:57.425 ","End":"11:01.460","Text":"all we\u0027re saying is that no heat is added."},{"Start":"11:01.460 ","End":"11:06.770","Text":"If we remember, heat is given the symbol Q."},{"Start":"11:06.770 ","End":"11:08.090","Text":"No heat is added,"},{"Start":"11:08.090 ","End":"11:11.510","Text":"so Q is equal to 0."},{"Start":"11:11.510 ","End":"11:16.190","Text":"Now what\u0027s important to note is that just because no heat is"},{"Start":"11:16.190 ","End":"11:20.150","Text":"added doesn\u0027t mean that the temperature cannot change."},{"Start":"11:20.150 ","End":"11:24.785","Text":"Because the temperature is given from the internal energy."},{"Start":"11:24.785 ","End":"11:29.090","Text":"Internal energy and temperature is connected."},{"Start":"11:29.090 ","End":"11:32.825","Text":"No heat is added but the temperature can change."},{"Start":"11:32.825 ","End":"11:41.240","Text":"That means that the temperature change isn\u0027t given by the surroundings of the system."},{"Start":"11:41.240 ","End":"11:48.575","Text":"We can say that in an adiabatic system is completely insulated,"},{"Start":"11:48.575 ","End":"11:51.845","Text":"no heat can be added or taken away from the system."},{"Start":"11:51.845 ","End":"11:53.840","Text":"The system is insulated."},{"Start":"11:53.840 ","End":"11:58.880","Text":"However, due to work done on the system"},{"Start":"11:58.880 ","End":"12:03.789","Text":"such as by increasing or decreasing the pressure or the volume,"},{"Start":"12:03.789 ","End":"12:06.093","Text":"so we\u0027re doing work on the system,"},{"Start":"12:06.093 ","End":"12:09.725","Text":"the temperature can increase."},{"Start":"12:09.725 ","End":"12:12.905","Text":"This 0 heat change,"},{"Start":"12:12.905 ","End":"12:16.670","Text":"so Q is equal to 0 is very important to remember,"},{"Start":"12:16.670 ","End":"12:19.640","Text":"and the equation which I\u0027m not going to derive right now,"},{"Start":"12:19.640 ","End":"12:24.425","Text":"we\u0027ll deal with this adiabatic process more in detail later on,"},{"Start":"12:24.425 ","End":"12:27.920","Text":"but the equation that you need to use as soon as you"},{"Start":"12:27.920 ","End":"12:31.760","Text":"see that it\u0027s an adiabatic process or that Q is equal to 0"},{"Start":"12:31.760 ","End":"12:38.195","Text":"is pressure multiplied by"},{"Start":"12:38.195 ","End":"12:45.215","Text":"volume to the power of Gamma is equal to some constant."},{"Start":"12:45.215 ","End":"12:47.014","Text":"This is the equation,"},{"Start":"12:47.014 ","End":"12:53.160","Text":"and this Gamma is the ratio of the specific heats."},{"Start":"12:53.650 ","End":"12:58.220","Text":"This is what\u0027s important to remember with the adiabatic process."},{"Start":"12:58.220 ","End":"13:03.080","Text":"Now what I want to do is I want to look at the graphs of these 3,"},{"Start":"13:03.080 ","End":"13:05.000","Text":"the first 3 processes."},{"Start":"13:05.000 ","End":"13:07.565","Text":"Let\u0027s look at Graph Number 1."},{"Start":"13:07.565 ","End":"13:09.605","Text":"Let\u0027s write 1 over here."},{"Start":"13:09.605 ","End":"13:12.290","Text":"This is the isothermal process."},{"Start":"13:12.290 ","End":"13:17.600","Text":"Over here, we\u0027re going to do a graph of our pressure on"},{"Start":"13:17.600 ","End":"13:25.190","Text":"the y-axes versus our volume on the x-axes."},{"Start":"13:25.190 ","End":"13:27.770","Text":"In an isothermal process,"},{"Start":"13:27.770 ","End":"13:31.670","Text":"the graph is going to look something like so."},{"Start":"13:31.670 ","End":"13:35.540","Text":"We can see that as our pressure decreases,"},{"Start":"13:35.540 ","End":"13:37.820","Text":"our volume is going to increase,"},{"Start":"13:37.820 ","End":"13:42.155","Text":"and we can see that this is a non-linear process."},{"Start":"13:42.155 ","End":"13:47.045","Text":"We can really see that from this equation over here."},{"Start":"13:47.045 ","End":"13:51.620","Text":"Now let\u0027s look at Graph Number 2."},{"Start":"13:51.620 ","End":"13:55.385","Text":"This is the isochoric process."},{"Start":"13:55.385 ","End":"13:58.400","Text":"Here we can see that our volume is constant."},{"Start":"13:58.400 ","End":"14:02.810","Text":"Again, we have pressure on the y and volume on the x."},{"Start":"14:02.810 ","End":"14:05.285","Text":"If our volume is constant,"},{"Start":"14:05.285 ","End":"14:07.520","Text":"our graph is going to look like this,"},{"Start":"14:07.520 ","End":"14:11.000","Text":"just a straight line. Why is this?"},{"Start":"14:11.000 ","End":"14:13.640","Text":"Our volume is at some constant value over"},{"Start":"14:13.640 ","End":"14:18.065","Text":"here and our temperature is changing or whatever,"},{"Start":"14:18.065 ","End":"14:21.110","Text":"resulting in a change in pressure."},{"Start":"14:21.110 ","End":"14:23.929","Text":"Graph Number 3, the third process,"},{"Start":"14:23.929 ","End":"14:25.985","Text":"is the isobaric process,"},{"Start":"14:25.985 ","End":"14:28.835","Text":"which means that our pressure is constant."},{"Start":"14:28.835 ","End":"14:34.280","Text":"We have our pressure on our y and our volume on our x."},{"Start":"14:34.280 ","End":"14:35.675","Text":"If we have a constant pressure,"},{"Start":"14:35.675 ","End":"14:42.140","Text":"we just have a flat line in this direction because we have a constant pressure value,"},{"Start":"14:42.140 ","End":"14:47.370","Text":"but then our temperature and our volume is changing."},{"Start":"14:48.250 ","End":"14:52.340","Text":"It\u0027s important to remember these graphs as well."},{"Start":"14:52.340 ","End":"14:54.815","Text":"Don\u0027t get confused between the processes."},{"Start":"14:54.815 ","End":"14:57.840","Text":"That is the end of this lesson."}],"ID":12355},{"Watched":false,"Name":"Work","Duration":"16m 58s","ChapterTopicVideoID":11927,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.755","Text":"Hello. In this lesson,"},{"Start":"00:01.755 ","End":"00:03.870","Text":"we\u0027re going to be dealing with work,"},{"Start":"00:03.870 ","End":"00:09.645","Text":"and we\u0027re going to see how a gas can do work."},{"Start":"00:09.645 ","End":"00:14.010","Text":"For instance, we can talk about mechanical work by raising"},{"Start":"00:14.010 ","End":"00:19.300","Text":"some mass from a certain height to a higher height."},{"Start":"00:19.300 ","End":"00:23.015","Text":"Let\u0027s imagine that we have this cylinder over here,"},{"Start":"00:23.015 ","End":"00:29.300","Text":"and we place some kind of lid on top of the gas that we\u0027ve placed in the cylinder."},{"Start":"00:29.300 ","End":"00:33.840","Text":"So let\u0027s say that the lid is over here."},{"Start":"00:33.920 ","End":"00:40.374","Text":"This lid, it contains the gas inside over here,"},{"Start":"00:40.374 ","End":"00:42.815","Text":"so here we can see the gas molecules."},{"Start":"00:42.815 ","End":"00:46.850","Text":"However, it can move up and down."},{"Start":"00:46.850 ","End":"00:49.354","Text":"We have our gas in here,"},{"Start":"00:49.354 ","End":"00:51.305","Text":"and it\u0027s at some pressure."},{"Start":"00:51.305 ","End":"00:56.559","Text":"Let\u0027s call this pressure P. It\u0027s at some volume."},{"Start":"00:56.559 ","End":"00:59.460","Text":"Let\u0027s call this volume V_1."},{"Start":"00:59.750 ","End":"01:06.470","Text":"Here we have some force pushing downwards on the gas."},{"Start":"01:06.470 ","End":"01:13.748","Text":"The movement of the gas molecules is holding this force over here."},{"Start":"01:13.748 ","End":"01:15.975","Text":"It\u0027s balancing it out."},{"Start":"01:15.975 ","End":"01:23.585","Text":"Now, let\u0027s imagine that we increase the temperature of the gas over here."},{"Start":"01:23.585 ","End":"01:26.900","Text":"So let\u0027s say we light a fire underneath it."},{"Start":"01:26.900 ","End":"01:28.775","Text":"This is a flame."},{"Start":"01:28.775 ","End":"01:31.730","Text":"Imagine we\u0027re burning wood over here."},{"Start":"01:31.730 ","End":"01:34.985","Text":"So we\u0027re increasing the temperature of the gas."},{"Start":"01:34.985 ","End":"01:37.882","Text":"Because the temperature is going to be increased,"},{"Start":"01:37.882 ","End":"01:44.033","Text":"so we started off also with T_1,"},{"Start":"01:44.033 ","End":"01:47.545","Text":"but now our temperature is increasing."},{"Start":"01:47.545 ","End":"01:49.100","Text":"So what\u0027s going to happen?"},{"Start":"01:49.100 ","End":"01:50.740","Text":"If the temperature is increasing,"},{"Start":"01:50.740 ","End":"01:55.940","Text":"we know that the temperature is connected to the internal energy of the system."},{"Start":"01:55.940 ","End":"02:02.855","Text":"The internal energy of the gas molecules is going to increase which means that"},{"Start":"02:02.855 ","End":"02:06.650","Text":"the gas molecules are going to be moving around much faster and"},{"Start":"02:06.650 ","End":"02:11.450","Text":"hitting the walls of the cylinder and the lid much more."},{"Start":"02:11.450 ","End":"02:16.525","Text":"So that means that we\u0027re going to have an increase in pressure."},{"Start":"02:16.525 ","End":"02:20.735","Text":"This increase in pressure is now going to have"},{"Start":"02:20.735 ","End":"02:27.200","Text":"a disproportionate force on this lid which is pushing down with some force, F,"},{"Start":"02:27.200 ","End":"02:34.424","Text":"which means that the lid is going to move up until we reach some equilibrium,"},{"Start":"02:34.424 ","End":"02:42.080","Text":"some balance between the force that the atmosphere is applying and the force of"},{"Start":"02:42.080 ","End":"02:45.095","Text":"the lid due to the mass of it and"},{"Start":"02:45.095 ","End":"02:51.595","Text":"the gas inside the cylinder which are pushing against this."},{"Start":"02:51.595 ","End":"02:57.825","Text":"Now we can say that the lid has moved up till over here."},{"Start":"02:57.825 ","End":"02:59.930","Text":"So we\u0027ve had a change in volume,"},{"Start":"02:59.930 ","End":"03:03.665","Text":"and our gas is going to fill out this empty space."},{"Start":"03:03.665 ","End":"03:07.595","Text":"Now what we can see is that our pressure"},{"Start":"03:07.595 ","End":"03:12.399","Text":"at the end of this process is going to be the same."},{"Start":"03:12.399 ","End":"03:14.184","Text":"Because we\u0027re at equilibrium,"},{"Start":"03:14.184 ","End":"03:15.514","Text":"the lid is resting,"},{"Start":"03:15.514 ","End":"03:18.080","Text":"so the pressure is exactly the same."},{"Start":"03:18.080 ","End":"03:25.940","Text":"However, our volume has increased from the lid being over here to over here."},{"Start":"03:25.940 ","End":"03:29.900","Text":"So we can see that the space of the cylinder being filled is larger."},{"Start":"03:29.900 ","End":"03:34.285","Text":"So we can say that now we have a different volume, V_2."},{"Start":"03:34.285 ","End":"03:39.050","Text":"We know that all of this happened because we increased the temperature."},{"Start":"03:39.050 ","End":"03:43.770","Text":"So now we\u0027re located at some different temperature, T_2."},{"Start":"03:43.910 ","End":"03:46.035","Text":"All of this happened,"},{"Start":"03:46.035 ","End":"03:50.945","Text":"and we can see that some mechanical work has occurred because"},{"Start":"03:50.945 ","End":"03:58.100","Text":"our mass that is resting on this lid has moved from over here upwards."},{"Start":"03:58.100 ","End":"04:03.180","Text":"So some work has been done."},{"Start":"04:03.470 ","End":"04:09.400","Text":"Let\u0027s take a look at how we can derive the equation for work."},{"Start":"04:09.400 ","End":"04:13.669","Text":"What we can see is that we were dealing with pressure."},{"Start":"04:13.669 ","End":"04:18.620","Text":"It\u0027s the pressure of the gas which is what caused our lid that"},{"Start":"04:18.620 ","End":"04:23.749","Text":"had some mass that was exerting some pressure back down on to the gas."},{"Start":"04:23.749 ","End":"04:27.610","Text":"So that pressure is what caused this work to be done."},{"Start":"04:27.610 ","End":"04:30.115","Text":"So what is the equation for pressure?"},{"Start":"04:30.115 ","End":"04:31.775","Text":"Pressure, as we know,"},{"Start":"04:31.775 ","End":"04:38.200","Text":"is equal to the force being applied divided by the cross-sectional area."},{"Start":"04:38.200 ","End":"04:42.241","Text":"So this is the area,"},{"Start":"04:42.241 ","End":"04:43.940","Text":"the area over here,"},{"Start":"04:43.940 ","End":"04:46.675","Text":"the cross-sectional area of our cylinder."},{"Start":"04:46.675 ","End":"04:51.095","Text":"As we know, if someone steps on your foot wearing trainers,"},{"Start":"04:51.095 ","End":"04:54.095","Text":"or sneakers, or some kind of sports shoe,"},{"Start":"04:54.095 ","End":"04:58.130","Text":"it will hurt a lot less than,"},{"Start":"04:58.130 ","End":"05:01.395","Text":"in the case of the same person with the same weight,"},{"Start":"05:01.395 ","End":"05:05.090","Text":"if that person steps on you wearing stiletto heels."},{"Start":"05:05.090 ","End":"05:06.560","Text":"Why is that?"},{"Start":"05:06.560 ","End":"05:09.275","Text":"Because of the difference in surface area between"},{"Start":"05:09.275 ","End":"05:13.435","Text":"a stiletto heel or the heel of your sports shoe."},{"Start":"05:13.435 ","End":"05:16.655","Text":"The stiletto heel has a much smaller surface area,"},{"Start":"05:16.655 ","End":"05:20.450","Text":"which means that the pressure exerted on your toe is far bigger,"},{"Start":"05:20.450 ","End":"05:24.060","Text":"and then it will hurt you a lot more."},{"Start":"05:24.140 ","End":"05:26.940","Text":"So this is our equation for pressure."},{"Start":"05:26.940 ","End":"05:29.870","Text":"We saw that the pressure at the beginning of"},{"Start":"05:29.870 ","End":"05:33.800","Text":"our process was the same as at the end of our process."},{"Start":"05:33.800 ","End":"05:35.810","Text":"That\u0027s how we got to this equilibrium that"},{"Start":"05:35.810 ","End":"05:41.180","Text":"the same force could remain balanced without moving up and down."},{"Start":"05:41.180 ","End":"05:44.855","Text":"So it just reaches some height and stays there."},{"Start":"05:44.855 ","End":"05:49.730","Text":"What we can see is that there was a change in the volume."},{"Start":"05:49.730 ","End":"05:54.740","Text":"We went from a volume over here of V_1 to a volume,"},{"Start":"05:54.740 ","End":"05:58.385","Text":"V_2, of all of this space over here."},{"Start":"05:58.385 ","End":"06:01.655","Text":"So we can see that we had a change of volume."},{"Start":"06:01.655 ","End":"06:06.590","Text":"Let\u0027s first write out our equation for volume,"},{"Start":"06:06.590 ","End":"06:09.755","Text":"and then we\u0027ll see how to write the change in volume."},{"Start":"06:09.755 ","End":"06:11.120","Text":"Volume, as we know,"},{"Start":"06:11.120 ","End":"06:16.820","Text":"is equal to the area multiplied by the length."},{"Start":"06:16.820 ","End":"06:27.460","Text":"Let\u0027s call this length L. So this is our equation for volume."},{"Start":"06:27.460 ","End":"06:30.010","Text":"But we saw that our volume changed."},{"Start":"06:30.010 ","End":"06:34.990","Text":"What we\u0027re going to do is we\u0027re going to represent this in derivative form,"},{"Start":"06:34.990 ","End":"06:36.895","Text":"so a change in volume,"},{"Start":"06:36.895 ","End":"06:39.770","Text":"dV, is equal to the area."},{"Start":"06:39.770 ","End":"06:41.565","Text":"The area stayed the same,"},{"Start":"06:41.565 ","End":"06:47.335","Text":"so we\u0027ll just keep it as S. But we can see that the length is what actually changed."},{"Start":"06:47.335 ","End":"06:51.390","Text":"We went from L_1 to L_2."},{"Start":"06:51.390 ","End":"06:57.790","Text":"So we can see that we have to say that a change in volume in this specific example"},{"Start":"06:57.790 ","End":"07:01.450","Text":"resulted from a change in length because"},{"Start":"07:01.450 ","End":"07:06.495","Text":"the cross-sectional area of the cylinder remained the same."},{"Start":"07:06.495 ","End":"07:10.520","Text":"So this is our equation for the change in volume."},{"Start":"07:10.520 ","End":"07:12.770","Text":"Now, as we know,"},{"Start":"07:12.770 ","End":"07:17.580","Text":"work is equal to"},{"Start":"07:18.280 ","End":"07:26.810","Text":"the force being applied multiplied by the distance that it was being applied to."},{"Start":"07:26.810 ","End":"07:30.770","Text":"So if we travel 100 meters applying this force,"},{"Start":"07:30.770 ","End":"07:32.840","Text":"we\u0027ll do a certain amount of work."},{"Start":"07:32.840 ","End":"07:35.945","Text":"But if we travel a 1000 meters,"},{"Start":"07:35.945 ","End":"07:37.535","Text":"applying the same force,"},{"Start":"07:37.535 ","End":"07:39.830","Text":"more work would have been done."},{"Start":"07:39.830 ","End":"07:44.675","Text":"So here we can see that we had a change in work, dW,"},{"Start":"07:44.675 ","End":"07:47.269","Text":"because first we were located here,"},{"Start":"07:47.269 ","End":"07:52.940","Text":"and then the work increased because we traveled this distance upwards with the lid."},{"Start":"07:52.940 ","End":"07:58.425","Text":"So we can see that this results in dW,"},{"Start":"07:58.425 ","End":"08:01.940","Text":"the change in work is equal to FdL in"},{"Start":"08:01.940 ","End":"08:06.933","Text":"this example over here because our length has changed."},{"Start":"08:06.933 ","End":"08:09.100","Text":"Now, just a little note,"},{"Start":"08:09.100 ","End":"08:13.870","Text":"sometimes dL can be represented by dx,"},{"Start":"08:13.870 ","End":"08:16.690","Text":"l can be given in units of x,"},{"Start":"08:16.690 ","End":"08:18.025","Text":"but it doesn\u0027t really matter."},{"Start":"08:18.025 ","End":"08:22.465","Text":"It\u0027s some change in unit length."},{"Start":"08:22.465 ","End":"08:27.790","Text":"Now how do we get to calculate the work?"},{"Start":"08:27.790 ","End":"08:30.015","Text":"This is 1 unit of work."},{"Start":"08:30.015 ","End":"08:34.990","Text":"It\u0027s given by applying this force to 1 unit of length."},{"Start":"08:34.990 ","End":"08:40.015","Text":"But if I want to find the total amount of work done during this entire length,"},{"Start":"08:40.015 ","End":"08:41.905","Text":"so I\u0027m going to integrate both sides,"},{"Start":"08:41.905 ","End":"08:47.110","Text":"and I\u0027ll get that the work is equal to the integral from"},{"Start":"08:47.110 ","End":"08:53.590","Text":"our original point to our end point of a force dL,"},{"Start":"08:53.590 ","End":"08:58.810","Text":"or dx, it doesn\u0027t matter whichever unit you\u0027re using."},{"Start":"08:58.810 ","End":"09:01.555","Text":"This, as we know,"},{"Start":"09:01.555 ","End":"09:04.270","Text":"is our equation for work."},{"Start":"09:04.270 ","End":"09:07.630","Text":"Now what we want to do is we want to see how to"},{"Start":"09:07.630 ","End":"09:12.340","Text":"derive our work equation when we\u0027re dealing with ideal gases."},{"Start":"09:12.340 ","End":"09:19.620","Text":"As we can see, force is equal to pressure times the surface area."},{"Start":"09:19.620 ","End":"09:23.805","Text":"Here we\u0027re going into ideal gas mode."},{"Start":"09:23.805 ","End":"09:26.730","Text":"Work is equal to the integral of force,"},{"Start":"09:26.730 ","End":"09:31.585","Text":"which is pressure times surface area,"},{"Start":"09:31.585 ","End":"09:36.310","Text":"and then we\u0027re multiplying this by dL."},{"Start":"09:36.310 ","End":"09:40.840","Text":"dL is equal to our change in volume,"},{"Start":"09:40.840 ","End":"09:49.405","Text":"so dV divided by our surface area,"},{"Start":"09:49.405 ","End":"09:50.965","Text":"is equal to dL."},{"Start":"09:50.965 ","End":"09:54.550","Text":"Now we can see that our surface areas cancel out,"},{"Start":"09:54.550 ","End":"09:57.400","Text":"and what we get is that work,"},{"Start":"09:57.400 ","End":"09:59.920","Text":"when dealing with ideal gases,"},{"Start":"09:59.920 ","End":"10:04.130","Text":"is equal to the integral of pressure,"},{"Start":"10:04.230 ","End":"10:09.789","Text":"PdV, multiplied by the change in volume."},{"Start":"10:09.789 ","End":"10:18.360","Text":"So this is the equation to calculate the work done by a system with an ideal gas."},{"Start":"10:18.360 ","End":"10:23.955","Text":"Let\u0027s just do a work example with this example that we were dealing with."},{"Start":"10:23.955 ","End":"10:29.580","Text":"Originally, our lid was located over here."},{"Start":"10:29.580 ","End":"10:33.780","Text":"Then it moved up until over here so let\u0027s say that"},{"Start":"10:33.780 ","End":"10:40.885","Text":"this length is L. On our lid,"},{"Start":"10:40.885 ","End":"10:46.199","Text":"of course, we have a force pressing downwards,"},{"Start":"10:46.199 ","End":"10:53.069","Text":"and this force is equal to the mass of the lid multiplied by gravity,"},{"Start":"10:53.069 ","End":"10:58.180","Text":"mg. Now we can, therefore,"},{"Start":"10:58.180 ","End":"11:06.196","Text":"say that the work done is equal to the integral on PdV."},{"Start":"11:06.196 ","End":"11:13.870","Text":"We can say that the work done is equal to the integral"},{"Start":"11:13.870 ","End":"11:17.830","Text":"from 0 up until L because"},{"Start":"11:17.830 ","End":"11:22.345","Text":"here we can say that we were at height 0 and here we are at height L, the lid."},{"Start":"11:22.345 ","End":"11:25.690","Text":"0 to L of P, pressure,"},{"Start":"11:25.690 ","End":"11:28.210","Text":"which is equal to force divided by area,"},{"Start":"11:28.210 ","End":"11:32.365","Text":"so that\u0027s equal to mg, that\u0027s the force,"},{"Start":"11:32.365 ","End":"11:33.850","Text":"divided by the area,"},{"Start":"11:33.850 ","End":"11:36.874","Text":"which is S, and of course all of this is changing,"},{"Start":"11:36.874 ","End":"11:39.280","Text":"so that\u0027s why our pressure is constant,"},{"Start":"11:39.280 ","End":"11:43.645","Text":"and then multiplied by dV."},{"Start":"11:43.645 ","End":"11:52.495","Text":"So we\u0027re integrating along V. So then we can say that this is equal to mg divided by S,"},{"Start":"11:52.495 ","End":"11:54.010","Text":"which is a constant."},{"Start":"11:54.010 ","End":"11:58.450","Text":"Then our V is the only thing that\u0027s changing so we\u0027ll add in our V over here,"},{"Start":"11:58.450 ","End":"12:03.805","Text":"and our bounds are from 0 till this height,"},{"Start":"12:03.805 ","End":"12:06.625","Text":"L. Then we can see that we have"},{"Start":"12:06.625 ","End":"12:13.160","Text":"mg/S multiplied by L minus mg/S multiplied by 0, which is 0,"},{"Start":"12:13.160 ","End":"12:16.960","Text":"so we\u0027ll just get that the work done in this simple process is"},{"Start":"12:16.960 ","End":"12:21.325","Text":"equal to mg/S multiplied by"},{"Start":"12:21.325 ","End":"12:30.565","Text":"L. This is the answer to how much work is done in this system of this gas."},{"Start":"12:30.565 ","End":"12:33.770","Text":"See how easy that is to solve."},{"Start":"12:33.770 ","End":"12:39.430","Text":"Now let\u0027s take a look at this information graphically."},{"Start":"12:39.620 ","End":"12:43.127","Text":"Here we have our graph,"},{"Start":"12:43.127 ","End":"12:49.830","Text":"and we have P on the y-axis and V on the x-axis."},{"Start":"12:49.830 ","End":"12:54.760","Text":"Let\u0027s say that we start at an original pressure,"},{"Start":"12:54.760 ","End":"12:57.850","Text":"and we end with a slightly higher pressure,"},{"Start":"12:57.850 ","End":"13:01.330","Text":"and we have this type of process happening."},{"Start":"13:01.330 ","End":"13:09.090","Text":"If we take our initial volume and our final volume,"},{"Start":"13:09.090 ","End":"13:17.190","Text":"so we can see that the area under the graph is equal to W. What\u0027s W?"},{"Start":"13:17.190 ","End":"13:23.430","Text":"It\u0027s our work. What we can see is that, from this equation,"},{"Start":"13:23.430 ","End":"13:27.960","Text":"we know why we often use this graph of"},{"Start":"13:27.960 ","End":"13:33.775","Text":"pressure as a function of volume because we know that if we integrate along this,"},{"Start":"13:33.775 ","End":"13:40.250","Text":"then the area under the graph is going to be equal to the work done by the system."},{"Start":"13:40.320 ","End":"13:45.430","Text":"Now notice it makes a difference in which direction we\u0027re integrating."},{"Start":"13:45.430 ","End":"13:50.650","Text":"If we start from this initial pressure,"},{"Start":"13:50.650 ","End":"13:52.795","Text":"and we finish at this final pressure,"},{"Start":"13:52.795 ","End":"13:56.848","Text":"which means that our system is moving in this direction,"},{"Start":"13:56.848 ","End":"14:00.908","Text":"so our pressure is increasing and our volume is increasing,"},{"Start":"14:00.908 ","End":"14:04.885","Text":"so our system is doing work."},{"Start":"14:04.885 ","End":"14:07.600","Text":"That is what we saw, let\u0027s say,"},{"Start":"14:07.600 ","End":"14:14.740","Text":"over here where our system did work in order to raise the lid that had some mass on it."},{"Start":"14:14.740 ","End":"14:18.885","Text":"However, so this is positive work,"},{"Start":"14:18.885 ","End":"14:21.490","Text":"if we\u0027re going from a higher pressure and"},{"Start":"14:21.490 ","End":"14:24.250","Text":"a higher volume to a lower pressure and a lower volume,"},{"Start":"14:24.250 ","End":"14:26.650","Text":"so in the opposite direction,"},{"Start":"14:26.650 ","End":"14:32.470","Text":"then that means that we can see that the lid is going downwards."},{"Start":"14:32.470 ","End":"14:33.640","Text":"So what does that mean?"},{"Start":"14:33.640 ","End":"14:36.280","Text":"Work is being done on the system."},{"Start":"14:36.280 ","End":"14:40.820","Text":"We have a negative value for our work."},{"Start":"14:41.640 ","End":"14:47.050","Text":"This positive or negative value of work will tell us if our system"},{"Start":"14:47.050 ","End":"14:52.405","Text":"has exerted energy or has taken energy in."},{"Start":"14:52.405 ","End":"14:54.610","Text":"If our work is positive,"},{"Start":"14:54.610 ","End":"14:56.725","Text":"that means that energy was exerted,"},{"Start":"14:56.725 ","End":"14:58.270","Text":"and if our work is a negative,"},{"Start":"14:58.270 ","End":"15:02.080","Text":"that means that energy was taken into the system."},{"Start":"15:02.080 ","End":"15:08.260","Text":"Now let\u0027s look at the 3 processes that we learned in the previous lesson."},{"Start":"15:08.260 ","End":"15:10.975","Text":"We\u0027re ignoring the adiabatic process."},{"Start":"15:10.975 ","End":"15:12.565","Text":"So we have the isothermal,"},{"Start":"15:12.565 ","End":"15:15.895","Text":"isochoric, and isobaric processes."},{"Start":"15:15.895 ","End":"15:19.375","Text":"The isothermal process we just saw now,"},{"Start":"15:19.375 ","End":"15:21.460","Text":"where we increased the temperature of the gas,"},{"Start":"15:21.460 ","End":"15:25.225","Text":"and therefore, our lid on our cylinder was raised."},{"Start":"15:25.225 ","End":"15:31.900","Text":"We already saw that in order to work out the work for such a process,"},{"Start":"15:31.900 ","End":"15:34.850","Text":"we integrate along PdV."},{"Start":"15:35.310 ","End":"15:39.805","Text":"What about when we\u0027re dealing with an isochoric process?"},{"Start":"15:39.805 ","End":"15:43.900","Text":"There we have that our volume is constant."},{"Start":"15:43.900 ","End":"15:46.225","Text":"What we can see is if my volume is constant,"},{"Start":"15:46.225 ","End":"15:50.290","Text":"our graph is just some line going straight up."},{"Start":"15:50.290 ","End":"15:57.085","Text":"We can see that there\u0027s no energy in the system because there\u0027s no area under the graph."},{"Start":"15:57.085 ","End":"15:59.155","Text":"So in this isochoric system,"},{"Start":"15:59.155 ","End":"16:02.680","Text":"we know that the work done is equal to 0."},{"Start":"16:02.680 ","End":"16:05.050","Text":"So there is no area under the graph."},{"Start":"16:05.050 ","End":"16:07.629","Text":"Now, in an isobaric process,"},{"Start":"16:07.629 ","End":"16:10.855","Text":"we know that our pressure is constant."},{"Start":"16:10.855 ","End":"16:15.610","Text":"What we can see is that we do have an area under the graph,"},{"Start":"16:15.610 ","End":"16:19.580","Text":"and what we can see is that the area is just the change in volume."},{"Start":"16:19.580 ","End":"16:24.560","Text":"So it\u0027s the final volume minus the initial volume."},{"Start":"16:24.560 ","End":"16:28.715","Text":"The work is simply equal to the pressure,"},{"Start":"16:28.715 ","End":"16:30.215","Text":"which is a constant value,"},{"Start":"16:30.215 ","End":"16:32.390","Text":"multiplied by Delta V,"},{"Start":"16:32.390 ","End":"16:33.980","Text":"the change in volume,"},{"Start":"16:33.980 ","End":"16:40.920","Text":"where here we don\u0027t need to integrate because our volume is just V_2 minus V_1."},{"Start":"16:40.920 ","End":"16:45.465","Text":"Here we have to integrate because we can see that there\u0027s no linear relationship."},{"Start":"16:45.465 ","End":"16:50.525","Text":"So these are equations that are useful to remember,"},{"Start":"16:50.525 ","End":"16:52.730","Text":"especially this equation, because, from this,"},{"Start":"16:52.730 ","End":"16:55.329","Text":"you can derive all of this."},{"Start":"16:55.329 ","End":"16:58.710","Text":"That is the end of this lesson."}],"ID":12356},{"Watched":false,"Name":"U=Q-W","Duration":"10m 54s","ChapterTopicVideoID":11928,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"Hello. In this lesson,"},{"Start":"00:02.175 ","End":"00:07.470","Text":"we\u0027re going to be learning how the internal energy of a system can be changed."},{"Start":"00:07.470 ","End":"00:09.390","Text":"Here we have an equation."},{"Start":"00:09.390 ","End":"00:14.850","Text":"Now, this equation is also known as the first law of thermodynamics."},{"Start":"00:14.850 ","End":"00:19.920","Text":"We can see that the U is equal to our internal energy,"},{"Start":"00:19.920 ","End":"00:23.220","Text":"and we can see that there are two things that can"},{"Start":"00:23.220 ","End":"00:27.090","Text":"increase or decrease the internal energy of a system."},{"Start":"00:27.090 ","End":"00:29.895","Text":"One is heat,"},{"Start":"00:29.895 ","End":"00:34.947","Text":"so the addition or removal of heat to affirm the system,"},{"Start":"00:34.947 ","End":"00:38.250","Text":"and the other is the work of the system."},{"Start":"00:38.250 ","End":"00:42.920","Text":"We already saw in the previous lesson the work can be some mechanical work,"},{"Start":"00:42.920 ","End":"00:48.319","Text":"lifting some kind of movable piston or expanding a balloon,"},{"Start":"00:48.319 ","End":"00:50.315","Text":"and so on and so forth."},{"Start":"00:50.315 ","End":"00:53.210","Text":"Let\u0027s see what this Q is."},{"Start":"00:53.210 ","End":"00:57.675","Text":"This Q is the heat added to the system,"},{"Start":"00:57.675 ","End":"01:01.160","Text":"so that means that we\u0027re not taking into account"},{"Start":"01:01.160 ","End":"01:06.350","Text":"the heat change due to different thermodynamic processes."},{"Start":"01:06.350 ","End":"01:10.490","Text":"This is how much heat was added to the system,"},{"Start":"01:10.490 ","End":"01:12.635","Text":"so if we heated it up,"},{"Start":"01:12.635 ","End":"01:20.210","Text":"and this W is the work done by this system."},{"Start":"01:22.160 ","End":"01:25.940","Text":"It\u0027s important to know that this is by the system,"},{"Start":"01:25.940 ","End":"01:31.843","Text":"that the system is doing the work on some other body and not the other way round,"},{"Start":"01:31.843 ","End":"01:36.533","Text":"not another system or another buddy doing work on our system."},{"Start":"01:36.533 ","End":"01:42.545","Text":"Our system is doing the work and that is important to remember with the minus sign."},{"Start":"01:42.545 ","End":"01:49.095","Text":"We\u0027ve seen other equations for internal energy that deal with ideal gases."},{"Start":"01:49.095 ","End":"01:54.510","Text":"There\u0027s this first one which is Beta n RT,"},{"Start":"01:54.510 ","End":"02:00.250","Text":"where Beta is the number of degrees of freedom divided by 2,"},{"Start":"02:00.250 ","End":"02:01.760","Text":"n is the number of moles,"},{"Start":"02:01.760 ","End":"02:06.904","Text":"R is the gas constant and T is the temperature given in Kelvin."},{"Start":"02:06.904 ","End":"02:12.590","Text":"The other equation that we saw is N C_v Delta T,"},{"Start":"02:12.590 ","End":"02:15.250","Text":"where N is the number of molecules,"},{"Start":"02:15.250 ","End":"02:19.280","Text":"C_v is the heat capacity at constant volume,"},{"Start":"02:19.280 ","End":"02:22.190","Text":"and Delta T is the change in temperature."},{"Start":"02:22.190 ","End":"02:26.765","Text":"Of course, because we know that the jumps in temperature between"},{"Start":"02:26.765 ","End":"02:34.200","Text":"1 degree Celsius and a jump of 1 degree Kelvin is the same size,"},{"Start":"02:34.200 ","End":"02:39.184","Text":"so our Delta T here can be calculated either in Kelvin or Celsius."},{"Start":"02:39.184 ","End":"02:41.330","Text":"It will be the same answer."},{"Start":"02:42.530 ","End":"02:46.230","Text":"Now let\u0027s look at an example."},{"Start":"02:46.230 ","End":"02:49.211","Text":"A lot of times we\u0027ll have a graph like this,"},{"Start":"02:49.211 ","End":"02:51.560","Text":"so pressure as a function of volume,"},{"Start":"02:51.560 ","End":"02:57.300","Text":"and we\u0027ll see what work is done by the system."},{"Start":"02:57.300 ","End":"03:02.465","Text":"Let\u0027s imagine that we\u0027re going in this clockwise direction."},{"Start":"03:02.465 ","End":"03:09.270","Text":"This is the positive direction of motion for each section of this graph."},{"Start":"03:09.270 ","End":"03:11.895","Text":"Let\u0027s look at this section of the graph."},{"Start":"03:11.895 ","End":"03:16.640","Text":"We\u0027re starting over here at this corner and we\u0027re ending over here at this corner."},{"Start":"03:16.640 ","End":"03:21.350","Text":"What we can see is that the volume is remaining constant,"},{"Start":"03:21.350 ","End":"03:23.130","Text":"keyword a constant volume,"},{"Start":"03:23.130 ","End":"03:27.400","Text":"so that means that we\u0027re dealing with an isochoric process."},{"Start":"03:27.400 ","End":"03:32.450","Text":"What we can see that it\u0027s happening here is that we have some kind of gas,"},{"Start":"03:32.450 ","End":"03:37.340","Text":"an ideal gas, which is in some kind of container of a fixed volume."},{"Start":"03:37.340 ","End":"03:42.675","Text":"But for some reason the pressure is increasing,"},{"Start":"03:42.675 ","End":"03:48.380","Text":"so we can say from our equation that if our volume is constant,"},{"Start":"03:48.380 ","End":"03:49.805","Text":"but our pressure is changing,"},{"Start":"03:49.805 ","End":"03:53.450","Text":"that means that our temperature must also be changing."},{"Start":"03:53.450 ","End":"03:55.700","Text":"Now, this temperature change,"},{"Start":"03:55.700 ","End":"03:58.085","Text":"we can also see from here the colors,"},{"Start":"03:58.085 ","End":"04:01.910","Text":"imagine there\u0027s a gradual change from blue to red,"},{"Start":"04:01.910 ","End":"04:05.270","Text":"so we\u0027re going from cooler ideal gas,"},{"Start":"04:05.270 ","End":"04:07.849","Text":"which is being heated up gradually."},{"Start":"04:07.849 ","End":"04:15.480","Text":"Because we see that the internal energy is connected to the temperature,"},{"Start":"04:15.480 ","End":"04:17.945","Text":"so as the temperature increases,"},{"Start":"04:17.945 ","End":"04:20.060","Text":"the internal energy increases,"},{"Start":"04:20.060 ","End":"04:24.600","Text":"which means that the pressure is increasing as well."},{"Start":"04:25.190 ","End":"04:29.540","Text":"What we can see is in the isochoric process,"},{"Start":"04:29.540 ","End":"04:34.475","Text":"we have no work being done because there\u0027s no area under the graph."},{"Start":"04:34.475 ","End":"04:37.145","Text":"Work is the area under the graph."},{"Start":"04:37.145 ","End":"04:40.040","Text":"We can see that in an isochoric process,"},{"Start":"04:40.040 ","End":"04:48.570","Text":"the internal energy U is only changed by the heat added to the system."},{"Start":"04:50.080 ","End":"04:55.000","Text":"I\u0027ve made a bit more space on the screen,"},{"Start":"04:55.000 ","End":"04:58.170","Text":"and I\u0027ve written that in an isochoric process,"},{"Start":"04:58.170 ","End":"05:01.280","Text":"so that\u0027s where our volume is constant."},{"Start":"05:01.280 ","End":"05:05.090","Text":"U the internal energy can only be changed"},{"Start":"05:05.090 ","End":"05:10.530","Text":"by adding or removing heat to or from the system."},{"Start":"05:11.140 ","End":"05:16.355","Text":"Now let\u0027s take a look at this area first before our constant temperature."},{"Start":"05:16.355 ","End":"05:20.840","Text":"Let\u0027s first look at the area of constant pressure."},{"Start":"05:20.840 ","End":"05:24.950","Text":"Here we can see that our system is moving from the right to the left direction,"},{"Start":"05:24.950 ","End":"05:31.290","Text":"and we have a straight line denoting a constant pressure over here."},{"Start":"05:31.290 ","End":"05:35.345","Text":"What we can see is that our pressure is constant, however,"},{"Start":"05:35.345 ","End":"05:39.060","Text":"our volume is starting from this initial volume,"},{"Start":"05:39.060 ","End":"05:41.450","Text":"and we\u0027re finishing at this volume,"},{"Start":"05:41.450 ","End":"05:43.595","Text":"which has a lower volume."},{"Start":"05:43.595 ","End":"05:52.085","Text":"We can see that in order to maintain a constant pressure,"},{"Start":"05:52.085 ","End":"05:55.130","Text":"however, to decrease the volume,"},{"Start":"05:55.130 ","End":"06:00.679","Text":"that means that we have to be changing the temperature of our system."},{"Start":"06:00.679 ","End":"06:03.913","Text":"Here we can see specifically that we\u0027re cooling the system."},{"Start":"06:03.913 ","End":"06:06.710","Text":"Number 1 from the color change in the graph,"},{"Start":"06:06.710 ","End":"06:09.230","Text":"we can see that we\u0027re going from red to blue,"},{"Start":"06:09.230 ","End":"06:10.670","Text":"so from hot to cold."},{"Start":"06:10.670 ","End":"06:15.395","Text":"But also if we can see that our volume is decreasing,"},{"Start":"06:15.395 ","End":"06:18.285","Text":"given our pressure is remaining the same,"},{"Start":"06:18.285 ","End":"06:23.045","Text":"then that means that the system must be cooling down."},{"Start":"06:23.045 ","End":"06:27.290","Text":"Why is that? We saw the temperature is connected to the internal energy,"},{"Start":"06:27.290 ","End":"06:31.655","Text":"and we know that our internal energy"},{"Start":"06:31.655 ","End":"06:38.750","Text":"causes a change in pressure or a change in volume."},{"Start":"06:38.750 ","End":"06:41.390","Text":"If our pressures are remaining the same,"},{"Start":"06:41.390 ","End":"06:47.240","Text":"then our volume has to be decreasing when we\u0027re cooling because our particles in"},{"Start":"06:47.240 ","End":"06:49.549","Text":"the gas have less energy to balance"},{"Start":"06:49.549 ","End":"06:54.355","Text":"on one another and against the walls of the container."},{"Start":"06:54.355 ","End":"06:57.665","Text":"In order to maintain the same pressure,"},{"Start":"06:57.665 ","End":"07:01.730","Text":"even though the particles are bouncing less,"},{"Start":"07:01.730 ","End":"07:04.210","Text":"is to decrease the volume."},{"Start":"07:04.210 ","End":"07:06.115","Text":"Now let\u0027s look at this section."},{"Start":"07:06.115 ","End":"07:09.820","Text":"Here we can see that our temperature is constant."},{"Start":"07:09.820 ","End":"07:14.040","Text":"We\u0027re being told and also we can see from the red line."},{"Start":"07:14.040 ","End":"07:18.250","Text":"What we can see is that if we\u0027re maintaining a constant temperature,"},{"Start":"07:18.250 ","End":"07:21.040","Text":"but we change our volume."},{"Start":"07:21.040 ","End":"07:25.900","Text":"Let\u0027s say that we move from an initial volume to a final volume,"},{"Start":"07:25.900 ","End":"07:27.760","Text":"which is double the initial volume,"},{"Start":"07:27.760 ","End":"07:30.490","Text":"but the temperature mean is the same."},{"Start":"07:30.490 ","End":"07:34.105","Text":"That means that the pressure is going to decrease from"},{"Start":"07:34.105 ","End":"07:38.035","Text":"an initial pressure to half of the initial pressure."},{"Start":"07:38.035 ","End":"07:43.530","Text":"The pressure is going to decrease if the volume is increasing."},{"Start":"07:43.530 ","End":"07:45.125","Text":"That makes sense."},{"Start":"07:45.125 ","End":"07:47.975","Text":"Now a quick note about this type of graph."},{"Start":"07:47.975 ","End":"07:52.070","Text":"This constant temperature or isothermal section of"},{"Start":"07:52.070 ","End":"07:56.835","Text":"the graph is similar to the function 1 divided by x."},{"Start":"07:56.835 ","End":"07:58.230","Text":"That\u0027s what it looks like,"},{"Start":"07:58.230 ","End":"08:02.555","Text":"and this represents an isothermal process."},{"Start":"08:02.555 ","End":"08:06.779","Text":"However, we also saw that there is an adiabatic process,"},{"Start":"08:06.779 ","End":"08:12.900","Text":"and the graph where the adiabatic process looks very similar to an isothermal system."},{"Start":"08:12.900 ","End":"08:14.990","Text":"It\u0027s also some kind of curve,"},{"Start":"08:14.990 ","End":"08:19.310","Text":"except it\u0027s not exactly the curve of 1 divided by x."},{"Start":"08:19.310 ","End":"08:25.620","Text":"We spoke a little bit about the adiabatic process in the previous lesson."},{"Start":"08:25.620 ","End":"08:29.195","Text":"We saw that in the adiabatic process,"},{"Start":"08:29.195 ","End":"08:33.965","Text":"both temperature, volume and pressure can change."},{"Start":"08:33.965 ","End":"08:38.060","Text":"But what is important to remember in the adiabatic process is that it is"},{"Start":"08:38.060 ","End":"08:42.455","Text":"isolated or insulated. What does that mean?"},{"Start":"08:42.455 ","End":"08:44.210","Text":"In an adiabatic process,"},{"Start":"08:44.210 ","End":"08:48.830","Text":"no heat can be added or removed to the system."},{"Start":"08:48.830 ","End":"08:51.365","Text":"The temperature can still change,"},{"Start":"08:51.365 ","End":"08:55.685","Text":"but no heat can be added or removed to the system."},{"Start":"08:55.685 ","End":"09:01.835","Text":"If we\u0027re going to write out this equation for internal energy for an adiabatic process,"},{"Start":"09:01.835 ","End":"09:09.440","Text":"we will simply get that U is equal to Q in an adiabatic process is equal to 0,"},{"Start":"09:09.440 ","End":"09:19.230","Text":"so U is equal to negative W. It\u0027s just equal to negative the work done by the system."},{"Start":"09:19.230 ","End":"09:24.440","Text":"A trick to know when we\u0027re dealing with an adiabatic or an isothermal process,"},{"Start":"09:24.440 ","End":"09:28.130","Text":"is that if the temperature between point A and"},{"Start":"09:28.130 ","End":"09:31.505","Text":"B is the same or is a constant temperature,"},{"Start":"09:31.505 ","End":"09:33.935","Text":"then we\u0027re dealing with an isothermal temperature."},{"Start":"09:33.935 ","End":"09:37.265","Text":"If the temperature at A and B is not the same,"},{"Start":"09:37.265 ","End":"09:41.015","Text":"then we\u0027re of course dealing with an adiabatic process."},{"Start":"09:41.015 ","End":"09:48.575","Text":"If we want to work out the total energy or the total work done on the system."},{"Start":"09:48.575 ","End":"09:54.125","Text":"We just have to work out this area inside the graph."},{"Start":"09:54.125 ","End":"09:57.350","Text":"Because we see in this first isochoric section,"},{"Start":"09:57.350 ","End":"10:00.515","Text":"we are doing no work on the system."},{"Start":"10:00.515 ","End":"10:04.220","Text":"Then in this isothermal section, we are doing work,"},{"Start":"10:04.220 ","End":"10:08.865","Text":"because there\u0027s all of this work or area under the graph."},{"Start":"10:08.865 ","End":"10:14.135","Text":"Then we\u0027re going to the isobaric section where our pressure is constant,"},{"Start":"10:14.135 ","End":"10:21.500","Text":"where we\u0027re taking away all of this white space over here of the area in the graph,"},{"Start":"10:21.500 ","End":"10:23.720","Text":"and then we\u0027re left with this area."},{"Start":"10:23.720 ","End":"10:27.590","Text":"This is the work done in this cyclic process."},{"Start":"10:27.590 ","End":"10:32.990","Text":"Now, of course, if our cyclic process was going in the opposite direction, for instance,"},{"Start":"10:32.990 ","End":"10:35.120","Text":"in this direction,"},{"Start":"10:35.120 ","End":"10:37.985","Text":"anticlockwise instead of clockwise,"},{"Start":"10:37.985 ","End":"10:42.395","Text":"we would still have this work being done by the system."},{"Start":"10:42.395 ","End":"10:46.325","Text":"However, it would have a negative sign in front of it."},{"Start":"10:46.325 ","End":"10:51.260","Text":"Work would be done on the system in that case."},{"Start":"10:51.260 ","End":"10:54.270","Text":"That\u0027s the end of this lesson."}],"ID":12357},{"Watched":false,"Name":"Adiabatic Process","Duration":"9m 3s","ChapterTopicVideoID":11929,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"Hello. In this lesson we\u0027re going to be speaking"},{"Start":"00:02.850 ","End":"00:05.940","Text":"about the adiabatic process in a little bit more detail."},{"Start":"00:05.940 ","End":"00:10.770","Text":"We spoke about it a little bit in previous lessons and we saw that"},{"Start":"00:10.770 ","End":"00:16.275","Text":"the adiabatic process involves some kind of insulation."},{"Start":"00:16.275 ","End":"00:21.390","Text":"That means that there\u0027s no heat transfer in an adiabatic process,"},{"Start":"00:21.390 ","End":"00:24.450","Text":"so Q is equal to 0."},{"Start":"00:24.450 ","End":"00:31.185","Text":"Now aside from the system having to be insulated in an adiabatic process,"},{"Start":"00:31.185 ","End":"00:37.370","Text":"another way that we can achieve this adiabatic process without our system being"},{"Start":"00:37.370 ","End":"00:46.160","Text":"insulated is if the work or the change in energy is so fast that in that time-frame,"},{"Start":"00:46.160 ","End":"00:51.800","Text":"there\u0027s not enough time for heat to be transferred to or from our system."},{"Start":"00:51.800 ","End":"00:53.780","Text":"In which case, again,"},{"Start":"00:53.780 ","End":"01:00.155","Text":"we have a situation of zero heat transfer and then we\u0027re again in an adiabatic process,"},{"Start":"01:00.155 ","End":"01:05.010","Text":"even though our system isn\u0027t necessarily insulated."},{"Start":"01:06.080 ","End":"01:11.120","Text":"There\u0027s a very good video on the Internet that demonstrates"},{"Start":"01:11.120 ","End":"01:16.035","Text":"the adiabatic process with some piston."},{"Start":"01:16.035 ","End":"01:20.405","Text":"The video you can search for it is called transparent fire piston,"},{"Start":"01:20.405 ","End":"01:27.500","Text":"explosive flash and what we can see is how the adiabatic process works over here."},{"Start":"01:27.500 ","End":"01:31.670","Text":"In this example, I\u0027m going to draw it very crudely so we have"},{"Start":"01:31.670 ","End":"01:35.640","Text":"some cylinder and in the cylinder,"},{"Start":"01:35.640 ","End":"01:37.365","Text":"there\u0027s some gas,"},{"Start":"01:37.365 ","End":"01:41.440","Text":"and here is the piston lid."},{"Start":"01:41.440 ","End":"01:45.860","Text":"Now, in order to make this an adiabatic process,"},{"Start":"01:45.860 ","End":"01:48.860","Text":"the cylinder actually isn\u0027t insulated,"},{"Start":"01:48.860 ","End":"01:53.450","Text":"but what we do is we push down very quickly on"},{"Start":"01:53.450 ","End":"02:03.859","Text":"this piston lid so that it goes downwards in this direction in a very fast motion."},{"Start":"02:03.859 ","End":"02:09.484","Text":"Now this fast motion causes a rapid change in the volume,"},{"Start":"02:09.484 ","End":"02:13.220","Text":"which means that we can see that there\u0027s work being"},{"Start":"02:13.220 ","End":"02:17.580","Text":"done and a lot of work in a very short period of time."},{"Start":"02:17.580 ","End":"02:21.650","Text":"Then we can see or you can see in the video,"},{"Start":"02:21.650 ","End":"02:25.790","Text":"that there\u0027s some kind of flash because the temperature"},{"Start":"02:25.790 ","End":"02:32.650","Text":"increases due to the volume change and pressure change."},{"Start":"02:32.960 ","End":"02:35.690","Text":"When we push the lid downwards,"},{"Start":"02:35.690 ","End":"02:38.330","Text":"the volume becomes much smaller,"},{"Start":"02:38.330 ","End":"02:43.265","Text":"which means that the pressure increases also very quickly and that means that it"},{"Start":"02:43.265 ","End":"02:48.625","Text":"manages to raise the temperature of the gas inside the piston."},{"Start":"02:48.625 ","End":"02:51.045","Text":"Then you can see that there\u0027s a flash."},{"Start":"02:51.045 ","End":"02:54.559","Text":"That is an example of an adiabatic process."},{"Start":"02:54.559 ","End":"03:02.150","Text":"We\u0027re just doing this change in volume and pressure very quickly,"},{"Start":"03:02.150 ","End":"03:06.995","Text":"which means that there\u0027s no time for heat transfer."},{"Start":"03:06.995 ","End":"03:10.850","Text":"Then that causes some flash."},{"Start":"03:10.850 ","End":"03:18.119","Text":"Now this works in a lot of different machinery and in different engines."},{"Start":"03:18.119 ","End":"03:24.215","Text":"What would happen in an engine is that this piston will move down, up and down,"},{"Start":"03:24.215 ","End":"03:26.015","Text":"up and down very quickly,"},{"Start":"03:26.015 ","End":"03:30.704","Text":"generating a big temperature change."},{"Start":"03:30.704 ","End":"03:35.200","Text":"Therefore, we get energy and work."},{"Start":"03:35.270 ","End":"03:39.350","Text":"Now let\u0027s go back to our equation of state."},{"Start":"03:39.350 ","End":"03:44.120","Text":"We remember that we had pressure multiplied by volume is equal"},{"Start":"03:44.120 ","End":"03:48.980","Text":"to lowercase n. The number of moles multiplied by R,"},{"Start":"03:48.980 ","End":"03:53.705","Text":"which is the gas constant or the universal gas constant multiplied by T,"},{"Start":"03:53.705 ","End":"03:57.135","Text":"which is temperature in Kelvin."},{"Start":"03:57.135 ","End":"04:00.005","Text":"We know that in the adiabatic process,"},{"Start":"04:00.005 ","End":"04:04.175","Text":"unlike the other 3 processes that we learned about,"},{"Start":"04:04.175 ","End":"04:05.900","Text":"in the adiabatic process,"},{"Start":"04:05.900 ","End":"04:08.090","Text":"all 3 of these variables,"},{"Start":"04:08.090 ","End":"04:10.670","Text":"P, V, and T, can change."},{"Start":"04:10.670 ","End":"04:12.980","Text":"In other processes we saw that either"},{"Start":"04:12.980 ","End":"04:15.785","Text":"the pressure is constant or the volume or the temperature."},{"Start":"04:15.785 ","End":"04:19.519","Text":"In the adiabatic, everything can change."},{"Start":"04:19.519 ","End":"04:23.450","Text":"Just like we saw in the example of the piston."},{"Start":"04:23.450 ","End":"04:30.655","Text":"Our pressure was changed and that increased the temperature."},{"Start":"04:30.655 ","End":"04:37.130","Text":"Now another time that we see an adiabatic process is when using a bicycle pump."},{"Start":"04:37.130 ","End":"04:42.620","Text":"Say you want to pump up the wheels of your bike and you\u0027re using one of those handheld,"},{"Start":"04:42.620 ","End":"04:48.215","Text":"old-fashioned pumps, so you\u0027ll feel that the pump itself is getting very hot."},{"Start":"04:48.215 ","End":"04:52.790","Text":"Now some people might think that this increase in temperature is due to the friction."},{"Start":"04:52.790 ","End":"04:59.170","Text":"However, it isn\u0027t due to the friction it\u0027s due to this adiabatic process."},{"Start":"05:00.100 ","End":"05:06.185","Text":"There we spoke about examples where objects or systems are heating up."},{"Start":"05:06.185 ","End":"05:08.810","Text":"But of course, the adiabatic process can"},{"Start":"05:08.810 ","End":"05:11.390","Text":"work in the opposite direction where things cool down."},{"Start":"05:11.390 ","End":"05:16.305","Text":"For instance, if you take your classic aerosol cans."},{"Start":"05:16.305 ","End":"05:20.750","Text":"The pressure inside the can is very large and when you"},{"Start":"05:20.750 ","End":"05:25.290","Text":"spray out your deodorant whichever aerosol you are using,"},{"Start":"05:25.290 ","End":"05:31.580","Text":"so suddenly the pressure is significantly reduced."},{"Start":"05:31.580 ","End":"05:36.085","Text":"The pressure inside the aerosol can is much higher than the pressure outside the can."},{"Start":"05:36.085 ","End":"05:39.230","Text":"When the pressure is reduced,"},{"Start":"05:39.230 ","End":"05:41.120","Text":"so you can feel that the gas has"},{"Start":"05:41.120 ","End":"05:43.790","Text":"cooled and that\u0027s why when you spray deodorant on yourself,"},{"Start":"05:43.790 ","End":"05:47.015","Text":"you might feel a cold sensation."},{"Start":"05:47.015 ","End":"05:49.790","Text":"That\u0027s because the temperature of the gas has been"},{"Start":"05:49.790 ","End":"05:52.730","Text":"reduced due to the pressure being reduced."},{"Start":"05:52.730 ","End":"05:56.555","Text":"Another example of lower pressure"},{"Start":"05:56.555 ","End":"06:01.190","Text":"and lower temperatures is if we go up in our atmosphere,"},{"Start":"06:01.190 ","End":"06:04.040","Text":"the higher that we go up,"},{"Start":"06:04.040 ","End":"06:08.330","Text":"say any mountain or if you\u0027re on Everest,"},{"Start":"06:08.330 ","End":"06:12.485","Text":"the temperatures will be much lower than down on earth and part"},{"Start":"06:12.485 ","End":"06:16.670","Text":"of that is also due to that change in pressure."},{"Start":"06:16.670 ","End":"06:19.940","Text":"Obviously at higher altitudes the pressure is much lower,"},{"Start":"06:19.940 ","End":"06:23.700","Text":"so the temperature is also lower."},{"Start":"06:23.700 ","End":"06:28.370","Text":"When we\u0027re using our equation of state in the adiabatic process,"},{"Start":"06:28.370 ","End":"06:34.260","Text":"it\u0027s a bit difficult to solve the equation because our P,"},{"Start":"06:34.260 ","End":"06:37.230","Text":"V, and T are variables."},{"Start":"06:37.230 ","End":"06:42.260","Text":"We have 3 unknowns and only one equation."},{"Start":"06:42.260 ","End":"06:46.400","Text":"How to solve an adiabatic process or a question"},{"Start":"06:46.400 ","End":"06:50.660","Text":"involving an adiabatic process is by using this equation,"},{"Start":"06:50.660 ","End":"06:54.020","Text":"PV to the power of Gamma,"},{"Start":"06:54.020 ","End":"06:57.055","Text":"soon we\u0027ll speak about what this Gamma is equal to,"},{"Start":"06:57.055 ","End":"07:01.560","Text":"is equal to a constant."},{"Start":"07:01.560 ","End":"07:07.400","Text":"Then from this, let\u0027s say we have a process and we have an initial pressure multiplied by"},{"Start":"07:07.400 ","End":"07:14.190","Text":"an initial volume to the power of Gamma is going to be equal to some constant."},{"Start":"07:14.260 ","End":"07:20.150","Text":"Then we can have that our final pressure multiplied by"},{"Start":"07:20.150 ","End":"07:27.080","Text":"our final volume to the same power Gamma is also equal to this constant."},{"Start":"07:27.080 ","End":"07:30.605","Text":"Then we can just use algebra in order to solve,"},{"Start":"07:30.605 ","End":"07:35.345","Text":"because now we have 3 unknowns and 3 equations."},{"Start":"07:35.345 ","End":"07:37.810","Text":"What is this Gamma?"},{"Start":"07:37.810 ","End":"07:41.420","Text":"Gamma is an equation that involves"},{"Start":"07:41.420 ","End":"07:46.980","Text":"the degrees of freedom of the gas molecules that we\u0027re using."},{"Start":"07:47.090 ","End":"07:57.085","Text":"The equation for Gamma is given by Beta plus 1 divided by Beta,"},{"Start":"07:57.085 ","End":"08:03.890","Text":"where Beta is dependent on the degrees of freedom of our gas."},{"Start":"08:03.890 ","End":"08:06.350","Text":"As we remember, Beta,"},{"Start":"08:06.350 ","End":"08:09.095","Text":"if we\u0027re dealing with a monatomic gas,"},{"Start":"08:09.095 ","End":"08:12.260","Text":"is equal to 3 divided by 2 and Beta,"},{"Start":"08:12.260 ","End":"08:14.255","Text":"if we\u0027re dealing with a diatomic gas,"},{"Start":"08:14.255 ","End":"08:17.760","Text":"is equal to 5 divided by 2."},{"Start":"08:17.760 ","End":"08:21.065","Text":"Depending if you\u0027re dealing with the monatomic or diatomic gas,"},{"Start":"08:21.065 ","End":"08:26.195","Text":"you\u0027ll substitute in one of these values for Beta into this equation, Gamma."},{"Start":"08:26.195 ","End":"08:29.520","Text":"Then you substitute it in over here."},{"Start":"08:30.140 ","End":"08:33.049","Text":"For the adiabatic process,"},{"Start":"08:33.049 ","End":"08:36.620","Text":"these are the 3 equations to remember and of course this isn\u0027t"},{"Start":"08:36.620 ","End":"08:40.639","Text":"the equation of state which is useful for all of the processes."},{"Start":"08:40.639 ","End":"08:45.200","Text":"The adiabatic process, there\u0027s no heat exchange."},{"Start":"08:45.200 ","End":"08:49.925","Text":"We use this equation PV to the power of Gamma is equal to a constant,"},{"Start":"08:49.925 ","End":"08:53.525","Text":"where Gamma is equal to Beta plus 1 divided by Beta,"},{"Start":"08:53.525 ","End":"08:55.880","Text":"where Beta is dependent on the degrees of"},{"Start":"08:55.880 ","End":"09:00.640","Text":"freedom if we\u0027re dealing with a monatomic or diatomic gas."},{"Start":"09:00.640 ","End":"09:03.520","Text":"That\u0027s the end of this lesson."}],"ID":12358},{"Watched":false,"Name":"Processes and Efficiency","Duration":"7m 42s","ChapterTopicVideoID":11930,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.835","Text":"Hello. Up until now,"},{"Start":"00:02.835 ","End":"00:06.525","Text":"we\u0027ve learned of these 4 thermodynamic processes."},{"Start":"00:06.525 ","End":"00:09.780","Text":"Isothermal, where our temperature is constant,"},{"Start":"00:09.780 ","End":"00:13.050","Text":"isochoric where our volume is constant,"},{"Start":"00:13.050 ","End":"00:15.615","Text":"isobaric, where our pressure is constant,"},{"Start":"00:15.615 ","End":"00:20.340","Text":"and adiabatic where we don\u0027t have any heat transfer."},{"Start":"00:20.340 ","End":"00:26.790","Text":"We saw the equations that relate to each one of these processes."},{"Start":"00:26.790 ","End":"00:35.020","Text":"These processes are what every single engine uses in order to generate energy."},{"Start":"00:35.960 ","End":"00:40.310","Text":"Now we\u0027re going to learn 2 more terms."},{"Start":"00:40.310 ","End":"00:44.570","Text":"One of them is called the energy conversion efficiency,"},{"Start":"00:44.570 ","End":"00:48.780","Text":"and it\u0027s denoted by the Greek letter Eta."},{"Start":"00:49.490 ","End":"00:54.080","Text":"This energy conversion efficiency, or Eta,"},{"Start":"00:54.080 ","End":"01:02.760","Text":"is the ratio between the useful output of some engine and the input."},{"Start":"01:02.760 ","End":"01:06.635","Text":"If we have to input a lot of energy into"},{"Start":"01:06.635 ","End":"01:10.928","Text":"an engine and it\u0027s output of energy is very low,"},{"Start":"01:10.928 ","End":"01:14.615","Text":"the energy conversion efficiency will be low."},{"Start":"01:14.615 ","End":"01:18.005","Text":"Whereas if we have to input only a small amount of energy,"},{"Start":"01:18.005 ","End":"01:24.600","Text":"but we get a lot of energy out then the energy conversion efficiency will be high."},{"Start":"01:24.600 ","End":"01:31.715","Text":"The inputs and outputs maybe in terms of electrical power,"},{"Start":"01:31.715 ","End":"01:39.675","Text":"mechanical work, or chemical energy and of course, many other examples."},{"Start":"01:39.675 ","End":"01:41.920","Text":"The equation for Eta,"},{"Start":"01:41.920 ","End":"01:46.580","Text":"for energy conversion efficiency is given as"},{"Start":"01:46.580 ","End":"01:53.520","Text":"the work divided by the heat put in, so Q_in."},{"Start":"01:55.610 ","End":"02:00.050","Text":"What we can see here is that we have an equation"},{"Start":"02:00.050 ","End":"02:04.505","Text":"that is going to be some percentage or some decimal point."},{"Start":"02:04.505 ","End":"02:08.220","Text":"We can say that from a guess,"},{"Start":"02:08.220 ","End":"02:12.740","Text":"our value for Eta is going to be somewhere between 0 and 1,"},{"Start":"02:12.740 ","End":"02:18.475","Text":"where 0 is an incredibly inefficient engine,"},{"Start":"02:18.475 ","End":"02:23.290","Text":"where we had to put in a lot of energy in order to only get some energy out and"},{"Start":"02:23.290 ","End":"02:29.665","Text":"1 is an incredibly efficient engine where almost 0 energy was put in,"},{"Start":"02:29.665 ","End":"02:34.970","Text":"but a lot of energy was produced by the machine."},{"Start":"02:34.970 ","End":"02:39.110","Text":"Now a scientists called Carnot said that,"},{"Start":"02:39.110 ","End":"02:45.610","Text":"that doesn\u0027t really make sense and we can\u0027t really get a value between 0 and 1."},{"Start":"02:45.610 ","End":"02:49.840","Text":"Specifically, we\u0027ll never have a machine that has"},{"Start":"02:49.840 ","End":"02:54.895","Text":"an efficiency of 1 where for a little bit of energy put in,"},{"Start":"02:54.895 ","End":"02:58.375","Text":"we get a lot of energy out."},{"Start":"02:58.375 ","End":"03:01.190","Text":"Carnot worked on a machine which is called"},{"Start":"03:01.190 ","End":"03:08.990","Text":"the Carnot engine and that is considered the most efficient engine in the world."},{"Start":"03:08.990 ","End":"03:11.000","Text":"It\u0027s a theoretical engine."},{"Start":"03:11.000 ","End":"03:14.930","Text":"No engine can be as efficient as the Carnot engine."},{"Start":"03:14.930 ","End":"03:21.275","Text":"Even that engine is less efficient than the number 1."},{"Start":"03:21.275 ","End":"03:27.370","Text":"Let\u0027s first of all put in here that this is always going to be smaller than the value 1."},{"Start":"03:27.370 ","End":"03:31.640","Text":"Carnot worked on the energy conversion efficiency for"},{"Start":"03:31.640 ","End":"03:38.475","Text":"the maximum conversion efficiency possible and he came up with this value."},{"Start":"03:38.475 ","End":"03:46.370","Text":"It\u0027s equal to the hot temperature minus the cold temperature."},{"Start":"03:46.370 ","End":"03:48.905","Text":"This can be either in degrees"},{"Start":"03:48.905 ","End":"03:53.270","Text":"Celsius or in Kelvin because the jumps in Kelvin and Celsius,"},{"Start":"03:53.270 ","End":"03:54.980","Text":"the increments are the same,"},{"Start":"03:54.980 ","End":"03:58.775","Text":"divided by the hot temperature."},{"Start":"03:58.775 ","End":"04:04.370","Text":"This is the maximum efficiency that any machine can have."},{"Start":"04:04.370 ","End":"04:13.080","Text":"All machines have a slightly lower or much lower value of efficiency than this."},{"Start":"04:13.370 ","End":"04:16.835","Text":"When we\u0027re speaking about the efficiency of a machine,"},{"Start":"04:16.835 ","End":"04:18.440","Text":"we can use this equation,"},{"Start":"04:18.440 ","End":"04:24.833","Text":"but if we want to find the maximum efficiency of the given engine,"},{"Start":"04:24.833 ","End":"04:27.350","Text":"this is the equation that we can use and this will give us"},{"Start":"04:27.350 ","End":"04:31.355","Text":"a theoretical value for the maximum efficiency."},{"Start":"04:31.355 ","End":"04:35.600","Text":"Now, we will get a value and of course,"},{"Start":"04:35.600 ","End":"04:37.805","Text":"in practice, in reality,"},{"Start":"04:37.805 ","End":"04:42.110","Text":"the efficiency that we will measure will also be lower than"},{"Start":"04:42.110 ","End":"04:47.250","Text":"this theoretical maximum possible efficiency."},{"Start":"04:47.780 ","End":"04:52.280","Text":"Let\u0027s give a little example and then let\u0027s also see why"},{"Start":"04:52.280 ","End":"04:56.285","Text":"this equation for Carnot is important and makes sense."},{"Start":"04:56.285 ","End":"05:03.275","Text":"Let\u0027s say we want to find the theoretical maximum efficiency for some machine where it\u0027s"},{"Start":"05:03.275 ","End":"05:10.900","Text":"hot temperature is 600 kelvin and it\u0027s cold temperature is 300 kelvin."},{"Start":"05:10.900 ","End":"05:15.560","Text":"We can see 300 kelvin is approximately room temperature and we can see that there\u0027s"},{"Start":"05:15.560 ","End":"05:18.450","Text":"a very big temperature change or"},{"Start":"05:18.450 ","End":"05:21.920","Text":"temperature difference between the hot temperature"},{"Start":"05:21.920 ","End":"05:23.900","Text":"and the cold temperature of the machine."},{"Start":"05:23.900 ","End":"05:31.750","Text":"This large temperature change is what actually usually increases the efficiency."},{"Start":"05:31.750 ","End":"05:33.890","Text":"According to this equation,"},{"Start":"05:33.890 ","End":"05:35.480","Text":"we substitute in our values."},{"Start":"05:35.480 ","End":"05:38.839","Text":"What we\u0027ll get is that the maximum possible efficiency,"},{"Start":"05:38.839 ","End":"05:40.940","Text":"obviously a theoretical result,"},{"Start":"05:40.940 ","End":"05:44.190","Text":"is going to be equal to 1/2."},{"Start":"05:44.190 ","End":"05:48.080","Text":"This is the best that this engine can do theoretically."},{"Start":"05:48.080 ","End":"05:50.165","Text":"In practice, of course,"},{"Start":"05:50.165 ","End":"05:52.895","Text":"we won\u0027t even get to this efficiency as we know,"},{"Start":"05:52.895 ","End":"05:56.900","Text":"because we\u0027re going to be losing energy to other things,"},{"Start":"05:56.900 ","End":"06:01.074","Text":"such as to friction, for example."},{"Start":"06:01.074 ","End":"06:04.685","Text":"Why is this equation very, very important?"},{"Start":"06:04.685 ","End":"06:07.669","Text":"Up until Carnot came up with this equation,"},{"Start":"06:07.669 ","End":"06:10.100","Text":"scientists and engineers of the day thought that"},{"Start":"06:10.100 ","End":"06:13.220","Text":"their engines weren\u0027t efficient because of"},{"Start":"06:13.220 ","End":"06:17.360","Text":"some planning problem that they had"},{"Start":"06:17.360 ","End":"06:22.970","Text":"in either how the engine works or the material that the engine was made out of,"},{"Start":"06:22.970 ","End":"06:24.670","Text":"and so on and so forth."},{"Start":"06:24.670 ","End":"06:28.430","Text":"They thought that they could improve the efficiency of their engine by a lot."},{"Start":"06:28.430 ","End":"06:34.175","Text":"You can imagine how many man hours went into trying to improve this efficiency."},{"Start":"06:34.175 ","End":"06:36.200","Text":"When Carnot came up with this equation,"},{"Start":"06:36.200 ","End":"06:38.300","Text":"they found out that it\u0027s"},{"Start":"06:38.300 ","End":"06:42.500","Text":"some fundamental reasoning why they cannot improve the efficiency,"},{"Start":"06:42.500 ","End":"06:44.660","Text":"it\u0027s not some mistake that they\u0027re making,"},{"Start":"06:44.660 ","End":"06:49.160","Text":"but there\u0027s absolutely no way possible to improve this efficiency."},{"Start":"06:49.160 ","End":"06:55.585","Text":"This was very important development in this engine theory."},{"Start":"06:55.585 ","End":"07:01.350","Text":"What is the idea behind this equation?"},{"Start":"07:01.350 ","End":"07:03.680","Text":"When those heat in the system,"},{"Start":"07:03.680 ","End":"07:06.590","Text":"we know that heat travels around the system from one place to another,"},{"Start":"07:06.590 ","End":"07:09.260","Text":"heating different areas of our system up."},{"Start":"07:09.260 ","End":"07:12.890","Text":"Now, in this process of heating a different area up,"},{"Start":"07:12.890 ","End":"07:16.390","Text":"we\u0027re losing some of this heat energy."},{"Start":"07:16.390 ","End":"07:22.790","Text":"That is where this energy is going and there\u0027s no possible way to move heat around,"},{"Start":"07:22.790 ","End":"07:29.105","Text":"thereby heating up other things without losing some of the energy."},{"Start":"07:29.105 ","End":"07:32.840","Text":"That is where this equation comes from and that is also why we"},{"Start":"07:32.840 ","End":"07:38.420","Text":"can\u0027t get to an efficiency of 1 or of 100 percent."},{"Start":"07:38.420 ","End":"07:42.430","Text":"Okay, that\u0027s the end of this lesson."}],"ID":12359},{"Watched":false,"Name":"Stirling Engine","Duration":"7m 39s","ChapterTopicVideoID":11931,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this lesson,"},{"Start":"00:01.950 ","End":"00:04.560","Text":"we\u0027re going to be learning about the Stirling engine."},{"Start":"00:04.560 ","End":"00:07.110","Text":"Now, the Stirling engine, unlike other engines,"},{"Start":"00:07.110 ","End":"00:10.695","Text":"doesn\u0027t use any type of fuel such as gas or petrol,"},{"Start":"00:10.695 ","End":"00:12.195","Text":"and all it does,"},{"Start":"00:12.195 ","End":"00:13.740","Text":"this is how it gets its energy,"},{"Start":"00:13.740 ","End":"00:16.859","Text":"is it uses this change in temperature."},{"Start":"00:16.859 ","End":"00:21.120","Text":"So you can imagine that this type of engine will be very useful"},{"Start":"00:21.120 ","End":"00:25.670","Text":"to people that live in far out places that are difficult to reach,"},{"Start":"00:25.670 ","End":"00:28.140","Text":"because that means that you never have to"},{"Start":"00:28.140 ","End":"00:32.700","Text":"transport some kind of fuel in order to keep the engine running."},{"Start":"00:32.700 ","End":"00:36.765","Text":"If you have a group of researchers located somewhere in Alaska,"},{"Start":"00:36.765 ","End":"00:38.745","Text":"you could bring them the Stirling engine,"},{"Start":"00:38.745 ","End":"00:42.315","Text":"place it somewhere next to some hot geyser,"},{"Start":"00:42.315 ","End":"00:47.060","Text":"and then the temperature difference between the hot geyser and the snow or cold there"},{"Start":"00:47.060 ","End":"00:53.160","Text":"outside will allow the Stirling engine to create electricity."},{"Start":"00:53.260 ","End":"00:57.230","Text":"That\u0027s the major advantage of the Stirling engine,"},{"Start":"00:57.230 ","End":"01:00.335","Text":"that it can be used in fireplaces and doesn\u0027t need fuel,"},{"Start":"01:00.335 ","End":"01:05.090","Text":"and it utilizes an energy source that wouldn\u0027t otherwise be utilized,"},{"Start":"01:05.090 ","End":"01:07.040","Text":"so we don\u0027t use petrol,"},{"Start":"01:07.040 ","End":"01:09.860","Text":"but instead we\u0027re utilizing this change in temperature."},{"Start":"01:09.860 ","End":"01:14.075","Text":"However, the 2 major drawbacks of the Stirling engine is that 1,"},{"Start":"01:14.075 ","End":"01:15.830","Text":"the engine is extremely heavy,"},{"Start":"01:15.830 ","End":"01:20.340","Text":"and 2, the engine is extremely inefficient."},{"Start":"01:20.420 ","End":"01:25.745","Text":"The Stirling engine relies on 2 thermodynamic process."},{"Start":"01:25.745 ","End":"01:28.340","Text":"One is the process at"},{"Start":"01:28.340 ","End":"01:33.235","Text":"a constant volume and the other is a process at a constant pressure."},{"Start":"01:33.235 ","End":"01:36.970","Text":"The Stirling engine is an example of a heat engine,"},{"Start":"01:36.970 ","End":"01:40.835","Text":"and of course, every single heat engine is going to work slightly differently."},{"Start":"01:40.835 ","End":"01:42.845","Text":"But we\u0027re going to look at the Stirling engine"},{"Start":"01:42.845 ","End":"01:46.980","Text":"because it\u0027s a relatively common engine to study."},{"Start":"01:47.930 ","End":"01:52.155","Text":"This is how the Stirling engine works,"},{"Start":"01:52.155 ","End":"01:56.765","Text":"so what we can see over here is that here we have our hot source for instance,"},{"Start":"01:56.765 ","End":"01:59.075","Text":"the hot water in the geyser,"},{"Start":"01:59.075 ","End":"02:04.150","Text":"and here these bars are the cooling rods which are going to cool"},{"Start":"02:04.150 ","End":"02:10.500","Text":"the gas and these bars are attached to the ice or the snow in the area."},{"Start":"02:11.620 ","End":"02:16.070","Text":"Here we have most of our gas in this section over here."},{"Start":"02:16.070 ","End":"02:19.880","Text":"But then there\u0027s also a little bit of gas over here."},{"Start":"02:19.880 ","End":"02:23.615","Text":"The heat from the geyser is going to"},{"Start":"02:23.615 ","End":"02:27.623","Text":"heat up the gas that is located over here, therefore,"},{"Start":"02:27.623 ","End":"02:33.800","Text":"resulting in the gas over here expanding or creating a larger pressure,"},{"Start":"02:33.800 ","End":"02:43.920","Text":"which is then going to push this blue paddle over here in the right direction."},{"Start":"02:44.240 ","End":"02:46.560","Text":"Let\u0027s see that in action,"},{"Start":"02:46.560 ","End":"02:50.030","Text":"so the piston is being pushed in the right direction,"},{"Start":"02:50.030 ","End":"02:55.005","Text":"and we can see that the piston is attached to this wheel over here."},{"Start":"02:55.005 ","End":"02:58.955","Text":"We can see that by the piston moving to 1 side,"},{"Start":"02:58.955 ","End":"03:02.450","Text":"the wheel is going to be rotating."},{"Start":"03:02.450 ","End":"03:05.075","Text":"Now we can see over here in green,"},{"Start":"03:05.075 ","End":"03:08.045","Text":"we have another piston and it too,"},{"Start":"03:08.045 ","End":"03:10.565","Text":"this green piston over here."},{"Start":"03:10.565 ","End":"03:14.050","Text":"The green piston is also attached to the wheel."},{"Start":"03:14.050 ","End":"03:17.435","Text":"As the blue piston moves in the right direction,"},{"Start":"03:17.435 ","End":"03:21.920","Text":"we can see that the green piston is going to move in the leftwards direction,"},{"Start":"03:21.920 ","End":"03:26.700","Text":"and soon we\u0027re going to speak about the importance of this second piston."},{"Start":"03:26.990 ","End":"03:29.310","Text":"Now let\u0027s look at this in action."},{"Start":"03:29.310 ","End":"03:33.560","Text":"Again, we can see that the blue piston is moving in"},{"Start":"03:33.560 ","End":"03:38.880","Text":"the right direction and the green piston is moving in the leftwards direction."},{"Start":"03:39.440 ","End":"03:46.885","Text":"Now, we\u0027ve seen the first step of how the Stirling engine works."},{"Start":"03:46.885 ","End":"03:54.150","Text":"It\u0027s this idea of the expansion of the gas when it\u0027s located above the heat source."},{"Start":"03:54.230 ","End":"03:58.820","Text":"It\u0027s this section, this step of the gas expansion,"},{"Start":"03:58.820 ","End":"04:03.455","Text":"which is really what drives the whole Stirling engine to"},{"Start":"04:03.455 ","End":"04:08.345","Text":"work because we know that we\u0027re moving this wheel over here."},{"Start":"04:08.345 ","End":"04:14.660","Text":"The wheel is going to want to carry on to rotate and to spin around due to the inertia."},{"Start":"04:14.660 ","End":"04:18.875","Text":"Therefore, due to the engine,"},{"Start":"04:18.875 ","End":"04:21.515","Text":"even though it\u0027s providing energy to the wheel,"},{"Start":"04:21.515 ","End":"04:24.800","Text":"we\u0027re going to have to take energy away in order to stop"},{"Start":"04:24.800 ","End":"04:29.910","Text":"the wheel from spinning around in the same direction continuously."},{"Start":"04:29.960 ","End":"04:33.730","Text":"Let\u0027s imagine that my Stirling engine is running,"},{"Start":"04:33.730 ","End":"04:36.970","Text":"and from this gas expansion over here,"},{"Start":"04:36.970 ","End":"04:38.245","Text":"at this stage over here,"},{"Start":"04:38.245 ","End":"04:43.225","Text":"I managed to produce 100 joules of energy."},{"Start":"04:43.225 ","End":"04:46.000","Text":"This 100 joules of energy goes into the wheel,"},{"Start":"04:46.000 ","End":"04:50.890","Text":"but then the wheel has to exert 70 joules of energy in order to"},{"Start":"04:50.890 ","End":"04:56.200","Text":"keep the system over here with the 2 pistons running."},{"Start":"04:56.200 ","End":"05:02.680","Text":"That means that only 30 joules of energy is going to go into the wheel that will then"},{"Start":"05:02.680 ","End":"05:05.890","Text":"produce my electricity or my useful energy that I want"},{"Start":"05:05.890 ","End":"05:09.370","Text":"to take out from the Stirling engine."},{"Start":"05:09.370 ","End":"05:15.440","Text":"We can see that a lot of energy that I get from this heating up and expansion"},{"Start":"05:15.440 ","End":"05:22.380","Text":"of the gas is then going back into the engine in order to carry on running it."},{"Start":"05:22.720 ","End":"05:29.390","Text":"The first step was the gas over here was heated up and therefore it expanded,"},{"Start":"05:29.390 ","End":"05:32.585","Text":"and now as our machine carries on working,"},{"Start":"05:32.585 ","End":"05:37.130","Text":"we can see that the gas that was in this space over here that was hot,"},{"Start":"05:37.130 ","End":"05:43.220","Text":"is therefore going to be moved to this space over here so the gas is moved,"},{"Start":"05:43.220 ","End":"05:44.735","Text":"is pushed over here,"},{"Start":"05:44.735 ","End":"05:47.870","Text":"where here it is left to cool."},{"Start":"05:47.870 ","End":"05:51.545","Text":"The second stage is this cooling of the gas."},{"Start":"05:51.545 ","End":"05:53.990","Text":"Now what happens when the gas is cooling?"},{"Start":"05:53.990 ","End":"05:58.200","Text":"It means that its volume is going to decrease."},{"Start":"05:58.430 ","End":"06:04.460","Text":"Now the gas is cooled and now we have this green piston pushing the gas back to"},{"Start":"06:04.460 ","End":"06:10.355","Text":"this section over here so over here on top of the heat source,"},{"Start":"06:10.355 ","End":"06:15.185","Text":"where the gas is going to be heated up again and therefore expand,"},{"Start":"06:15.185 ","End":"06:19.170","Text":"and then the whole process is repeated again."},{"Start":"06:19.490 ","End":"06:22.365","Text":"Let\u0027s go over this again."},{"Start":"06:22.365 ","End":"06:24.875","Text":"Here we can see the engine working."},{"Start":"06:24.875 ","End":"06:29.555","Text":"So over here we have a little bit of gas which is heated up by the geyser."},{"Start":"06:29.555 ","End":"06:35.290","Text":"It expands, therefore causing the blue piston to move backwards."},{"Start":"06:35.290 ","End":"06:38.225","Text":"Then the hot gas which has expanded,"},{"Start":"06:38.225 ","End":"06:45.214","Text":"moves into this area over here next to the cooling blocks,"},{"Start":"06:45.214 ","End":"06:51.304","Text":"where it is then cooled and therefore the gas reduces its volume."},{"Start":"06:51.304 ","End":"06:53.585","Text":"Then we have this green piston,"},{"Start":"06:53.585 ","End":"06:56.810","Text":"which comes at it and pushes this cooled gas"},{"Start":"06:56.810 ","End":"07:01.610","Text":"back over here to the area above the hot section,"},{"Start":"07:01.610 ","End":"07:06.290","Text":"where again the gas heats up and therefore its volume increases,"},{"Start":"07:06.290 ","End":"07:09.830","Text":"and we carry on doing this over and over and over again,"},{"Start":"07:09.830 ","End":"07:14.970","Text":"and this is how the Stirling engine works and how it produces energy."},{"Start":"07:15.950 ","End":"07:20.720","Text":"This is the basic of how the Stirling engine"},{"Start":"07:20.720 ","End":"07:24.994","Text":"works so it\u0027s just working on the idea of gas,"},{"Start":"07:24.994 ","End":"07:27.125","Text":"which is heated and cooled,"},{"Start":"07:27.125 ","End":"07:30.620","Text":"therefore changing its volume and pumped up and back"},{"Start":"07:30.620 ","End":"07:36.395","Text":"down due to 2 pistons that are working in opposite directions."},{"Start":"07:36.395 ","End":"07:39.480","Text":"That\u0027s the end of this lesson."}],"ID":12360},{"Watched":false,"Name":"Exercise 4","Duration":"19m 9s","ChapterTopicVideoID":11918,"CourseChapterTopicPlaylistID":84746,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.130","Text":"Hello. In this lesson,"},{"Start":"00:02.130 ","End":"00:04.260","Text":"we\u0027re going to be answering a question,"},{"Start":"00:04.260 ","End":"00:07.515","Text":"which is going to take us a little bit of time."},{"Start":"00:07.515 ","End":"00:10.980","Text":"However, this is a very important question because this really"},{"Start":"00:10.980 ","End":"00:16.200","Text":"summarizes what we\u0027ve learned in this whole chapter of thermodynamics."},{"Start":"00:16.200 ","End":"00:19.755","Text":"The first thing that we\u0027re given"},{"Start":"00:19.755 ","End":"00:23.595","Text":"is we have this graph and we can see that we have point A over here,"},{"Start":"00:23.595 ","End":"00:25.230","Text":"point B over here,"},{"Start":"00:25.230 ","End":"00:26.955","Text":"and point C over here."},{"Start":"00:26.955 ","End":"00:30.735","Text":"Our question is, at each stage from A to B,"},{"Start":"00:30.735 ","End":"00:31.950","Text":"from B to C,"},{"Start":"00:31.950 ","End":"00:35.600","Text":"and from C to D, we\u0027re going to be asked something"},{"Start":"00:35.600 ","End":"00:39.735","Text":"about our heat or our change in heat in each stage."},{"Start":"00:39.735 ","End":"00:45.220","Text":"We\u0027re going to be asked about our internal energy or a change in internal energy,"},{"Start":"00:45.220 ","End":"00:50.600","Text":"and we\u0027re going to be asked about the work done by the system."},{"Start":"00:50.600 ","End":"01:00.160","Text":"These are the 3 questions that are usually asked given a graph of pressure versus volume."},{"Start":"01:00.160 ","End":"01:05.500","Text":"What do we have in front of us is our pressure-volume graph."},{"Start":"01:05.500 ","End":"01:07.910","Text":"We can see that we have these 3 stages."},{"Start":"01:07.910 ","End":"01:13.355","Text":"What\u0027s important to note in these 3 stages is the direction of our arrows."},{"Start":"01:13.355 ","End":"01:19.245","Text":"We can see that our system is working in this clockwise direction."},{"Start":"01:19.245 ","End":"01:21.690","Text":"That\u0027s an important point to notice."},{"Start":"01:21.690 ","End":"01:24.080","Text":"Then we can also see at each point A,"},{"Start":"01:24.080 ","End":"01:27.170","Text":"B, and C, what the temperature is."},{"Start":"01:27.170 ","End":"01:31.445","Text":"Now, what\u0027s important to note is that between point B and point C,"},{"Start":"01:31.445 ","End":"01:33.545","Text":"the temperature is different."},{"Start":"01:33.545 ","End":"01:38.520","Text":"We can see that this curve represents which process?"},{"Start":"01:38.520 ","End":"01:41.650","Text":"It represents the adiabatic process,"},{"Start":"01:41.650 ","End":"01:47.920","Text":"not isothermal because the temperatures are different."},{"Start":"01:48.530 ","End":"01:53.340","Text":"Let\u0027s look at our process from A to B."},{"Start":"01:53.340 ","End":"01:57.830","Text":"First of all, we can see that we\u0027re at a constant volume."},{"Start":"01:57.830 ","End":"02:01.520","Text":"We\u0027re dealing with an isochoric process,"},{"Start":"02:01.520 ","End":"02:05.250","Text":"constant volume, and what we can see is on our graph,"},{"Start":"02:05.250 ","End":"02:06.290","Text":"we\u0027re given the temperature,"},{"Start":"02:06.290 ","End":"02:08.300","Text":"so we can see that there\u0027s a temperature change,"},{"Start":"02:08.300 ","End":"02:10.879","Text":"and we\u0027re going from a colder temperature,"},{"Start":"02:10.879 ","End":"02:14.240","Text":"which is increasing to a hotter temperature."},{"Start":"02:14.240 ","End":"02:16.195","Text":"This is what we can see."},{"Start":"02:16.195 ","End":"02:19.310","Text":"What we have really is given in this diagram."},{"Start":"02:19.310 ","End":"02:24.725","Text":"We have some container with a gas inside and the gas is being heated."},{"Start":"02:24.725 ","End":"02:27.245","Text":"Because the gas is contained in this container,"},{"Start":"02:27.245 ","End":"02:29.450","Text":"the volume is remaining the same."},{"Start":"02:29.450 ","End":"02:31.805","Text":"However, due to the temperature change"},{"Start":"02:31.805 ","End":"02:35.035","Text":"which we know is connected to the internal energy,"},{"Start":"02:35.035 ","End":"02:36.560","Text":"so the energy is going up,"},{"Start":"02:36.560 ","End":"02:38.450","Text":"which means that all of the gas particles are"},{"Start":"02:38.450 ","End":"02:41.810","Text":"knocking against the walls of the container,"},{"Start":"02:41.810 ","End":"02:45.480","Text":"which is therefore increasing the pressure."},{"Start":"02:45.480 ","End":"02:49.080","Text":"That is exactly what we can see over here."},{"Start":"02:49.760 ","End":"02:54.395","Text":"That\u0027s some information of our process from A to B."},{"Start":"02:54.395 ","End":"02:56.270","Text":"What we can see is that one,"},{"Start":"02:56.270 ","End":"02:58.250","Text":"we\u0027re dealing with an isochoric process,"},{"Start":"02:58.250 ","End":"03:00.635","Text":"which automatically we should remember,"},{"Start":"03:00.635 ","End":"03:05.300","Text":"means that the work being done is equal to 0."},{"Start":"03:05.300 ","End":"03:08.840","Text":"Another way if we don\u0027t remember that this is called"},{"Start":"03:08.840 ","End":"03:13.715","Text":"an isochoric process or automatically remember that that means that work is equal to 0,"},{"Start":"03:13.715 ","End":"03:17.780","Text":"we remember that the work is equal to the area under the graph."},{"Start":"03:17.780 ","End":"03:20.805","Text":"That means that that\u0027s here,"},{"Start":"03:20.805 ","End":"03:24.100","Text":"which is just zero work."},{"Start":"03:24.320 ","End":"03:27.665","Text":"Now, let\u0027s see what else we can tell."},{"Start":"03:27.665 ","End":"03:30.605","Text":"We know that our internal energy,"},{"Start":"03:30.605 ","End":"03:33.860","Text":"or a change in internal energy Delta U,"},{"Start":"03:33.860 ","End":"03:37.100","Text":"is given by an equation of Q,"},{"Start":"03:37.100 ","End":"03:41.950","Text":"the heat minus the work."},{"Start":"03:41.950 ","End":"03:44.970","Text":"Where we know that Q, the heat,"},{"Start":"03:44.970 ","End":"03:48.015","Text":"is the heat added to the system,"},{"Start":"03:48.015 ","End":"03:51.815","Text":"and the work is the work done by the system."},{"Start":"03:51.815 ","End":"03:56.875","Text":"Now, because we know that our work over here in this isochoric process is equal to 0,"},{"Start":"03:56.875 ","End":"03:58.455","Text":"so we can cross that off."},{"Start":"03:58.455 ","End":"04:08.940","Text":"We know that our change in internal energy is due to the heat given to the system."},{"Start":"04:10.100 ","End":"04:13.135","Text":"What is our equation for Q,"},{"Start":"04:13.135 ","End":"04:16.220","Text":"which is also going to be our equation for Delta U?"},{"Start":"04:16.220 ","End":"04:22.355","Text":"Then we\u0027ve answered all of what we needed to for this section AB."},{"Start":"04:22.355 ","End":"04:31.585","Text":"We know that another equation for our Delta U is going to be simply equal to n,"},{"Start":"04:31.585 ","End":"04:33.250","Text":"the number of moles,"},{"Start":"04:33.250 ","End":"04:35.425","Text":"multiplied by C_v,"},{"Start":"04:35.425 ","End":"04:39.430","Text":"which is the heat capacity for constant volume."},{"Start":"04:39.430 ","End":"04:45.910","Text":"Which as we know, because the gases in some container and using an isochoric process,"},{"Start":"04:45.910 ","End":"04:48.588","Text":"so we know the volume is constant,"},{"Start":"04:48.588 ","End":"04:55.780","Text":"and this is multiplied by Delta T. Let\u0027s imagine that we\u0027re using 1 mole,"},{"Start":"04:55.780 ","End":"05:00.070","Text":"so we\u0027ll just substitute for this lowercase n, the number 1."},{"Start":"05:00.070 ","End":"05:01.540","Text":"If we were using 10 moles,"},{"Start":"05:01.540 ","End":"05:04.060","Text":"we\u0027d substitute in here the number 10."},{"Start":"05:04.060 ","End":"05:05.380","Text":"But just for ease of calculation,"},{"Start":"05:05.380 ","End":"05:09.655","Text":"let\u0027s say we have 1 mole of gas inside this container."},{"Start":"05:09.655 ","End":"05:13.385","Text":"What is the heat capacity at constant volume?"},{"Start":"05:13.385 ","End":"05:20.000","Text":"C_v at constant volume is simply equal to 3 divided by 2."},{"Start":"05:20.000 ","End":"05:22.580","Text":"If we\u0027re not dealing with a constant volume,"},{"Start":"05:22.580 ","End":"05:26.165","Text":"the coefficient is 5 divided by 2 multiplied by R,"},{"Start":"05:26.165 ","End":"05:31.030","Text":"which is the universal gas constant and is equal to 8.31."},{"Start":"05:31.030 ","End":"05:34.455","Text":"Now, let\u0027s start substituting in our numbers."},{"Start":"05:34.455 ","End":"05:38.900","Text":"The number of moles that we\u0027re using is 1 mole, let\u0027s say."},{"Start":"05:38.900 ","End":"05:43.065","Text":"We have, let\u0027s move this on over to here,"},{"Start":"05:43.065 ","End":"05:46.245","Text":"we\u0027ll have 1 multiplied by C_v."},{"Start":"05:46.245 ","End":"05:53.700","Text":"Which we said is 3 divided by 2R multiplied by, where are R?"},{"Start":"05:53.700 ","End":"05:55.485","Text":"Let\u0027s write that in actually,"},{"Start":"05:55.485 ","End":"06:02.430","Text":"where R is equal to 8.31 multiplied by our Delta"},{"Start":"06:02.430 ","End":"06:09.470","Text":"T. Our final temperature is 600 degrees Kelvin minus our initial temperature,"},{"Start":"06:09.470 ","End":"06:12.555","Text":"which is 300 degrees Kelvin."},{"Start":"06:12.555 ","End":"06:16.235","Text":"Then, once we plug all of this into a calculator,"},{"Start":"06:16.235 ","End":"06:18.800","Text":"we get that the change in internal energy,"},{"Start":"06:18.800 ","End":"06:23.345","Text":"which is also equal to the heat put into the system,"},{"Start":"06:23.345 ","End":"06:30.660","Text":"is equal to 3,740 joules."},{"Start":"06:30.880 ","End":"06:36.365","Text":"Now, something to note is that our value is a positive value."},{"Start":"06:36.365 ","End":"06:44.015","Text":"Which means that our system between points A and B,"},{"Start":"06:44.015 ","End":"06:47.585","Text":"the energy of the system is increasing,"},{"Start":"06:47.585 ","End":"06:52.065","Text":"and also the heat put into the system is also positive."},{"Start":"06:52.065 ","End":"06:56.810","Text":"Heat was put into the system rather than removed from the system."},{"Start":"06:56.810 ","End":"06:59.790","Text":"That we can see from the positive, which makes sense."},{"Start":"06:59.790 ","End":"07:02.225","Text":"We know that if heat is put into a system,"},{"Start":"07:02.225 ","End":"07:05.875","Text":"the internal energy will go up."},{"Start":"07:05.875 ","End":"07:10.010","Text":"Now, let\u0027s look at stage B to C. We"},{"Start":"07:10.010 ","End":"07:13.700","Text":"already saw that it looks a bit like an isothermal curve."},{"Start":"07:13.700 ","End":"07:17.600","Text":"However, we can see that there is a temperature change."},{"Start":"07:17.600 ","End":"07:19.414","Text":"Which means that it\u0027s not isothermal,"},{"Start":"07:19.414 ","End":"07:21.380","Text":"but rather, adiabatic."},{"Start":"07:21.380 ","End":"07:26.845","Text":"Let\u0027s write adiabatic."},{"Start":"07:26.845 ","End":"07:30.680","Text":"The second that we notice that something is adiabatic,"},{"Start":"07:30.680 ","End":"07:34.985","Text":"that means that straightaway our Q is equal to 0."},{"Start":"07:34.985 ","End":"07:39.380","Text":"This is why you always have to remember when dealing with an adiabatic process,"},{"Start":"07:39.380 ","End":"07:44.640","Text":"the heat put into the system or taken out of the system is equal to 0."},{"Start":"07:45.050 ","End":"07:50.405","Text":"Now, our next equation is for our energy."},{"Start":"07:50.405 ","End":"07:55.685","Text":"We know that our change in internal energy is equal to, save over here,"},{"Start":"07:55.685 ","End":"08:02.210","Text":"Q minus, so the heat put in to the system minus the work done by the system."},{"Start":"08:02.210 ","End":"08:04.250","Text":"We know that the heat put into the system because"},{"Start":"08:04.250 ","End":"08:06.740","Text":"it\u0027s an adiabatic process is equal to 0."},{"Start":"08:06.740 ","End":"08:11.600","Text":"We can see that the change in energy of the system between points B and"},{"Start":"08:11.600 ","End":"08:18.460","Text":"C is equal to simply negative of the work done."},{"Start":"08:19.610 ","End":"08:24.260","Text":"Let\u0027s write out our equation again for Delta U."},{"Start":"08:24.260 ","End":"08:26.060","Text":"It\u0027s just this, and again,"},{"Start":"08:26.060 ","End":"08:27.875","Text":"we\u0027re using 1 mole of gas,"},{"Start":"08:27.875 ","End":"08:30.905","Text":"so we\u0027ll have 1 multiplied by our C_v,"},{"Start":"08:30.905 ","End":"08:34.610","Text":"which is 3 divided by 2 multiplied by R,"},{"Start":"08:34.610 ","End":"08:38.025","Text":"which is equal to 8.31,"},{"Start":"08:38.025 ","End":"08:41.590","Text":"and this is multiplied by the change in temperature."},{"Start":"08:41.590 ","End":"08:48.635","Text":"Our final temperature is 455 degrees Kelvin minus our initial temperature,"},{"Start":"08:48.635 ","End":"08:54.830","Text":"which is 600 degrees Kelvin using the exact same equation."},{"Start":"08:54.830 ","End":"08:59.180","Text":"Then we\u0027ll get that our change in energy is simply equal"},{"Start":"08:59.180 ","End":"09:07.350","Text":"to negative 1,807 joules."},{"Start":"09:07.660 ","End":"09:12.275","Text":"We can see that our system has lost energy."},{"Start":"09:12.275 ","End":"09:17.575","Text":"We\u0027ve lost the internal energy and we know that Delta U is equal to negative W,"},{"Start":"09:17.575 ","End":"09:19.360","Text":"the negative of the work."},{"Start":"09:19.360 ","End":"09:22.475","Text":"Now, because here we have a negative and here we have a negative,"},{"Start":"09:22.475 ","End":"09:24.370","Text":"they can cancel out,"},{"Start":"09:24.370 ","End":"09:31.250","Text":"and so we\u0027ll get therefore that the work done by the system is equal to 1,807."},{"Start":"09:33.930 ","End":"09:38.800","Text":"This is the information that we need for our adiabatic process from B to"},{"Start":"09:38.800 ","End":"09:43.525","Text":"C. Let\u0027s just talk about what is happening in this adiabatic process."},{"Start":"09:43.525 ","End":"09:46.900","Text":"A gas was put into this container,"},{"Start":"09:46.900 ","End":"09:53.170","Text":"and in this process between B and C we can see that our volume is increasing,"},{"Start":"09:53.170 ","End":"09:55.870","Text":"so our container has gotten bigger."},{"Start":"09:55.870 ","End":"09:58.090","Text":"Because our container has gotten bigger but"},{"Start":"09:58.090 ","End":"10:00.685","Text":"the amount of gas inside has stayed the same,"},{"Start":"10:00.685 ","End":"10:03.535","Text":"so our pressure has decreased."},{"Start":"10:03.535 ","End":"10:07.990","Text":"That\u0027s what we can see here and also because we know that it\u0027s"},{"Start":"10:07.990 ","End":"10:14.176","Text":"an adiabatic process that the temperature is also decreasing."},{"Start":"10:14.176 ","End":"10:16.000","Text":"This is really what happens;"},{"Start":"10:16.000 ","End":"10:20.605","Text":"we spoke about this a few lessons ago in the lesson about the adiabatic process,"},{"Start":"10:20.605 ","End":"10:26.230","Text":"is that when a gas is put under high pressure and then"},{"Start":"10:26.230 ","End":"10:32.245","Text":"it exits that area where it was in high pressure to an area of lower pressure,"},{"Start":"10:32.245 ","End":"10:35.545","Text":"you will feel that the gas has cooled down."},{"Start":"10:35.545 ","End":"10:37.210","Text":"The gas actually it does cool down,"},{"Start":"10:37.210 ","End":"10:39.370","Text":"its temperature decreases,"},{"Start":"10:39.370 ","End":"10:41.290","Text":"and that\u0027s exactly what we can see."},{"Start":"10:41.290 ","End":"10:46.850","Text":"We start from a higher temperature and its temperature then decreases when it gets here."},{"Start":"10:46.890 ","End":"10:54.940","Text":"Now let\u0027s look on our last process which is from a point C to point A."},{"Start":"10:54.940 ","End":"11:00.627","Text":"What we can see over here is that we\u0027re dealing with a constant pressure."},{"Start":"11:00.627 ","End":"11:04.670","Text":"This is the isobaric process."},{"Start":"11:06.270 ","End":"11:10.597","Text":"What we can see is that in this isobaric process"},{"Start":"11:10.597 ","End":"11:16.990","Text":"our volume is decreasing but we\u0027re keeping a constant pressure,"},{"Start":"11:16.990 ","End":"11:21.775","Text":"and we can also see that we have a change in temperature over here as well."},{"Start":"11:21.775 ","End":"11:29.185","Text":"The first thing that we want to do is we want to find the change in the internal energy."},{"Start":"11:29.185 ","End":"11:33.078","Text":"Now we know that when we\u0027re dealing with a cyclic process"},{"Start":"11:33.078 ","End":"11:37.467","Text":"once we are finishing the cycle and we get back to our starting point;"},{"Start":"11:37.467 ","End":"11:41.350","Text":"so remember we started from point A and we went up to B and"},{"Start":"11:41.350 ","End":"11:45.420","Text":"now we\u0027re going from C back to our starting point,"},{"Start":"11:45.420 ","End":"11:48.850","Text":"we know that our system can\u0027t increase in it energy,"},{"Start":"11:49.010 ","End":"11:52.830","Text":"can\u0027t collect energy from its cycle otherwise this will"},{"Start":"11:52.830 ","End":"11:56.220","Text":"go to infinity and that just isn\u0027t a physical possibility."},{"Start":"11:56.220 ","End":"12:05.950","Text":"We know that the energy from point C to A has to be the same as the energy"},{"Start":"12:05.950 ","End":"12:10.180","Text":"or minus the energy of from A to B plus from B to"},{"Start":"12:10.180 ","End":"12:15.715","Text":"C. Then when we add up the energy from A to B plus the energy change from B to C,"},{"Start":"12:15.715 ","End":"12:18.085","Text":"plus the energy change from C to A,"},{"Start":"12:18.085 ","End":"12:20.455","Text":"we\u0027re meant to get 0."},{"Start":"12:20.455 ","End":"12:25.195","Text":"That means that the energy change from C to A"},{"Start":"12:25.195 ","End":"12:30.415","Text":"is going to be negative the energy from A to C."},{"Start":"12:30.415 ","End":"12:37.705","Text":"That\u0027s going to amount negative for our energy change from A to B"},{"Start":"12:37.705 ","End":"12:45.325","Text":"plus our energy change from B to C. These are values that we\u0027ve worked out,"},{"Start":"12:45.325 ","End":"12:52.630","Text":"so therefore we\u0027ll get that our energy change from C to A if we plug in these values,"},{"Start":"12:52.630 ","End":"13:00.610","Text":"is going to simply be equal to negative 1,932 Joules."},{"Start":"13:00.610 ","End":"13:03.162","Text":"Once you plug this into the calculator,"},{"Start":"13:03.162 ","End":"13:07.555","Text":"once you plug in negative 1,932,"},{"Start":"13:07.555 ","End":"13:14.110","Text":"negative 1,807 plus 3,740,"},{"Start":"13:14.110 ","End":"13:19.940","Text":"you should get a total change in energy of 0."},{"Start":"13:19.950 ","End":"13:27.655","Text":"Now what we want to find out is what is our change in heat?"},{"Start":"13:27.655 ","End":"13:31.930","Text":"What we want to find out is what is our Delta Q?"},{"Start":"13:31.930 ","End":"13:36.640","Text":"I\u0027m going to use the exact same equation that we used before."},{"Start":"13:36.640 ","End":"13:42.070","Text":"My change in internal energy is equal to the number of moles multiplied by"},{"Start":"13:42.070 ","End":"13:48.415","Text":"my heat capacity except now instead of using the heat capacity for a constant volume,"},{"Start":"13:48.415 ","End":"13:52.750","Text":"I\u0027m using the heat capacity for a constant pressure because I can see on the graph that"},{"Start":"13:52.750 ","End":"13:55.360","Text":"my pressure is remaining the same and then"},{"Start":"13:55.360 ","End":"13:58.390","Text":"again multiplied by the temperature difference."},{"Start":"13:58.390 ","End":"14:00.460","Text":"Again, I\u0027m using 1 mole."},{"Start":"14:00.460 ","End":"14:05.380","Text":"Now, my heat capacity at constant pressure is simply 5 divided"},{"Start":"14:05.380 ","End":"14:10.960","Text":"by 2 multiplied by R where R is equal to 8.31."},{"Start":"14:10.960 ","End":"14:14.455","Text":"Let\u0027s plug that in. Multiplied by 5 divided by 2,"},{"Start":"14:14.455 ","End":"14:19.090","Text":"multiplied by 8.31 multiplied by the temperature change."},{"Start":"14:19.090 ","End":"14:23.005","Text":"My final temperature is 300 degrees Kelvin,"},{"Start":"14:23.005 ","End":"14:32.515","Text":"and my initial temperature over here was 455 degrees Kelvin."},{"Start":"14:32.515 ","End":"14:37.000","Text":"Then, once we plug this into the calculator,"},{"Start":"14:37.000 ","End":"14:46.717","Text":"we\u0027ll get that Delta Q is equal to negative 3,220 Joules."},{"Start":"14:46.717 ","End":"14:55.185","Text":"Now we know that our equation Delta U is equal to Q minus W,"},{"Start":"14:55.185 ","End":"14:57.915","Text":"and we have our Delta U and we have our Q."},{"Start":"14:57.915 ","End":"15:00.555","Text":"What do we want to find out is what is our W?"},{"Start":"15:00.555 ","End":"15:02.250","Text":"We\u0027re going to isolate our W,"},{"Start":"15:02.250 ","End":"15:08.050","Text":"and it\u0027s simply equal to Q minus Delta U."},{"Start":"15:08.050 ","End":"15:11.470","Text":"When we plug in our value for Q which is negative"},{"Start":"15:11.470 ","End":"15:15.657","Text":"3,220 Joules and plug in our value for Delta U"},{"Start":"15:15.657 ","End":"15:18.719","Text":"which is negative 1,932 and"},{"Start":"15:18.719 ","End":"15:22.987","Text":"notice here there\u0027s a minus sign and here there\u0027s a minus sign so they cancel out,"},{"Start":"15:22.987 ","End":"15:32.510","Text":"we\u0027ll get that our work is equal to negative 1,288 Joules."},{"Start":"15:33.690 ","End":"15:41.185","Text":"Now we have all the information that we wanted for each of the 3 stages."},{"Start":"15:41.185 ","End":"15:48.625","Text":"Now the question that they can ask is how much energy is generated or"},{"Start":"15:48.625 ","End":"15:57.830","Text":"how much energy is secreted by the system in 1 cycle?"},{"Start":"15:58.140 ","End":"16:01.960","Text":"Let\u0027s scroll down. What we want to do is we want to see"},{"Start":"16:01.960 ","End":"16:07.235","Text":"how much energy was generated or how much work was done by the system,"},{"Start":"16:07.235 ","End":"16:09.940","Text":"and this is very important when we\u0027re dealing with"},{"Start":"16:09.940 ","End":"16:13.810","Text":"things such as the efficiency of the engine."},{"Start":"16:13.810 ","End":"16:17.305","Text":"As we know, the total work done is simply going to be"},{"Start":"16:17.305 ","End":"16:22.960","Text":"this area inside or contained within the lines of the graph."},{"Start":"16:22.960 ","End":"16:28.195","Text":"How do we do that? We know that the area under the graph is simply equal to the work,"},{"Start":"16:28.195 ","End":"16:33.010","Text":"so all we have to do is we have to sum up all of our values for work."},{"Start":"16:33.010 ","End":"16:41.950","Text":"Then, what we\u0027ll get is because here we have some value for work we\u0027re getting work."},{"Start":"16:41.950 ","End":"16:43.840","Text":"But then in this stage,"},{"Start":"16:43.840 ","End":"16:48.130","Text":"we\u0027re minusing all of the work in this empty space under this line for"},{"Start":"16:48.130 ","End":"16:50.740","Text":"the isobaric process and that will leave us with"},{"Start":"16:50.740 ","End":"16:54.640","Text":"the work done by the system which is the area inside the graph."},{"Start":"16:54.640 ","End":"17:00.640","Text":"The total work done by the system or the total energy is simply"},{"Start":"17:00.640 ","End":"17:06.175","Text":"equal to the sum of the work done by each process."},{"Start":"17:06.175 ","End":"17:07.900","Text":"We have in the first process 0,"},{"Start":"17:07.900 ","End":"17:13.105","Text":"in the second process 1,807,"},{"Start":"17:13.105 ","End":"17:18.700","Text":"and then the third process negative 1,288."},{"Start":"17:18.700 ","End":"17:24.610","Text":"The total work therefore is simply going to be"},{"Start":"17:24.610 ","End":"17:30.890","Text":"equal to 519 Joules."},{"Start":"17:31.560 ","End":"17:34.960","Text":"Another question that can be asked is,"},{"Start":"17:34.960 ","End":"17:40.690","Text":"how much heat is needed in order to run the system or to run this engine?"},{"Start":"17:40.690 ","End":"17:42.520","Text":"What are we doing?"},{"Start":"17:42.520 ","End":"17:43.570","Text":"How much heat?"},{"Start":"17:43.570 ","End":"17:47.545","Text":"That means that we want to know the total Q,"},{"Start":"17:47.545 ","End":"17:51.895","Text":"so all we\u0027re going to do is we\u0027re going to add up all of our Qs."},{"Start":"17:51.895 ","End":"17:54.220","Text":"Here, in the first process,"},{"Start":"17:54.220 ","End":"17:58.645","Text":"we have 3,740 Joules."},{"Start":"17:58.645 ","End":"18:00.483","Text":"In our second process,"},{"Start":"18:00.483 ","End":"18:02.995","Text":"we have 0 because it\u0027s adiabatic,"},{"Start":"18:02.995 ","End":"18:12.355","Text":"and in our third process our Q is equal to negative 3,220."},{"Start":"18:12.355 ","End":"18:15.055","Text":"Then when we plug this into the calculator,"},{"Start":"18:15.055 ","End":"18:20.640","Text":"we\u0027ll get the total heat put into the system is"},{"Start":"18:20.640 ","End":"18:27.390","Text":"equal to 520 Joules."},{"Start":"18:28.260 ","End":"18:33.460","Text":"We can see that we\u0027ve got around about the same number because we"},{"Start":"18:33.460 ","End":"18:39.334","Text":"rounded a value for the universal gas constant and so on."},{"Start":"18:39.334 ","End":"18:42.670","Text":"So we can see that this is around about the same answer."},{"Start":"18:42.670 ","End":"18:45.490","Text":"What we can see that in 1 cycle of"},{"Start":"18:45.490 ","End":"18:49.945","Text":"this engine which is working on thermodynamic processes,"},{"Start":"18:49.945 ","End":"18:57.580","Text":"we can see that our system did 519 Joules of work and it"},{"Start":"18:57.580 ","End":"19:00.870","Text":"took approximately the same amount of"},{"Start":"19:00.870 ","End":"19:06.825","Text":"Joules of heat in order to run the system in the first place."},{"Start":"19:06.825 ","End":"19:10.030","Text":"That\u0027s the end of this lesson."}],"ID":12361}],"Thumbnail":null,"ID":84746}]

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