Basic Concepts
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Heat
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Work
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First Law of Thermodynamics
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Enthalpy
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- Definition of Enthalpy
- Enthalpy Change in a Chemical Reaction
- Exercise 1
- Exercise 2
- Enthalpy of Vaporization and Fusion
- Exercise 3
- Standard States and Standard Enthalpy Changes
- Hess's Law
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Exercise 8
- Standard Enthalpy of Formation
- Standard Enthalpy of Reaction
- Exercise 9
- Exercise 10
- Exercise 11
- Exercise 12

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[{"Name":"Basic Concepts","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Systems and their Surroundings","Duration":"4m 8s","ChapterTopicVideoID":18668,"CourseChapterTopicPlaylistID":81298,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/18668.jpeg","UploadDate":"2019-04-14T09:57:26.3430000","DurationForVideoObject":"PT4M8S","Description":null,"MetaTitle":"Systems and their Surroundings: Video + Workbook | Proprep","MetaDescription":"Thermochemistry - Basic Concepts. 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/general-chemistry/thermochemistry/basic-concepts/vid19392","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.640","Text":"In this video, we will learn about some basic terminology used in thermochemistry."},{"Start":"00:05.640 ","End":"00:09.285","Text":"The first question is, what is thermochemistry?"},{"Start":"00:09.285 ","End":"00:14.505","Text":"Thermochemistry is a branch of chemistry concerned with the quantities of heat."},{"Start":"00:14.505 ","End":"00:19.890","Text":"Heat, thermo, you know it\u0027s heat absorbed or produced in"},{"Start":"00:19.890 ","End":"00:23.710","Text":"chemical reactions or physical changes of state so"},{"Start":"00:23.710 ","End":"00:28.100","Text":"either in chemical reactions or physical changes."},{"Start":"00:28.100 ","End":"00:32.285","Text":"It\u0027s part of the branch of physical science called thermodynamics,"},{"Start":"00:32.285 ","End":"00:34.370","Text":"which you\u0027ve no doubt heard about."},{"Start":"00:34.370 ","End":"00:36.485","Text":"So what\u0027s thermodynamics?"},{"Start":"00:36.485 ","End":"00:41.600","Text":"Thermodynamics is a branch of physical science, chemistry, physics, engineering,"},{"Start":"00:41.600 ","End":"00:48.335","Text":"biology that deals with heat and temperature and their relation to energy and work."},{"Start":"00:48.335 ","End":"00:51.515","Text":"Let\u0027s take 2 examples of thermochemistry."},{"Start":"00:51.515 ","End":"00:54.275","Text":"The first one is that, say, a base,"},{"Start":"00:54.275 ","End":"00:58.715","Text":"sodium hydroxide, reacts with an acid, like hydrochloric acid."},{"Start":"00:58.715 ","End":"01:03.175","Text":"Here\u0027s the reaction. HCl plus NaOH,"},{"Start":"01:03.175 ","End":"01:05.330","Text":"hydrochloric acid plus sodium hydroxide,"},{"Start":"01:05.330 ","End":"01:07.280","Text":"react to give us a salt,"},{"Start":"01:07.280 ","End":"01:08.975","Text":"sodium chloride, and water."},{"Start":"01:08.975 ","End":"01:12.425","Text":"We met this when we talked about acids and bases."},{"Start":"01:12.425 ","End":"01:15.290","Text":"Now, not only does the chemical reaction occur,"},{"Start":"01:15.290 ","End":"01:17.060","Text":"but heat is produced,"},{"Start":"01:17.060 ","End":"01:20.235","Text":"and we call this an exothermic reaction."},{"Start":"01:20.235 ","End":"01:22.285","Text":"Exo is outwards,"},{"Start":"01:22.285 ","End":"01:29.600","Text":"and thermic means heat so heat is released out of the system, exothermic reaction."},{"Start":"01:29.600 ","End":"01:32.030","Text":"We can also have a physical process."},{"Start":"01:32.030 ","End":"01:35.585","Text":"For example, when alcohol is rubbed on your skin,"},{"Start":"01:35.585 ","End":"01:39.185","Text":"as in before an injection, it evaporates."},{"Start":"01:39.185 ","End":"01:43.910","Text":"What is evaporating, changes in state from liquid to gas."},{"Start":"01:43.910 ","End":"01:47.180","Text":"Then it absorbs heat from your skin because it"},{"Start":"01:47.180 ","End":"01:50.720","Text":"requires heat to change from a liquid to a gas,"},{"Start":"01:50.720 ","End":"01:54.065","Text":"and then your skin feels colder."},{"Start":"01:54.065 ","End":"01:57.320","Text":"So this is another example of thermochemistry."},{"Start":"01:57.320 ","End":"01:59.795","Text":"Now, we need a few more definitions."},{"Start":"01:59.795 ","End":"02:01.565","Text":"What is a system?"},{"Start":"02:01.565 ","End":"02:04.955","Text":"A system is a part of the universe chosen for study."},{"Start":"02:04.955 ","End":"02:06.515","Text":"It can be very small,"},{"Start":"02:06.515 ","End":"02:08.260","Text":"or it can be very large."},{"Start":"02:08.260 ","End":"02:10.160","Text":"What are surroundings?"},{"Start":"02:10.160 ","End":"02:14.750","Text":"Surrounding is everything in the universe apart from the system being studied."},{"Start":"02:14.750 ","End":"02:20.270","Text":"We\u0027re dividing the universe into a system and surroundings."},{"Start":"02:20.270 ","End":"02:23.700","Text":"We\u0027re going to talk about 3 systems,"},{"Start":"02:23.700 ","End":"02:25.520","Text":"open, closed,"},{"Start":"02:25.520 ","End":"02:28.700","Text":"and isolated systems. Here\u0027s a picture."},{"Start":"02:28.700 ","End":"02:31.355","Text":"In this picture, we have an open beaker,"},{"Start":"02:31.355 ","End":"02:32.660","Text":"a closed flask,"},{"Start":"02:32.660 ","End":"02:35.075","Text":"it\u0027s supposed to have a stopper at the top,"},{"Start":"02:35.075 ","End":"02:37.159","Text":"and a vacuum flask."},{"Start":"02:37.159 ","End":"02:39.830","Text":"These are supposed to illustrate an open,"},{"Start":"02:39.830 ","End":"02:42.485","Text":"closed, and isolated systems."},{"Start":"02:42.485 ","End":"02:43.850","Text":"Let\u0027s take the first one."},{"Start":"02:43.850 ","End":"02:45.720","Text":"We have a liquid inside."},{"Start":"02:45.720 ","End":"02:48.380","Text":"Supposing the liquid is hot,"},{"Start":"02:48.380 ","End":"02:50.188","Text":"then energy will escape."},{"Start":"02:50.188 ","End":"02:51.590","Text":"Heat will escape."},{"Start":"02:51.590 ","End":"02:54.230","Text":"Let\u0027s say heat."},{"Start":"02:54.230 ","End":"02:57.020","Text":"Because it\u0027s open at the top,"},{"Start":"02:57.020 ","End":"03:00.060","Text":"matter will also escape."},{"Start":"03:00.140 ","End":"03:03.465","Text":"Liquid will be evaporating."},{"Start":"03:03.465 ","End":"03:05.620","Text":"In the second picture,"},{"Start":"03:05.620 ","End":"03:07.810","Text":"we have a stoppered flask."},{"Start":"03:07.810 ","End":"03:12.560","Text":"Here, if we have a hot liquid, heat can leave."},{"Start":"03:12.590 ","End":"03:15.715","Text":"But because it stoppered at the top,"},{"Start":"03:15.715 ","End":"03:17.365","Text":"no matter can leave."},{"Start":"03:17.365 ","End":"03:20.770","Text":"The third possibility is a vacuum flask,"},{"Start":"03:20.770 ","End":"03:24.400","Text":"where neither heat can leave nor matter."},{"Start":"03:24.400 ","End":"03:27.550","Text":"Because it\u0027s, of course it\u0027s completely insulated."},{"Start":"03:27.550 ","End":"03:30.204","Text":"This we call an open system,"},{"Start":"03:30.204 ","End":"03:32.755","Text":"a closed system,"},{"Start":"03:32.755 ","End":"03:36.380","Text":"and an isolated system."},{"Start":"03:37.620 ","End":"03:40.210","Text":"Let\u0027s summarize."},{"Start":"03:40.210 ","End":"03:45.845","Text":"We have an open system where heat and matter transfer to the surroundings."},{"Start":"03:45.845 ","End":"03:52.040","Text":"We have a closed system where heat but not matter is transferred to the surroundings."},{"Start":"03:52.040 ","End":"03:54.800","Text":"We have an isolated system where"},{"Start":"03:54.800 ","End":"03:58.445","Text":"neither heat nor matter are transferred to the surroundings."},{"Start":"03:58.445 ","End":"04:03.440","Text":"In this video, we talked about a few concepts in thermochemistry."},{"Start":"04:03.440 ","End":"04:08.910","Text":"In particular, we talked about systems and their surroundings."}],"ID":19392},{"Watched":false,"Name":"Energy and Work","Duration":"5m ","ChapterTopicVideoID":18669,"CourseChapterTopicPlaylistID":81298,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.635","Text":"In the previous video,"},{"Start":"00:01.635 ","End":"00:04.380","Text":"we talked about systems in their surroundings."},{"Start":"00:04.380 ","End":"00:09.975","Text":"In this video, we will introduce some more concepts required to study thermodynamics."},{"Start":"00:09.975 ","End":"00:12.870","Text":"We\u0027ll talk about energy and work."},{"Start":"00:12.870 ","End":"00:14.475","Text":"What\u0027s energy?"},{"Start":"00:14.475 ","End":"00:19.080","Text":"Energy is the capacity to do work or transfer heat."},{"Start":"00:19.080 ","End":"00:23.010","Text":"Later, we\u0027ll see that energy is either kinetic or potential."},{"Start":"00:23.010 ","End":"00:24.435","Text":"What about work?"},{"Start":"00:24.435 ","End":"00:26.580","Text":"Work is a form of energy."},{"Start":"00:26.580 ","End":"00:32.430","Text":"For example, a force F does work when it causes an object to move"},{"Start":"00:32.430 ","End":"00:39.410","Text":"a distance d. Work with small w is equal to force times distance."},{"Start":"00:39.410 ","End":"00:41.645","Text":"Force, according to Newton\u0027s law,"},{"Start":"00:41.645 ","End":"00:44.285","Text":"can be written as mass times acceleration."},{"Start":"00:44.285 ","End":"00:48.155","Text":"So work is mass times acceleration times distance."},{"Start":"00:48.155 ","End":"00:49.655","Text":"What are the units?"},{"Start":"00:49.655 ","End":"00:53.270","Text":"Units? If we take the units according to SI system,"},{"Start":"00:53.270 ","End":"00:58.775","Text":"mass is kilograms, acceleration is meter per second squared,"},{"Start":"00:58.775 ","End":"01:00.695","Text":"and distance is meter."},{"Start":"01:00.695 ","End":"01:02.480","Text":"If we multiply all that together,"},{"Start":"01:02.480 ","End":"01:07.640","Text":"we get kilograms times meters squared per second squared."},{"Start":"01:07.640 ","End":"01:10.834","Text":"As we learned in the previous subject of gases."},{"Start":"01:10.834 ","End":"01:13.580","Text":"That is precisely the Joule."},{"Start":"01:13.580 ","End":"01:15.530","Text":"Let\u0027s take an example."},{"Start":"01:15.530 ","End":"01:19.610","Text":"If a ball is dropped from a height d. Here we have a ball,"},{"Start":"01:19.610 ","End":"01:26.165","Text":"is going to be dropped from a height d. The force is the mass times g,"},{"Start":"01:26.165 ","End":"01:28.465","Text":"the acceleration due to gravity."},{"Start":"01:28.465 ","End":"01:34.500","Text":"The work will be mass times acceleration due to gravity times the distance."},{"Start":"01:34.500 ","End":"01:38.030","Text":"Here is the distance d. What about heat?"},{"Start":"01:38.030 ","End":"01:41.150","Text":"Heat is the energy transferred between a system and"},{"Start":"01:41.150 ","End":"01:44.749","Text":"its surroundings as a result of temperature difference."},{"Start":"01:44.749 ","End":"01:48.050","Text":"For example, if I want a nice hot cup of tea,"},{"Start":"01:48.050 ","End":"01:49.900","Text":"and it\u0027s a very cold day."},{"Start":"01:49.900 ","End":"01:55.970","Text":"The heat would leave my cup of coffee or tea and go into the surroundings."},{"Start":"01:55.970 ","End":"01:58.460","Text":"It won\u0027t make much difference as surroundings,"},{"Start":"01:58.460 ","End":"02:00.950","Text":"but my cup will get cold."},{"Start":"02:00.950 ","End":"02:07.190","Text":"Now we said before that energy can be divided into kinetic or potential."},{"Start":"02:07.190 ","End":"02:10.310","Text":"Our kinetic energy is the energy of motion."},{"Start":"02:10.310 ","End":"02:13.175","Text":"We learned about it when we studied gases."},{"Start":"02:13.175 ","End":"02:18.990","Text":"Kinetic energy E_k is 1/2 times the mass times the velocity squared."},{"Start":"02:18.990 ","End":"02:20.855","Text":"So what are the units?"},{"Start":"02:20.855 ","End":"02:23.045","Text":"For mass, we have kilograms."},{"Start":"02:23.045 ","End":"02:26.090","Text":"For velocity, we have meter per second."},{"Start":"02:26.090 ","End":"02:27.985","Text":"We have to square that."},{"Start":"02:27.985 ","End":"02:33.860","Text":"We have altogether kilograms times meters squared times seconds,"},{"Start":"02:33.860 ","End":"02:35.540","Text":"the power of minus 2."},{"Start":"02:35.540 ","End":"02:39.800","Text":"So we have kilograms times meters squared per second squared."},{"Start":"02:39.800 ","End":"02:44.540","Text":"As we saw before, that\u0027s precisely joule."},{"Start":"02:44.540 ","End":"02:48.470","Text":"Work and energy have the same units."},{"Start":"02:48.470 ","End":"02:52.970","Text":"They\u0027re both measured in joules according to the SI system."},{"Start":"02:52.970 ","End":"02:57.469","Text":"When we studied molecular kinetic theory in gases,"},{"Start":"02:57.469 ","End":"03:04.880","Text":"we saw that the average kinetic energy of a whole mole of molecules is 3/2RT,"},{"Start":"03:04.880 ","End":"03:06.635","Text":"where R is the gas constant,"},{"Start":"03:06.635 ","End":"03:09.425","Text":"T is temperature in Kelvin,"},{"Start":"03:09.425 ","End":"03:14.040","Text":"and 3/2 is for the 3 distances x, y, and z."},{"Start":"03:14.040 ","End":"03:20.120","Text":"This kinetic energy is called thermal energy and arises because"},{"Start":"03:20.120 ","End":"03:26.705","Text":"of the random motion of the various molecule in different directions."},{"Start":"03:26.705 ","End":"03:28.864","Text":"That\u0027s thermal energy."},{"Start":"03:28.864 ","End":"03:31.235","Text":"Now what\u0027s potential energy?"},{"Start":"03:31.235 ","End":"03:35.885","Text":"We distinguish between potential energy and kinetic energy."},{"Start":"03:35.885 ","End":"03:38.480","Text":"Energy is either kinetic or potential."},{"Start":"03:38.480 ","End":"03:44.330","Text":"Potential means it is stored energy that can be turned into kinetic energy."},{"Start":"03:44.330 ","End":"03:49.160","Text":"Supposing a motionless ball is placed on the top of a hill."},{"Start":"03:49.160 ","End":"03:50.750","Text":"Here we have a hill."},{"Start":"03:50.750 ","End":"03:53.045","Text":"We have a motionless ball."},{"Start":"03:53.045 ","End":"03:56.390","Text":"At the top it has potential energy."},{"Start":"03:56.390 ","End":"04:01.010","Text":"It has the capacity to produce kinetic energy, but it\u0027s motionless."},{"Start":"04:01.010 ","End":"04:03.275","Text":"The kinetic energy is 0,"},{"Start":"04:03.275 ","End":"04:06.780","Text":"here the kinetic energy is 0,"},{"Start":"04:06.780 ","End":"04:09.270","Text":"and it has potential energy."},{"Start":"04:09.270 ","End":"04:12.580","Text":"Now it starts to roll down the hill."},{"Start":"04:12.640 ","End":"04:15.020","Text":"As it rolls down,"},{"Start":"04:15.020 ","End":"04:17.105","Text":"it gets faster and faster,"},{"Start":"04:17.105 ","End":"04:21.575","Text":"so its kinetic energy is increasing,"},{"Start":"04:21.575 ","End":"04:26.590","Text":"and its potential energy is decreasing."},{"Start":"04:26.590 ","End":"04:31.805","Text":"If we have a system which doesn\u0027t lose any energy,"},{"Start":"04:31.805 ","End":"04:41.420","Text":"then we can say that the potential energy plus the kinetic energy will be a constant."},{"Start":"04:41.420 ","End":"04:44.720","Text":"The potential energy would turn into kinetic energy,"},{"Start":"04:44.720 ","End":"04:47.704","Text":"and the sum of them will always be constant."},{"Start":"04:47.704 ","End":"04:52.910","Text":"That\u0027s for a system that obeys conservation of energy."},{"Start":"04:52.910 ","End":"04:56.140","Text":"When system that does not lose energy."},{"Start":"04:56.140 ","End":"05:00.870","Text":"So in this video we learned about energy and work."}],"ID":19393}],"Thumbnail":null,"ID":81298},{"Name":"Heat","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Heat Transfer","Duration":"9m 48s","ChapterTopicVideoID":18671,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.760","Text":"In this video, we\u0027ll talk about the amount of heat"},{"Start":"00:02.760 ","End":"00:05.970","Text":"required to heat a substance or an object."},{"Start":"00:05.970 ","End":"00:12.075","Text":"Lets begin with the quantity of heat required to raise the temperature of a substance."},{"Start":"00:12.075 ","End":"00:14.955","Text":"The equation we use is q,"},{"Start":"00:14.955 ","End":"00:16.500","Text":"that\u0027s the quantity of heat,"},{"Start":"00:16.500 ","End":"00:21.480","Text":"equal to the mass of the substance times the specific heat times"},{"Start":"00:21.480 ","End":"00:24.810","Text":"Delta T. Where s is the specific heat of"},{"Start":"00:24.810 ","End":"00:28.470","Text":"the substance and Delta T is the difference in temperature,"},{"Start":"00:28.470 ","End":"00:32.400","Text":"the final temperature minus the initial temperature."},{"Start":"00:32.400 ","End":"00:36.330","Text":"Now, different substances have different specific heat."},{"Start":"00:36.330 ","End":"00:38.265","Text":"Let\u0027s take an example."},{"Start":"00:38.265 ","End":"00:40.575","Text":"This is one you\u0027re very familiar with."},{"Start":"00:40.575 ","End":"00:45.180","Text":"When we cook, we stir with wooden rather than metal spoons,"},{"Start":"00:45.180 ","End":"00:48.531","Text":"since these have a higher specific heat."},{"Start":"00:48.531 ","End":"00:53.325","Text":"It takes more energy to heat a wooden spoon than a metal spoon."},{"Start":"00:53.325 ","End":"00:59.070","Text":"You\u0027re very aware that when you put a metal spoon in to stir something hot,"},{"Start":"00:59.070 ","End":"01:04.035","Text":"the metal spoon rapidly becomes hot and burns your hand."},{"Start":"01:04.035 ","End":"01:09.960","Text":"This is because of the difference in specific heat of wood and metal."},{"Start":"01:09.960 ","End":"01:14.384","Text":"Now, the specific heat of water is very important."},{"Start":"01:14.384 ","End":"01:19.965","Text":"The specific heat of liquid water is different depending on the phase of the water."},{"Start":"01:19.965 ","End":"01:22.335","Text":"At 25 degrees Celsius,"},{"Start":"01:22.335 ","End":"01:27.960","Text":"it\u0027s 4.18 Joules per gram per degree Celsius."},{"Start":"01:27.960 ","End":"01:30.555","Text":"It\u0027s the highest for all liquids."},{"Start":"01:30.555 ","End":"01:33.300","Text":"Now, before Joules became popular,"},{"Start":"01:33.300 ","End":"01:34.845","Text":"or became the official unit,"},{"Start":"01:34.845 ","End":"01:36.600","Text":"we used Calorie,"},{"Start":"01:36.600 ","End":"01:41.640","Text":"1 Calorie was then defined as the amount of heat required to"},{"Start":"01:41.640 ","End":"01:46.890","Text":"raise the temperature for 1 gram of water by 1 degree Celsius."},{"Start":"01:46.890 ","End":"01:49.898","Text":"The specific heat of water, this unit,"},{"Start":"01:49.898 ","End":"01:56.490","Text":"was 1 Calorie per gram per degrees Celsius."},{"Start":"01:56.490 ","End":"01:58.665","Text":"The modern definition is,"},{"Start":"01:58.665 ","End":"02:04.005","Text":"1 Calorie is precisely equal to 4.184 Joules."},{"Start":"02:04.005 ","End":"02:06.735","Text":"That\u0027s how we work it out now."},{"Start":"02:06.735 ","End":"02:09.960","Text":"Calories are often used for food."},{"Start":"02:09.960 ","End":"02:12.765","Text":"When you say you eat a lot of calories,"},{"Start":"02:12.765 ","End":"02:15.480","Text":"what you really mean are kilocalories."},{"Start":"02:15.480 ","End":"02:19.530","Text":"You are talking about 1,000 or more calories."},{"Start":"02:19.530 ","End":"02:23.190","Text":"Often it\u0027s written with a capital C. Now,"},{"Start":"02:23.190 ","End":"02:26.100","Text":"sometimes we just don\u0027t have just one substance."},{"Start":"02:26.100 ","End":"02:29.415","Text":"We may have a whole array of substances,"},{"Start":"02:29.415 ","End":"02:34.680","Text":"or we may say a metal box and inside it is water and other things."},{"Start":"02:34.680 ","End":"02:39.090","Text":"Then, with heat capacity of the apparatus,"},{"Start":"02:39.090 ","End":"02:43.230","Text":"the heat capacity of the whole thing."},{"Start":"02:43.230 ","End":"02:45.615","Text":"We use the heat capacity."},{"Start":"02:45.615 ","End":"02:49.890","Text":"Instead of using q=m times s times Delta T,"},{"Start":"02:49.890 ","End":"02:58.290","Text":"we take all the masses times specific heat and call it C. That\u0027s the heat capacity."},{"Start":"02:58.290 ","End":"03:04.470","Text":"If we have something consisting of several units or several substances,"},{"Start":"03:04.470 ","End":"03:07.260","Text":"we write q=C times"},{"Start":"03:07.260 ","End":"03:14.160","Text":"Delta T. This C is usually determined experimentally for the apparatus,"},{"Start":"03:14.160 ","End":"03:18.910","Text":"and then the apparatus is used many times over for other substances."},{"Start":"03:18.910 ","End":"03:24.425","Text":"The units are then joules per degrees Celsius."},{"Start":"03:24.425 ","End":"03:29.570","Text":"Let us take an example of how we use the specific heat."},{"Start":"03:29.570 ","End":"03:32.735","Text":"How much heat is required to raise the temperature"},{"Start":"03:32.735 ","End":"03:37.160","Text":"of 10 grams of water by 20 degrees Celsius,"},{"Start":"03:37.160 ","End":"03:40.205","Text":"q is equal to m, the mass,"},{"Start":"03:40.205 ","End":"03:44.150","Text":"times specific heat, times the difference in temperature."},{"Start":"03:44.150 ","End":"03:49.310","Text":"You should note that if it\u0027s a difference in temperature,"},{"Start":"03:49.310 ","End":"03:53.390","Text":"it doesn\u0027t matter whether it\u0027s degrees Celsius or Kelvin."},{"Start":"03:53.390 ","End":"03:55.855","Text":"That\u0027s precisely the same difference."},{"Start":"03:55.855 ","End":"04:00.570","Text":"Here, it\u0027s 10 grams because I have 10 grams of water."},{"Start":"04:00.570 ","End":"04:02.540","Text":"Here\u0027s the specific heat of water,"},{"Start":"04:02.540 ","End":"04:06.305","Text":"4.18 joules per gram per degrees Celsius."},{"Start":"04:06.305 ","End":"04:09.725","Text":"The temperature difference is 10 degrees Celsius."},{"Start":"04:09.725 ","End":"04:11.330","Text":"First, let\u0027s look at the units,"},{"Start":"04:11.330 ","End":"04:15.090","Text":"we have grams times grams to power minus 1."},{"Start":"04:15.090 ","End":"04:19.845","Text":"We have Celsius times Celsius to power minus one."},{"Start":"04:19.845 ","End":"04:22.815","Text":"All we\u0027re left with is the joules."},{"Start":"04:22.815 ","End":"04:24.510","Text":"When we work that out,"},{"Start":"04:24.510 ","End":"04:27.620","Text":"it comes to 418 joules."},{"Start":"04:27.620 ","End":"04:35.035","Text":"The amount of heat required to raise the temperature of the water is 418 joules."},{"Start":"04:35.035 ","End":"04:40.940","Text":"Now, there\u0027s a very important law called the law of conservation of energy."},{"Start":"04:40.940 ","End":"04:44.420","Text":"When a system interacts with its surroundings,"},{"Start":"04:44.420 ","End":"04:49.250","Text":"remember we divided everything into a system and its surroundings."},{"Start":"04:49.250 ","End":"04:52.130","Text":"The total energy remains constant,"},{"Start":"04:52.130 ","End":"04:54.215","Text":"there is no change in the total energy."},{"Start":"04:54.215 ","End":"04:58.640","Text":"It could go from the system to the surroundings or from the surroundings to the system,"},{"Start":"04:58.640 ","End":"05:01.775","Text":"but the total is the same."},{"Start":"05:01.775 ","End":"05:04.624","Text":"Now in heat, we usually write it like this,"},{"Start":"05:04.624 ","End":"05:10.474","Text":"q system plus q surroundings is equal to 0."},{"Start":"05:10.474 ","End":"05:15.725","Text":"That means that q system is equal to minus q surroundings."},{"Start":"05:15.725 ","End":"05:18.080","Text":"If we can calculate q surroundings,"},{"Start":"05:18.080 ","End":"05:19.280","Text":"we know q system."},{"Start":"05:19.280 ","End":"05:20.960","Text":"If we calculate q system,"},{"Start":"05:20.960 ","End":"05:22.990","Text":"we know q surroundings."},{"Start":"05:22.990 ","End":"05:25.205","Text":"If we\u0027re not quite sure, we\u0027re mixing,"},{"Start":"05:25.205 ","End":"05:27.366","Text":"for example, hot water and cold water."},{"Start":"05:27.366 ","End":"05:29.420","Text":"We\u0027re not quite sure which is the system,"},{"Start":"05:29.420 ","End":"05:30.875","Text":"which is the surroundings,"},{"Start":"05:30.875 ","End":"05:35.090","Text":"we just write q1 plus q2 equals to 0."},{"Start":"05:35.090 ","End":"05:39.380","Text":"Now, the signs of q are very important."},{"Start":"05:39.380 ","End":"05:43.025","Text":"If q system is greater than 0, what does it mean?"},{"Start":"05:43.025 ","End":"05:46.370","Text":"It means that Delta T must be greater than 0 because remember,"},{"Start":"05:46.370 ","End":"05:53.940","Text":"q is mass times s times Delta T. If q is positive,"},{"Start":"05:53.940 ","End":"05:55.650","Text":"Delta T is positive."},{"Start":"05:55.650 ","End":"06:00.170","Text":"That means that the final temperature is greater than the initial temperature."},{"Start":"06:00.170 ","End":"06:03.895","Text":"The system becomes hotter, gains heat."},{"Start":"06:03.895 ","End":"06:09.365","Text":"On the other hand, if q system is less than 0,"},{"Start":"06:09.365 ","End":"06:11.645","Text":"Delta T will be negative,"},{"Start":"06:11.645 ","End":"06:15.830","Text":"that T final will be less than T initial."},{"Start":"06:15.830 ","End":"06:18.440","Text":"That means the system loses heat."},{"Start":"06:18.440 ","End":"06:22.000","Text":"The temperature decreases, it loses heat."},{"Start":"06:22.000 ","End":"06:27.665","Text":"Negative means the system loses heat or gives off heat."},{"Start":"06:27.665 ","End":"06:32.525","Text":"If it\u0027s positive, the system absorbs or gains heat."},{"Start":"06:32.525 ","End":"06:34.345","Text":"That is very important,"},{"Start":"06:34.345 ","End":"06:36.290","Text":"we use that extensively."},{"Start":"06:36.290 ","End":"06:38.530","Text":"Let\u0027s take an example."},{"Start":"06:38.530 ","End":"06:42.125","Text":"Supposing 100 grams of water at"},{"Start":"06:42.125 ","End":"06:46.732","Text":"80 degrees Celsius is mixed in a thermally insulated flask."},{"Start":"06:46.732 ","End":"06:51.460","Text":"That means no heat escapes from the flask."},{"Start":"06:51.460 ","End":"06:55.285","Text":"With 200 grams of water at 20 degrees Celsius."},{"Start":"06:55.285 ","End":"06:57.670","Text":"What is the final temperature?"},{"Start":"06:57.670 ","End":"07:00.910","Text":"Now you all know from a common experience,"},{"Start":"07:00.910 ","End":"07:03.400","Text":"if you mix something hot and cold,"},{"Start":"07:03.400 ","End":"07:05.905","Text":"the hot one becomes cooler."},{"Start":"07:05.905 ","End":"07:10.120","Text":"The cool one becomes hotter until they reach the same temperature."},{"Start":"07:10.120 ","End":"07:13.970","Text":"They reach what we call equilibrium."},{"Start":"07:20.660 ","End":"07:24.680","Text":"Here it\u0027s not at all obvious which is the system,"},{"Start":"07:24.680 ","End":"07:25.745","Text":"which is the surroundings."},{"Start":"07:25.745 ","End":"07:29.465","Text":"We just write q1 plus q2 equal to 0."},{"Start":"07:29.465 ","End":"07:35.150","Text":"We have 100 grams times the specific heat of water,"},{"Start":"07:35.150 ","End":"07:39.050","Text":"4.18 joules per gram per degrees Celsius."},{"Start":"07:39.050 ","End":"07:43.950","Text":"Then T final, this is our Delta T,"},{"Start":"07:43.950 ","End":"07:46.670","Text":"minus the initial temperature,"},{"Start":"07:46.670 ","End":"07:49.444","Text":"which is 80 degrees Celsius."},{"Start":"07:49.444 ","End":"07:52.775","Text":"Then we have 200 grams of water."},{"Start":"07:52.775 ","End":"07:56.060","Text":"Remember that\u0027s 200 grams times the specific heat,"},{"Start":"07:56.060 ","End":"07:59.930","Text":"4.18, times T final."},{"Start":"07:59.930 ","End":"08:05.130","Text":"This is a Delta T for the water that starts at 20 degrees Celsius,"},{"Start":"08:05.130 ","End":"08:09.060","Text":"T_f minus 20 degrees Celsius."},{"Start":"08:09.060 ","End":"08:13.415","Text":"The T_f is the same in each case because they\u0027re in equilibrium."},{"Start":"08:13.415 ","End":"08:16.115","Text":"We know that sum is equal to 0."},{"Start":"08:16.115 ","End":"08:18.020","Text":"Now, to simplify matters,"},{"Start":"08:18.020 ","End":"08:21.879","Text":"we can divide by 4.18 throughout."},{"Start":"08:21.879 ","End":"08:24.810","Text":"Multiplying 100 times T_f is 100,"},{"Start":"08:24.810 ","End":"08:29.355","Text":"T_f minus 100 times 80, which is 8,000."},{"Start":"08:29.355 ","End":"08:38.810","Text":"Then we have 200 times T_f minus 20 times 200,"},{"Start":"08:38.810 ","End":"08:41.305","Text":"which gives us minus 4,000."},{"Start":"08:41.305 ","End":"08:43.820","Text":"That sum is equal to 0."},{"Start":"08:43.820 ","End":"08:48.080","Text":"We can write that, collect things multiplying T_f,"},{"Start":"08:48.080 ","End":"08:51.545","Text":"it\u0027s 100 plus 200, that\u0027s 300 T_f."},{"Start":"08:51.545 ","End":"08:57.120","Text":"We have 300 T_f minus 12,000,"},{"Start":"08:57.120 ","End":"08:59.590","Text":"minus 8,000 plus 4,000."},{"Start":"08:59.590 ","End":"09:01.280","Text":"If we bring that to the other side,"},{"Start":"09:01.280 ","End":"09:04.670","Text":"we have 300 T_f equal to 12,000,"},{"Start":"09:04.670 ","End":"09:07.770","Text":"so T_f is equal to 40,"},{"Start":"09:07.770 ","End":"09:09.830","Text":"and we\u0027re working in degrees Celsius."},{"Start":"09:09.830 ","End":"09:13.175","Text":"T_f is 40 degrees Celsius."},{"Start":"09:13.175 ","End":"09:18.470","Text":"We get to a temperature that\u0027s intermediate between 20 and 80."},{"Start":"09:18.470 ","End":"09:20.314","Text":"It\u0027s not in the middle,"},{"Start":"09:20.314 ","End":"09:23.945","Text":"which would be 50, because we have different masses."},{"Start":"09:23.945 ","End":"09:30.230","Text":"We have 100 grams of the very hot water and 200 grams of the cooler water."},{"Start":"09:30.230 ","End":"09:33.695","Text":"It gets closer to the cooler temperature."},{"Start":"09:33.695 ","End":"09:40.100","Text":"We get 40, which is closer to 20 than it is to 80."},{"Start":"09:40.100 ","End":"09:47.999","Text":"In this video, we talked about heat and its transfer from hot objects to cold objects."}],"ID":34061},{"Watched":false,"Name":"Heats of Reaction and Calorimetry","Duration":"2m 8s","ChapterTopicVideoID":18672,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"In the previous video,"},{"Start":"00:02.040 ","End":"00:07.035","Text":"we talked about thermal heat that could raise or lower the temperature of a system."},{"Start":"00:07.035 ","End":"00:09.990","Text":"In this video, we\u0027ll talk about the heat produced or"},{"Start":"00:09.990 ","End":"00:13.125","Text":"gained when a chemical reaction occurs."},{"Start":"00:13.125 ","End":"00:16.635","Text":"We\u0027re going to talk about the heat of reaction."},{"Start":"00:16.635 ","End":"00:21.930","Text":"A reaction can either be exothermic or endothermic."},{"Start":"00:21.930 ","End":"00:25.675","Text":"Exothermic is when the heat goes out of the system."},{"Start":"00:25.675 ","End":"00:29.420","Text":"So the official definition is an exothermic reaction"},{"Start":"00:29.420 ","End":"00:34.055","Text":"produces a temperature increase in an isolated system."},{"Start":"00:34.055 ","End":"00:35.990","Text":"In a non-isolated system,"},{"Start":"00:35.990 ","End":"00:38.750","Text":"it gives off heat to the surroundings,"},{"Start":"00:38.750 ","End":"00:43.092","Text":"so Q reaction is less than 0."},{"Start":"00:43.092 ","End":"00:45.330","Text":"It\u0027s negative."},{"Start":"00:45.330 ","End":"00:46.940","Text":"In the previous video,"},{"Start":"00:46.940 ","End":"00:53.030","Text":"we gave the reaction of a strong acid and base as an example of an exothermic reaction."},{"Start":"00:53.030 ","End":"00:57.595","Text":"Heat is produced where we mix the acid and the base."},{"Start":"00:57.595 ","End":"01:02.195","Text":"You should note that most chemical reactions are exothermic."},{"Start":"01:02.195 ","End":"01:04.700","Text":"What about the opposite of exothermic?"},{"Start":"01:04.700 ","End":"01:09.025","Text":"Endothermic, where the system absorbs heat."},{"Start":"01:09.025 ","End":"01:15.230","Text":"An endothermic reaction produces a temperature decrease in an isolated system."},{"Start":"01:15.230 ","End":"01:17.500","Text":"In a non-isolated system,"},{"Start":"01:17.500 ","End":"01:20.060","Text":"it gains heat from the surroundings."},{"Start":"01:20.060 ","End":"01:24.199","Text":"A Q reaction is greater than 0."},{"Start":"01:24.199 ","End":"01:26.540","Text":"We\u0027re writing rxn,"},{"Start":"01:26.540 ","End":"01:29.225","Text":"Q reaction, and here, it\u0027s positive."},{"Start":"01:29.225 ","End":"01:36.725","Text":"It\u0027s negative for exothermic and positive for endothermic reactions."},{"Start":"01:36.725 ","End":"01:38.585","Text":"In the previous video,"},{"Start":"01:38.585 ","End":"01:42.350","Text":"we gave the evaporation of alcohol as an example."},{"Start":"01:42.350 ","End":"01:47.420","Text":"When alcohol evaporates, it absorbs heat from the surroundings,"},{"Start":"01:47.420 ","End":"01:49.970","Text":"say from your arm or something like that."},{"Start":"01:49.970 ","End":"01:52.730","Text":"Your arm then feels cold."},{"Start":"01:52.730 ","End":"01:54.920","Text":"What\u0027s a calorimeter?"},{"Start":"01:54.920 ","End":"01:59.420","Text":"Heats of reaction are measured experimentally in a calorimeter."},{"Start":"01:59.420 ","End":"02:01.040","Text":"In the next video,"},{"Start":"02:01.040 ","End":"02:07.950","Text":"we\u0027ll talk about 2 sorts: a bomb calorimeter and a coffee cup calorimeter."}],"ID":34062},{"Watched":false,"Name":"Exercise 1","Duration":"6m 51s","ChapterTopicVideoID":22962,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.445","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.445 ","End":"00:05.094","Text":"Calculate the amount of heat in kilojoules required."},{"Start":"00:05.094 ","End":"00:11.595","Text":"A, to raise the temperature of 267 kilograms of iron by 16.8 degree Celsius."},{"Start":"00:11.595 ","End":"00:17.454","Text":"The specific heat capacity of iron equals 0.449 joule per gram degree Celsius."},{"Start":"00:17.454 ","End":"00:20.390","Text":"B, to lower the temperature of 5.85 liters of"},{"Start":"00:20.390 ","End":"00:23.660","Text":"water from 45 degree Celsius to 31 degree Celsius."},{"Start":"00:23.660 ","End":"00:29.225","Text":"The specific heat capacity of water equals 4.18 joule per gram degree Celsius."},{"Start":"00:29.225 ","End":"00:32.900","Text":"We\u0027re going to start with a, and we want to calculate the amount of heat."},{"Start":"00:32.900 ","End":"00:35.270","Text":"We\u0027re going to use the equation q,"},{"Start":"00:35.270 ","End":"00:42.464","Text":"which is the amount or the quantity of heat equals ms Delta T. Now,"},{"Start":"00:42.464 ","End":"00:44.375","Text":"m is the mass of the substance."},{"Start":"00:44.375 ","End":"00:46.865","Text":"Just going to write mass."},{"Start":"00:46.865 ","End":"00:53.840","Text":"s is the specific heat or the specific heat capacity of the substance,"},{"Start":"00:53.840 ","End":"00:57.180","Text":"and T is the temperature change."},{"Start":"01:00.950 ","End":"01:03.665","Text":"Now just a small note,"},{"Start":"01:03.665 ","End":"01:07.940","Text":"the specific heat capacity is usually given the symbol c,"},{"Start":"01:07.940 ","End":"01:11.870","Text":"lower case c. Just going to write that here but in our case,"},{"Start":"01:11.870 ","End":"01:14.105","Text":"we\u0027re going to use the symbol S,"},{"Start":"01:14.105 ","End":"01:17.990","Text":"just to avoid confusion with heat capacity,"},{"Start":"01:17.990 ","End":"01:20.360","Text":"which is given an uppercase C. Here,"},{"Start":"01:20.360 ","End":"01:23.165","Text":"we\u0027re going to use the lowercase s for this specific heat."},{"Start":"01:23.165 ","End":"01:26.690","Text":"Now q. In a,"},{"Start":"01:26.690 ","End":"01:36.470","Text":"we can see that the mass of the iron is 267 kilograms times the specific heat of iron,"},{"Start":"01:36.470 ","End":"01:46.290","Text":"which is given 0.449 joule per gram degree Celsius."},{"Start":"01:46.290 ","End":"01:52.230","Text":"The temperature change here is given and equals 16.8 degree Celsius."},{"Start":"01:54.640 ","End":"01:57.530","Text":"We can see that we have kilograms and grams,"},{"Start":"01:57.530 ","End":"01:59.360","Text":"so we\u0027re going to multiply by a conversion factor."},{"Start":"01:59.360 ","End":"02:05.175","Text":"Again, we\u0027re going to start 267 kilograms times,"},{"Start":"02:05.175 ","End":"02:07.200","Text":"we know that in every 1 kilogram,"},{"Start":"02:07.200 ","End":"02:09.640","Text":"we have 1,000 grams,"},{"Start":"02:10.820 ","End":"02:17.370","Text":"times 0.449 joule per"},{"Start":"02:17.370 ","End":"02:24.100","Text":"gram degree Celsius times 16.8 degree Celsius."},{"Start":"02:25.910 ","End":"02:29.525","Text":"Our degree Celsius cancels out,"},{"Start":"02:29.525 ","End":"02:33.100","Text":"and our kilograms and grams cancel out."},{"Start":"02:33.100 ","End":"02:41.780","Text":"This equals 2,014,034.4 joules."},{"Start":"02:42.720 ","End":"02:46.100","Text":"Now, we were asked to give the quantity of heat in kilojoules,"},{"Start":"02:46.100 ","End":"02:48.925","Text":"so we\u0027re just going to multiply by a conversion factor."},{"Start":"02:48.925 ","End":"02:53.980","Text":"In 1 kilojoule, we have 1,000 joules."},{"Start":"02:53.980 ","End":"02:56.890","Text":"The joules cancel out,"},{"Start":"02:56.890 ","End":"03:04.100","Text":"and this equals 2,014 kilojoules."},{"Start":"03:04.100 ","End":"03:10.420","Text":"The quantity of heat of the iron equals 2,014 kilojoules."},{"Start":"03:10.420 ","End":"03:13.030","Text":"That is our final answer for a."},{"Start":"03:13.030 ","End":"03:18.310","Text":"Now, I just want to note that when the quantity of heat is positive,"},{"Start":"03:18.310 ","End":"03:20.860","Text":"it means that heat was absorbed."},{"Start":"03:20.860 ","End":"03:23.350","Text":"If the quantity of heat is negative,"},{"Start":"03:23.350 ","End":"03:24.805","Text":"it means that heat was released."},{"Start":"03:24.805 ","End":"03:26.555","Text":"In our case, heat was absorbed."},{"Start":"03:26.555 ","End":"03:29.710","Text":"B, to lower the temperature of 5.85 liters of"},{"Start":"03:29.710 ","End":"03:33.835","Text":"water from 45 degree Celsius to 31 degree Celsius."},{"Start":"03:33.835 ","End":"03:39.100","Text":"The specific heat capacity of water equals 4.18 joule per gram degree Celsius."},{"Start":"03:39.100 ","End":"03:41.590","Text":"We solved a. Now, we\u0027re going to go on to b."},{"Start":"03:41.590 ","End":"03:43.413","Text":"Now we want to calculate the amount of heat,"},{"Start":"03:43.413 ","End":"03:44.890","Text":"so we\u0027re going to use the equation q,"},{"Start":"03:44.890 ","End":"03:46.854","Text":"which is the quantity or amount of heat,"},{"Start":"03:46.854 ","End":"03:51.336","Text":"equals ms Delta T. Now,"},{"Start":"03:51.336 ","End":"03:52.815","Text":"m is the mass of the substance,"},{"Start":"03:52.815 ","End":"03:55.430","Text":"s is the specific heat of the substance,"},{"Start":"03:55.430 ","End":"03:58.235","Text":"and Delta T is the temperature change."},{"Start":"03:58.235 ","End":"03:59.990","Text":"Now, first of all, we\u0027re going to calculate the mass of"},{"Start":"03:59.990 ","End":"04:01.610","Text":"the water because we know that the volume of"},{"Start":"04:01.610 ","End":"04:06.120","Text":"the water equals 5.85 liters."},{"Start":"04:07.030 ","End":"04:10.309","Text":"Now remember that d, which is the density,"},{"Start":"04:10.309 ","End":"04:15.320","Text":"equals m divided by V. m is the mass of the substance and V is the volume,"},{"Start":"04:15.320 ","End":"04:17.360","Text":"so you want to calculate the mass."},{"Start":"04:17.360 ","End":"04:18.980","Text":"The mass equals,"},{"Start":"04:18.980 ","End":"04:21.440","Text":"just going to multiply both sides by the volume,"},{"Start":"04:21.440 ","End":"04:29.305","Text":"d times V. Now the density of water is 1 gram per milliliter,"},{"Start":"04:29.305 ","End":"04:33.290","Text":"and V is 5.85 liters."},{"Start":"04:37.390 ","End":"04:39.905","Text":"Now, since we have milliliter and liters,"},{"Start":"04:39.905 ","End":"04:41.962","Text":"we\u0027re going to multiply by a conversion factor."},{"Start":"04:41.962 ","End":"04:44.300","Text":"Remember that in every 1 liter,"},{"Start":"04:44.300 ","End":"04:46.130","Text":"we have 1,000 milliliters,"},{"Start":"04:46.130 ","End":"04:49.060","Text":"so 1,000 mil divided by 1 liter."},{"Start":"04:49.060 ","End":"04:50.910","Text":"Liters cancel out,"},{"Start":"04:50.910 ","End":"04:52.605","Text":"our milliliters cancel out,"},{"Start":"04:52.605 ","End":"04:58.010","Text":"and this equals 5,850 grams."},{"Start":"04:58.010 ","End":"05:01.010","Text":"Our mass equals 5,850 grams."},{"Start":"05:01.010 ","End":"05:02.660","Text":"Back to our equation q,"},{"Start":"05:02.660 ","End":"05:08.960","Text":"the quantity of heat equals ms Delta T. The mass equals"},{"Start":"05:08.960 ","End":"05:15.300","Text":"5,850 grams times s,"},{"Start":"05:15.300 ","End":"05:16.695","Text":"which is the specific heat,"},{"Start":"05:16.695 ","End":"05:24.747","Text":"it\u0027s 4.18 joule per gram degree Celsius times Delta T. Now,"},{"Start":"05:24.747 ","End":"05:26.605","Text":"Delta T, I\u0027m going to write that here,"},{"Start":"05:26.605 ","End":"05:30.564","Text":"is T_final minus T_initial,"},{"Start":"05:30.564 ","End":"05:33.685","Text":"meaning, the final temperature minus the initial temperature."},{"Start":"05:33.685 ","End":"05:43.219","Text":"The final temperature in our case is 31 degree Celsius minus 45 degree Celsius."},{"Start":"05:44.430 ","End":"05:50.350","Text":"This equals, again, 5,850 grams times"},{"Start":"05:50.350 ","End":"06:00.150","Text":"4.18 joule per grams degree Celsius times minus 14 degree Celsius."},{"Start":"06:00.150 ","End":"06:02.030","Text":"The degree Celsius cancel out,"},{"Start":"06:02.030 ","End":"06:03.305","Text":"the grams cancel out,"},{"Start":"06:03.305 ","End":"06:11.735","Text":"and this equals minus 342, 342 joules."},{"Start":"06:11.735 ","End":"06:15.060","Text":"Then the question we were asked to calculate the kilojoules,"},{"Start":"06:15.060 ","End":"06:17.480","Text":"we\u0027re going to multiply it by a conversion factor."},{"Start":"06:17.480 ","End":"06:19.205","Text":"In 1 kilojoule,"},{"Start":"06:19.205 ","End":"06:24.770","Text":"we have 1,000 joules and the joules cancel out,"},{"Start":"06:24.770 ","End":"06:26.690","Text":"and we\u0027re left with kilojoules in our unit."},{"Start":"06:26.690 ","End":"06:34.340","Text":"This equals minus 342.34 kilojoules."},{"Start":"06:34.340 ","End":"06:38.970","Text":"Now, note that the quantity came out negative,"},{"Start":"06:39.440 ","End":"06:42.795","Text":"meaning, that in this case, heat was released."},{"Start":"06:42.795 ","End":"06:48.244","Text":"Our answer for b is q equals minus 342.34 kilojoules."},{"Start":"06:48.244 ","End":"06:49.520","Text":"That is our final answer."},{"Start":"06:49.520 ","End":"06:52.170","Text":"Thank you very much for watching."}],"ID":34063},{"Watched":false,"Name":"Exercise 2","Duration":"5m 4s","ChapterTopicVideoID":22963,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.774","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.774 ","End":"00:06.720","Text":"A 289 gram piece of aluminum is removed from an oven and"},{"Start":"00:06.720 ","End":"00:11.070","Text":"placed into 525 grams of water in an insulated container."},{"Start":"00:11.070 ","End":"00:16.500","Text":"The temperature of the water increases from 25 degrees Celsius to 64 degrees Celsius."},{"Start":"00:16.500 ","End":"00:19.265","Text":"What is the original temperature of the aluminum?"},{"Start":"00:19.265 ","End":"00:24.665","Text":"The specific heat capacity of water equals 4.18 joules per gram degrees Celsius."},{"Start":"00:24.665 ","End":"00:29.885","Text":"The specific heat of aluminum is 0.903 joule per gram degrees Celsius."},{"Start":"00:29.885 ","End":"00:32.795","Text":"Here we want to calculate the original temperature of the aluminum,"},{"Start":"00:32.795 ","End":"00:36.320","Text":"meaning the T initial, initial temperature."},{"Start":"00:36.320 ","End":"00:39.470","Text":"For this case we\u0027re going to use the equation Q equals"},{"Start":"00:39.470 ","End":"00:43.550","Text":"MS Delta T. Q is the quantity of heat,"},{"Start":"00:43.550 ","End":"00:45.110","Text":"M is the mass of the substance,"},{"Start":"00:45.110 ","End":"00:47.030","Text":"S is the specific heat of the substance,"},{"Start":"00:47.030 ","End":"00:52.060","Text":"and Delta T is the T final minus T initial."},{"Start":"00:52.060 ","End":"00:55.460","Text":"Now if we look at the question and we can see that the heat that is"},{"Start":"00:55.460 ","End":"00:58.910","Text":"lost from the aluminum is actually absorbed by the water."},{"Start":"00:58.910 ","End":"01:06.080","Text":"Therefore, Q aluminum plus Q water,"},{"Start":"01:06.080 ","End":"01:09.080","Text":"meaning the quantity of the heat of the aluminum"},{"Start":"01:09.080 ","End":"01:12.530","Text":"plus the quantity of heat of the water equals 0."},{"Start":"01:12.530 ","End":"01:16.745","Text":"Since the quantity of heat lost by the aluminum is absorbed by the water."},{"Start":"01:16.745 ","End":"01:22.610","Text":"Now again, Q the quantity of heat equals MS Delta T. Let\u0027s look at our aluminum."},{"Start":"01:22.610 ","End":"01:26.720","Text":"We can see that the mass of the aluminum is 289 grams, so Q,"},{"Start":"01:26.720 ","End":"01:31.490","Text":"M, the mass 289 grams,"},{"Start":"01:31.490 ","End":"01:34.745","Text":"times the specific heat of aluminum which is given here,"},{"Start":"01:34.745 ","End":"01:40.040","Text":"0.903 joule per gram"},{"Start":"01:40.040 ","End":"01:45.710","Text":"degrees Celsius times Delta T. Now in this case,"},{"Start":"01:45.710 ","End":"01:49.670","Text":"we know that the final temperature of the water was 64 degrees Celsius,"},{"Start":"01:49.670 ","End":"01:52.790","Text":"meaning that the final temperature also belongs to the aluminum."},{"Start":"01:52.790 ","End":"01:56.630","Text":"The final temperature of the aluminum is also 64 degrees Celsius."},{"Start":"01:56.630 ","End":"01:58.890","Text":"Now I have T final minus T initial,"},{"Start":"01:58.890 ","End":"02:03.170","Text":"so we have 64 degrees Celsius minus T initial,"},{"Start":"02:03.170 ","End":"02:06.515","Text":"which we don\u0027t know and we\u0027re trying to find out."},{"Start":"02:06.515 ","End":"02:09.335","Text":"Now that\u0027s the quantity of heat of the aluminum."},{"Start":"02:09.335 ","End":"02:12.605","Text":"Now, we\u0027re going to add to it the quantity of heat of the water."},{"Start":"02:12.605 ","End":"02:16.265","Text":"Now the mass of the water is 525 grams"},{"Start":"02:16.265 ","End":"02:23.660","Text":"given times the specific heat or specific heat capacity of water,"},{"Start":"02:23.660 ","End":"02:31.760","Text":"which is 4.18 joules per gram degrees Celsius times Delta T,"},{"Start":"02:31.760 ","End":"02:34.190","Text":"which is T final minus T initial of the water,"},{"Start":"02:34.190 ","End":"02:38.240","Text":"which is 64 degrees Celsius minus 25 degrees Celsius,"},{"Start":"02:38.240 ","End":"02:44.795","Text":"so 64 degrees Celsius minus 25 degrees Celsius."},{"Start":"02:44.795 ","End":"02:50.540","Text":"Now, the quantity of heat of the aluminum plus the quantity of heat of water equals 0."},{"Start":"02:50.540 ","End":"02:52.685","Text":"I\u0027m just writing that here because I don\u0027t have a place here."},{"Start":"02:52.685 ","End":"02:57.440","Text":"Here, 289 times 0.903"},{"Start":"02:57.440 ","End":"03:04.110","Text":"times 64 equals 16,701.89."},{"Start":"03:04.110 ","End":"03:05.900","Text":"If we look at our units,"},{"Start":"03:05.900 ","End":"03:14.599","Text":"we can see that we have grams times joule per gram degrees Celsius times degrees Celsius."},{"Start":"03:14.599 ","End":"03:19.295","Text":"Degrees Celsius cancel out and grams also cancel out and we\u0027re left with joule."},{"Start":"03:19.295 ","End":"03:28.295","Text":"This is minus 260.97."},{"Start":"03:28.295 ","End":"03:31.160","Text":"Here we have in our units,"},{"Start":"03:31.160 ","End":"03:37.230","Text":"we can see that we have again grams times joule per gram degrees Celsius."},{"Start":"03:38.300 ","End":"03:41.570","Text":"That\u0027s it. The grams cancel out and we\u0027re left with joule"},{"Start":"03:41.570 ","End":"03:44.500","Text":"per degrees Celsius, T_i of course."},{"Start":"03:44.500 ","End":"03:55.290","Text":"We have to multiply it by our initial temperature plus 85,585.5."},{"Start":"03:57.980 ","End":"04:01.135","Text":"Again, here we\u0027re left with joule,"},{"Start":"04:01.135 ","End":"04:03.590","Text":"since we have grams times joule per gram"},{"Start":"04:03.590 ","End":"04:07.100","Text":"degrees Celsius times degrees Celsius. We\u0027re left with joule."},{"Start":"04:07.100 ","End":"04:10.009","Text":"All of this equals 0."},{"Start":"04:10.009 ","End":"04:12.765","Text":"We\u0027re going to add the joules together,"},{"Start":"04:12.765 ","End":"04:21.660","Text":"and this comes to 102,287.39 joules equals."},{"Start":"04:21.660 ","End":"04:25.815","Text":"We\u0027re going to take our T_i to the other side of the equation."},{"Start":"04:25.815 ","End":"04:32.300","Text":"This equals 260.97 and"},{"Start":"04:32.300 ","End":"04:36.650","Text":"this is joule per degree Celsius times T_i."},{"Start":"04:36.650 ","End":"04:41.150","Text":"First of all, we can see that our joules cancel out and our T_i equals,"},{"Start":"04:41.150 ","End":"04:45.230","Text":"we\u0027re going to divide both sides by 260.97."},{"Start":"04:45.230 ","End":"04:50.305","Text":"This comes to 391.95."},{"Start":"04:50.305 ","End":"04:54.995","Text":"We\u0027re going to multiply by a degree of Celsius."},{"Start":"04:54.995 ","End":"05:00.650","Text":"The initial temperature equals 391.95 degrees Celsius."},{"Start":"05:00.650 ","End":"05:01.820","Text":"That is our final answer."},{"Start":"05:01.820 ","End":"05:04.230","Text":"Thank you very much for watching."}],"ID":34064},{"Watched":false,"Name":"Exercise 3","Duration":"5m 53s","ChapterTopicVideoID":22964,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.595","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.595 ","End":"00:06.600","Text":"A 850 gram piece of copper at 58 degrees Celsius is added to"},{"Start":"00:06.600 ","End":"00:12.330","Text":"500 milliliters of water maintained at 35 degrees Celsius in an insulated container."},{"Start":"00:12.330 ","End":"00:16.305","Text":"What will be the final temperature of the Cu-H_2O mixture?"},{"Start":"00:16.305 ","End":"00:21.420","Text":"The specific heat capacity of water equals 4.18 Joule per gram degrees Celsius."},{"Start":"00:21.420 ","End":"00:27.270","Text":"The specific heat capacity of copper equals 0.385 Joules per gram degrees Celsius."},{"Start":"00:27.270 ","End":"00:30.120","Text":"We can see from the question that the mass of the copper,"},{"Start":"00:30.120 ","End":"00:35.895","Text":"just going to write that here, equals 850 grams."},{"Start":"00:35.895 ","End":"00:40.898","Text":"We also know that the initial temperature of the copper equals 58 degrees Celsius,"},{"Start":"00:40.898 ","End":"00:45.560","Text":"so Ti equals 58 degrees Celsius."},{"Start":"00:45.560 ","End":"00:47.450","Text":"Now we know that the copper is added to water."},{"Start":"00:47.450 ","End":"00:49.280","Text":"It\u0027s added to 500 milliliters of water."},{"Start":"00:49.280 ","End":"00:54.330","Text":"The volume of the water equals 500 milliliters."},{"Start":"00:54.790 ","End":"01:00.005","Text":"We also knew that the initial temperature of the water is 35 degrees Celsius."},{"Start":"01:00.005 ","End":"01:07.775","Text":"T initial of water equals 35 degrees Celsius and before it was a T initial of copper."},{"Start":"01:07.775 ","End":"01:13.340","Text":"Now we\u0027re looking for the final temperature of the copper water mixture."},{"Start":"01:13.340 ","End":"01:14.870","Text":"In order to solve this problem,"},{"Start":"01:14.870 ","End":"01:16.685","Text":"we\u0027re going to use the equation q."},{"Start":"01:16.685 ","End":"01:20.945","Text":"The quantity of heat equals ms Delta T,"},{"Start":"01:20.945 ","End":"01:22.895","Text":"m is the mass of the substance,"},{"Start":"01:22.895 ","End":"01:26.690","Text":"s is the specific heat of the substance."},{"Start":"01:26.690 ","End":"01:31.850","Text":"Now, the symbol of the specific heat is usually given a lowercase c. In our case,"},{"Start":"01:31.850 ","End":"01:35.900","Text":"we\u0027re using an s just to avoid confusion with heat capacity,"},{"Start":"01:35.900 ","End":"01:37.655","Text":"which is an uppercase C,"},{"Start":"01:37.655 ","End":"01:40.234","Text":"we\u0027re going to see that in later problems,"},{"Start":"01:40.234 ","End":"01:41.240","Text":"times Delta T,"},{"Start":"01:41.240 ","End":"01:45.335","Text":"Delta T equals T final minus T initial."},{"Start":"01:45.335 ","End":"01:48.635","Text":"Now the quantity of heat which is lost by the copper,"},{"Start":"01:48.635 ","End":"01:50.000","Text":"is absorbed by the water."},{"Start":"01:50.000 ","End":"01:52.580","Text":"Therefore, the quantity of heat of the copper plus"},{"Start":"01:52.580 ","End":"01:57.105","Text":"the quantity of heat of the water equals 0."},{"Start":"01:57.105 ","End":"02:02.295","Text":"Again, q equals ms Delta T. The mass of the copper,"},{"Start":"02:02.295 ","End":"02:04.910","Text":"times the specific heat of the copper,"},{"Start":"02:04.910 ","End":"02:08.300","Text":"times Delta T of the copper,"},{"Start":"02:08.300 ","End":"02:13.640","Text":"plus the mass of the water times the specific heat of"},{"Start":"02:13.640 ","End":"02:20.065","Text":"the water, times Delta T of the water equals 0."},{"Start":"02:20.065 ","End":"02:21.950","Text":"The mass of the copper, we said,"},{"Start":"02:21.950 ","End":"02:28.325","Text":"is 850 grams, times the specific heat of copper,"},{"Start":"02:28.325 ","End":"02:33.770","Text":"which is 0.385 Joule"},{"Start":"02:33.770 ","End":"02:38.750","Text":"per gram degrees Celsius times Delta T. Delta T is T final,"},{"Start":"02:38.750 ","End":"02:40.175","Text":"which we don\u0027t know."},{"Start":"02:40.175 ","End":"02:42.800","Text":"That\u0027s what we want to find out, minus T initial,"},{"Start":"02:42.800 ","End":"02:46.770","Text":"which equals 58 degrees Celsius."},{"Start":"02:47.440 ","End":"02:50.120","Text":"That\u0027s the quantity of heat,"},{"Start":"02:50.120 ","End":"02:51.785","Text":"the q of the copper."},{"Start":"02:51.785 ","End":"02:54.380","Text":"Now we\u0027re adding the quantity of heat of the water."},{"Start":"02:54.380 ","End":"02:57.200","Text":"The quantity of heat of the water is the mass of the water."},{"Start":"02:57.200 ","End":"03:00.770","Text":"Now, remember the water is given in volume."},{"Start":"03:00.770 ","End":"03:03.170","Text":"The volume of the water equals 500 milliliters."},{"Start":"03:03.170 ","End":"03:06.230","Text":"We\u0027re just quickly going to calculate the mass."},{"Start":"03:06.230 ","End":"03:09.919","Text":"Remember that d, the density,"},{"Start":"03:09.919 ","End":"03:11.720","Text":"equals m divided by v,"},{"Start":"03:11.720 ","End":"03:14.060","Text":"which is the mass divided by the volume."},{"Start":"03:14.060 ","End":"03:15.320","Text":"Here we\u0027re looking for the mass, we\u0027re going to"},{"Start":"03:15.320 ","End":"03:17.480","Text":"multiply both sides by the volume, so m,"},{"Start":"03:17.480 ","End":"03:22.645","Text":"the mass equals d times v, density times volume."},{"Start":"03:22.645 ","End":"03:27.590","Text":"In our case, the mass of the water equals the density of water,"},{"Start":"03:27.590 ","End":"03:30.485","Text":"which is 1 gram per milliliter,"},{"Start":"03:30.485 ","End":"03:32.690","Text":"times the volume of the water,"},{"Start":"03:32.690 ","End":"03:34.160","Text":"which is 500 milliliters,"},{"Start":"03:34.160 ","End":"03:36.619","Text":"so times 500 milliliters."},{"Start":"03:36.619 ","End":"03:41.640","Text":"The milliliters cancel out and we\u0027re left with 500 grams."},{"Start":"03:42.070 ","End":"03:48.005","Text":"The mass of the water is 500 grams."},{"Start":"03:48.005 ","End":"03:50.720","Text":"This is times the specific heat of the water,"},{"Start":"03:50.720 ","End":"03:53.735","Text":"which is 4.18 Joule per gram degrees Celsius"},{"Start":"03:53.735 ","End":"04:01.590","Text":"times Delta T. Again,"},{"Start":"04:01.590 ","End":"04:03.890","Text":"Delta T of the water is T final, which we don\u0027t know."},{"Start":"04:03.890 ","End":"04:05.240","Text":"That\u0027s what we\u0027re trying to find out,"},{"Start":"04:05.240 ","End":"04:07.720","Text":"minus the initial temperature of the water,"},{"Start":"04:07.720 ","End":"04:09.870","Text":"which is 35 degrees Celsius,"},{"Start":"04:09.870 ","End":"04:12.734","Text":"so minus 35 degrees Celsius."},{"Start":"04:12.734 ","End":"04:15.405","Text":"All of this equals 0."},{"Start":"04:15.405 ","End":"04:18.050","Text":"We can see that the grams cancel out here."},{"Start":"04:18.050 ","End":"04:19.580","Text":"We\u0027re multiplying in here,"},{"Start":"04:19.580 ","End":"04:20.926","Text":"they also cancel out."},{"Start":"04:20.926 ","End":"04:23.210","Text":"Now we will begin with multiplication."},{"Start":"04:23.210 ","End":"04:31.010","Text":"This comes out to 327.25 Joule divided by degrees Celsius because that\u0027s"},{"Start":"04:31.010 ","End":"04:32.360","Text":"what we have leftover times"},{"Start":"04:32.360 ","End":"04:41.630","Text":"the T final minus 18,980.5."},{"Start":"04:41.630 ","End":"04:42.980","Text":"In here, in our second component,"},{"Start":"04:42.980 ","End":"04:47.525","Text":"we\u0027re only left with Joules because the degrees Celsius and degrees Celsius cancel out."},{"Start":"04:47.525 ","End":"04:56.240","Text":"Plus 2090 Joule per degrees Celsius times Tf minus"},{"Start":"04:56.240 ","End":"05:01.830","Text":"73,150 Joule."},{"Start":"05:01.830 ","End":"05:05.010","Text":"All of this equals 0. We\u0027re going,"},{"Start":"05:05.010 ","End":"05:07.770","Text":"to sum up, the components of the T final,"},{"Start":"05:07.770 ","End":"05:14.575","Text":"and this comes to 2417.25"},{"Start":"05:14.575 ","End":"05:19.175","Text":"Joule divided by degrees Celsius times Tf."},{"Start":"05:19.175 ","End":"05:22.140","Text":"On the other side, we have the Joule components,"},{"Start":"05:22.140 ","End":"05:30.720","Text":"which comes out to 92,130 Joule."},{"Start":"05:30.720 ","End":"05:32.890","Text":"The Joules cancel out,"},{"Start":"05:32.990 ","End":"05:41.045","Text":"and Tf comes out to 38.11 degrees Celsius."},{"Start":"05:41.045 ","End":"05:44.390","Text":"This is approximately 38 degrees Celsius."},{"Start":"05:44.390 ","End":"05:49.760","Text":"The final temperature that we calculated equals 38 degrees Celsius."},{"Start":"05:49.760 ","End":"05:51.020","Text":"That is our final answer."},{"Start":"05:51.020 ","End":"05:53.700","Text":"Thank you very much for watching."}],"ID":34065},{"Watched":false,"Name":"Exercise 4","Duration":"5m 6s","ChapterTopicVideoID":22965,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.535","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:02.535 ","End":"00:08.940","Text":"In 128.5 grams sample of magnesium at 127.9 degrees Celsius is"},{"Start":"00:08.940 ","End":"00:15.525","Text":"added to 236 grams of water at 25 degrees Celsius in an insulated vessel."},{"Start":"00:15.525 ","End":"00:19.410","Text":"The final temperature equals 37.1 degrees Celsius."},{"Start":"00:19.410 ","End":"00:25.080","Text":"The specific heat capacity of water equals 4.18 joule per gram degrees Celsius."},{"Start":"00:25.080 ","End":"00:27.915","Text":"What is the molar heat capacity of magnesium?"},{"Start":"00:27.915 ","End":"00:30.750","Text":"To find the molar heat capacity of magnesium first,"},{"Start":"00:30.750 ","End":"00:34.090","Text":"we\u0027re going to find the specific heat of magnesium."},{"Start":"00:34.090 ","End":"00:36.649","Text":"For this purpose, we\u0027re going to use the equation Q,"},{"Start":"00:36.649 ","End":"00:42.470","Text":"the quantity of heat equals ms Delta T m is the mass of the substance,"},{"Start":"00:42.470 ","End":"00:44.915","Text":"s is the specific heat of the substance instead,"},{"Start":"00:44.915 ","End":"00:47.860","Text":"T equals T final minus T initial,"},{"Start":"00:47.860 ","End":"00:49.940","Text":"and that\u0027s the change in temperature."},{"Start":"00:49.940 ","End":"00:53.570","Text":"We know that the quantity of heat which is lost by the magnesium,"},{"Start":"00:53.570 ","End":"00:54.920","Text":"is absorbed by the water."},{"Start":"00:54.920 ","End":"01:02.875","Text":"Therefore, q of magnesium plus q of water equals 0."},{"Start":"01:02.875 ","End":"01:04.560","Text":"If we look at again,"},{"Start":"01:04.560 ","End":"01:05.940","Text":"q equals ms Delta T,"},{"Start":"01:05.940 ","End":"01:07.665","Text":"so q of magnesium."},{"Start":"01:07.665 ","End":"01:12.890","Text":"The mass of magnesium times the specific heat of magnesium times the Delta T,"},{"Start":"01:12.890 ","End":"01:15.065","Text":"which is a change in temperature of the magnesium,"},{"Start":"01:15.065 ","End":"01:17.210","Text":"plus the quantity of heat of water,"},{"Start":"01:17.210 ","End":"01:19.460","Text":"which is close the mass of water,"},{"Start":"01:19.460 ","End":"01:21.740","Text":"times the specific heat of water,"},{"Start":"01:21.740 ","End":"01:26.540","Text":"times the change in temperature of water equals 0."},{"Start":"01:26.540 ","End":"01:32.735","Text":"We can see here that the mass of the magnesium equals 128.5 grams."},{"Start":"01:32.735 ","End":"01:39.230","Text":"So 128.5 grams times the specific heat of magnesium,"},{"Start":"01:39.230 ","End":"01:41.210","Text":"which is what we\u0027re looking for,"},{"Start":"01:41.210 ","End":"01:44.180","Text":"so times the specific heat of magnesium,"},{"Start":"01:44.180 ","End":"01:47.125","Text":"times the change in temperature and magnesium."},{"Start":"01:47.125 ","End":"01:50.350","Text":"It\u0027s T final minus T initial."},{"Start":"01:50.350 ","End":"01:54.650","Text":"We know that the final temperature is 37.1 degrees Celsius and"},{"Start":"01:54.650 ","End":"01:59.240","Text":"the initial temperature of the magnesium is 127.9 degrees Celsius."},{"Start":"01:59.240 ","End":"02:07.594","Text":"Times 37.1 degrees Celsius minus 127.9 degrees Celsius."},{"Start":"02:07.594 ","End":"02:10.700","Text":"That\u0027s the quantity of the heat of the magnesium."},{"Start":"02:10.700 ","End":"02:12.950","Text":"We\u0027re going to add to this the quantity of heat of the water,"},{"Start":"02:12.950 ","End":"02:14.555","Text":"which is the mass of the water,"},{"Start":"02:14.555 ","End":"02:18.000","Text":"which equals 236 grams,"},{"Start":"02:19.090 ","End":"02:21.590","Text":"times the specific heat of water,"},{"Start":"02:21.590 ","End":"02:27.850","Text":"which is 4.18 joule per gram degrees Celsius,"},{"Start":"02:28.020 ","End":"02:30.820","Text":"times the change in temperature of the water,"},{"Start":"02:30.820 ","End":"02:36.500","Text":"which is 37.1 degrees Celsius minus 25 degrees Celsius."},{"Start":"02:41.840 ","End":"02:45.345","Text":"All of this equals 0."},{"Start":"02:45.345 ","End":"02:50.050","Text":"After multiplying, we can see that we have minus"},{"Start":"02:50.050 ","End":"02:58.225","Text":"11,667.8 gram degrees Celsius times"},{"Start":"02:58.225 ","End":"03:00.985","Text":"the specific heat of the magnesium."},{"Start":"03:00.985 ","End":"03:08.705","Text":"You see the units are grams times degrees Celsius plus the quantity of heat of the water,"},{"Start":"03:08.705 ","End":"03:15.050","Text":"which comes to 11,936.41 joules."},{"Start":"03:15.050 ","End":"03:17.960","Text":"This equals 0. We\u0027re going to move"},{"Start":"03:17.960 ","End":"03:20.720","Text":"the component with a specific heat to the other side of the equation,"},{"Start":"03:20.720 ","End":"03:27.950","Text":"and we get 11,936.41 joule equals"},{"Start":"03:27.950 ","End":"03:38.495","Text":"11,667.8 grams degrees Celsius times the specific heat of the magnesium."},{"Start":"03:38.495 ","End":"03:42.155","Text":"The specific heat of the magnesium equals,"},{"Start":"03:42.155 ","End":"03:49.110","Text":"after dividing we can see 1.023 joule per gram degrees Celsius."},{"Start":"03:49.130 ","End":"03:53.915","Text":"That\u0027s the specific heat of the magnesium in joule per gram degrees Celsius."},{"Start":"03:53.915 ","End":"03:58.790","Text":"However, we were asked to find the molar heat capacity of the magnesium."},{"Start":"03:58.790 ","End":"04:01.520","Text":"Right now we have the specific heat capacity of the magnesium"},{"Start":"04:01.520 ","End":"04:04.745","Text":"and we want to find the molar heat capacity of the magnesium."},{"Start":"04:04.745 ","End":"04:07.655","Text":"To find the molar heat capacity of the magnesium,"},{"Start":"04:07.655 ","End":"04:11.075","Text":"we\u0027re going to call it C molar."},{"Start":"04:11.075 ","End":"04:18.845","Text":"This equals the specific heat of the magnesium times the molar mass of the magnesium."},{"Start":"04:18.845 ","End":"04:25.040","Text":"This equals 1.023 joule per gram"},{"Start":"04:25.040 ","End":"04:33.950","Text":"degrees Celsius times 24.31 gram per mole."},{"Start":"04:33.950 ","End":"04:36.470","Text":"That\u0027s the molar mass of magnesium,"},{"Start":"04:36.470 ","End":"04:38.215","Text":"so the grams cancel out."},{"Start":"04:38.215 ","End":"04:48.080","Text":"This equals 24.87 joule per mole degrees Celsius."},{"Start":"04:48.080 ","End":"04:54.425","Text":"That\u0027s the molar heat capacity of the magnesium, which we found."},{"Start":"04:54.425 ","End":"05:02.510","Text":"The molar heat capacity of the magnesium equals 24.87 joule per mole degrees Celsius."},{"Start":"05:02.510 ","End":"05:04.145","Text":"That is our final answer."},{"Start":"05:04.145 ","End":"05:06.750","Text":"Thank you very much for watching."}],"ID":34066},{"Watched":false,"Name":"Exercise 5","Duration":"2m 28s","ChapterTopicVideoID":22966,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.240","Text":"Hi, we\u0027re going to solve the following exercise."},{"Start":"00:03.240 ","End":"00:07.260","Text":"125.4 kilojoules in the form of heat is transferred to"},{"Start":"00:07.260 ","End":"00:12.535","Text":"a 2-liter sample of water density equals 1 gram per mil at 25 degrees Celsius."},{"Start":"00:12.535 ","End":"00:15.750","Text":"The final temperature of the water is 40 degrees Celsius."},{"Start":"00:15.750 ","End":"00:19.080","Text":"What is the heat capacity of water in joule per Kelvin?"},{"Start":"00:19.080 ","End":"00:22.440","Text":"I want you to note that here we need to find the heat capacity,"},{"Start":"00:22.440 ","End":"00:23.910","Text":"not specific heat capacity."},{"Start":"00:23.910 ","End":"00:26.370","Text":"Therefore, we\u0027re going to use the equation q,"},{"Start":"00:26.370 ","End":"00:31.620","Text":"the amount of heat equals c Delta T. C is the heat capacity,"},{"Start":"00:31.620 ","End":"00:34.045","Text":"Delta T is the difference in the temperature,"},{"Start":"00:34.045 ","End":"00:38.285","Text":"which is the final temperature minus the initial temperature."},{"Start":"00:38.285 ","End":"00:40.310","Text":"Since we\u0027re looking for c, the heat capacity,"},{"Start":"00:40.310 ","End":"00:43.250","Text":"we\u0027re going to divide both sides by Delta T. This is c,"},{"Start":"00:43.250 ","End":"00:51.260","Text":"the heat capacity equals q divided by Delta T. Q is the amount of heat which equals"},{"Start":"00:51.260 ","End":"01:00.430","Text":"125.4 kilojoules divided by Delta T,"},{"Start":"01:00.430 ","End":"01:01.850","Text":"which is the difference in temperature,"},{"Start":"01:01.850 ","End":"01:04.050","Text":"which equals T final minus T initial."},{"Start":"01:04.050 ","End":"01:06.140","Text":"The final temperature of water is 40 degrees"},{"Start":"01:06.140 ","End":"01:12.380","Text":"Celsius minus the initial temperature of the water,"},{"Start":"01:12.380 ","End":"01:15.120","Text":"which is 25 degrees Celsius."},{"Start":"01:16.400 ","End":"01:20.415","Text":"This equals 8.36,"},{"Start":"01:20.415 ","End":"01:26.360","Text":"and if we look at our units, we have kilojoules per degrees Celsius."},{"Start":"01:26.360 ","End":"01:30.130","Text":"Now, we want the heat capacity to be in joules per Kelvin."},{"Start":"01:30.130 ","End":"01:32.380","Text":"We want our heat capacity to be in joules per Kelvin."},{"Start":"01:32.380 ","End":"01:34.120","Text":"First of all, we\u0027re going to take"},{"Start":"01:34.120 ","End":"01:36.985","Text":"our kilojoules and we\u0027re going to multiply it by a conversion factor."},{"Start":"01:36.985 ","End":"01:45.795","Text":"There are 1,000 joules in 1 kilojoule,"},{"Start":"01:45.795 ","End":"01:47.835","Text":"the kilojoules cancel out,"},{"Start":"01:47.835 ","End":"01:52.680","Text":"and this equals 8.36 times 1,000,"},{"Start":"01:52.680 ","End":"01:56.789","Text":"which is 10^3 joules per degree Celsius."},{"Start":"01:56.789 ","End":"02:00.450","Text":"Now, since here the degrees Celsius came from our Delta T,"},{"Start":"02:00.450 ","End":"02:01.905","Text":"the difference in temperature,"},{"Start":"02:01.905 ","End":"02:04.460","Text":"we can change our degrees Celsius to Kelvin,"},{"Start":"02:04.460 ","End":"02:06.110","Text":"since we\u0027re talking about the difference in temperature,"},{"Start":"02:06.110 ","End":"02:08.305","Text":"1 Kelvin equals 1 degree Celsius."},{"Start":"02:08.305 ","End":"02:15.775","Text":"This equals 8.36 times 10^3 joule per Kelvin."},{"Start":"02:15.775 ","End":"02:19.368","Text":"The heat capacity that we found equals, again,"},{"Start":"02:19.368 ","End":"02:24.435","Text":"8.36 times 10^3 joules per Kelvin."},{"Start":"02:24.435 ","End":"02:25.890","Text":"That is our final answer,"},{"Start":"02:25.890 ","End":"02:28.360","Text":"thank you very much for watching."}],"ID":34067},{"Watched":false,"Name":"Bomb Calorimeter","Duration":"7m 2s","ChapterTopicVideoID":18673,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"In the previous video,"},{"Start":"00:01.650 ","End":"00:03.854","Text":"we talked about heats of reaction."},{"Start":"00:03.854 ","End":"00:07.920","Text":"In this video, we\u0027ll describe how to measure the heat of reaction of"},{"Start":"00:07.920 ","End":"00:12.390","Text":"a combustion reaction using a bomb calorimeter."},{"Start":"00:12.390 ","End":"00:15.660","Text":"We\u0027re going to talk about a bomb calorimeter."},{"Start":"00:15.660 ","End":"00:17.610","Text":"Here\u0027s a picture of it."},{"Start":"00:17.610 ","End":"00:20.070","Text":"We have a steel container,"},{"Start":"00:20.070 ","End":"00:22.665","Text":"very strong steel container."},{"Start":"00:22.665 ","End":"00:24.900","Text":"Inside we have water,"},{"Start":"00:24.900 ","End":"00:28.470","Text":"here is the water and here is the bomb itself."},{"Start":"00:28.470 ","End":"00:32.550","Text":"It\u0027s made of steel and it\u0027s very strong."},{"Start":"00:32.550 ","End":"00:37.290","Text":"Inside we place the substance we want to burn."},{"Start":"00:37.290 ","End":"00:39.420","Text":"We\u0027re talking about combustion reaction."},{"Start":"00:39.420 ","End":"00:42.550","Text":"The substances we will want to burn is placed here."},{"Start":"00:42.550 ","End":"00:45.125","Text":"Inside the bomb there\u0027s oxygen,"},{"Start":"00:45.125 ","End":"00:48.500","Text":"so that the combustion reaction can take place."},{"Start":"00:48.500 ","End":"00:52.285","Text":"When the reaction takes place, heat is emitted,"},{"Start":"00:52.285 ","End":"00:56.630","Text":"comes very hot and surrounded by water,"},{"Start":"00:56.630 ","End":"01:00.155","Text":"which is stirred, so you get to a uniform temperature."},{"Start":"01:00.155 ","End":"01:05.000","Text":"Of course, there is a thermometer to measure how high the temperature goes."},{"Start":"01:05.000 ","End":"01:07.910","Text":"There\u0027s a wire for igniting the reaction,"},{"Start":"01:07.910 ","End":"01:09.860","Text":"for starting off the reaction."},{"Start":"01:09.860 ","End":"01:13.205","Text":"How can we work out the heat of reaction?"},{"Start":"01:13.205 ","End":"01:18.050","Text":"The first thing we worked out is the heat gained by the calorimeter."},{"Start":"01:18.050 ","End":"01:24.355","Text":"We need everything except the reaction itself."},{"Start":"01:24.355 ","End":"01:26.815","Text":"Everything surrounding the bomb,"},{"Start":"01:26.815 ","End":"01:29.975","Text":"the water, the outer case, everything."},{"Start":"01:29.975 ","End":"01:33.560","Text":"We\u0027ll work that out by multiplying the heat capacity,"},{"Start":"01:33.560 ","End":"01:37.739","Text":"which is something that has to be measured for the steel bomb times"},{"Start":"01:37.739 ","End":"01:43.915","Text":"the rise in temperature delta t. The final temperature minus the initial temperature."},{"Start":"01:43.915 ","End":"01:46.175","Text":"Once we know that,"},{"Start":"01:46.175 ","End":"01:50.735","Text":"we can work out the heat of reaction,"},{"Start":"01:50.735 ","End":"01:55.100","Text":"because that\u0027s equal to minus q of the calorimeter."},{"Start":"01:55.100 ","End":"01:58.115","Text":"This is a closed isolated system."},{"Start":"01:58.115 ","End":"01:59.735","Text":"All the heats inside."},{"Start":"01:59.735 ","End":"02:02.165","Text":"It doesn\u0027t heat up the environment."},{"Start":"02:02.165 ","End":"02:04.160","Text":"Let\u0027s take an example."},{"Start":"02:04.160 ","End":"02:08.210","Text":"The combustion of 1 gram of sucrose, sugar."},{"Start":"02:08.210 ","End":"02:12.680","Text":"We talked about the combustion of sucrose very early videos."},{"Start":"02:12.680 ","End":"02:17.090","Text":"In a bomb calorimeter causes the temperature of the water to rise from"},{"Start":"02:17.090 ","End":"02:21.985","Text":"25 degrees Celsius to 28.3 degrees Celsius."},{"Start":"02:21.985 ","End":"02:29.225","Text":"The heat capacity of the calorimeter is 5.00 kilojoules per degrees Celsius."},{"Start":"02:29.225 ","End":"02:35.470","Text":"Calculate the heat of combustion of sucrose per gram and also per mole."},{"Start":"02:35.470 ","End":"02:42.650","Text":"We begin by working out the amount of heat absorbed by the calorimeter."},{"Start":"02:42.650 ","End":"02:47.995","Text":"That\u0027s equal to the heat capacity times the rise in temperature."},{"Start":"02:47.995 ","End":"02:54.500","Text":"That\u0027s 5.00 kilojoules per degrees Celsius, that\u0027s heat capacity."},{"Start":"02:54.500 ","End":"02:58.100","Text":"Then we multiply that by the change in temperature."},{"Start":"02:58.100 ","End":"03:01.155","Text":"The final temperature was 28.3,"},{"Start":"03:01.155 ","End":"03:03.700","Text":"the initial temperature is 25.0,"},{"Start":"03:03.700 ","End":"03:06.560","Text":"so here\u0027s the difference between the two,"},{"Start":"03:06.560 ","End":"03:13.335","Text":"gives us delta t. We can see that\u0027s 3.3 degrees Celsius."},{"Start":"03:13.335 ","End":"03:15.575","Text":"Now, we multiply out,"},{"Start":"03:15.575 ","End":"03:19.415","Text":"degrees Celsius the power minus 1 times degrees Celsius."},{"Start":"03:19.415 ","End":"03:22.805","Text":"It\u0027s just 1 and we\u0027re left just with kilojoules."},{"Start":"03:22.805 ","End":"03:24.080","Text":"We multiply the numbers,"},{"Start":"03:24.080 ","End":"03:27.065","Text":"we get 16.5 kilojoules."},{"Start":"03:27.065 ","End":"03:33.110","Text":"That means the heat of reaction is equal to minus q of the calorimeter,"},{"Start":"03:33.110 ","End":"03:36.785","Text":"so that\u0027s minus 16.5 kilojoules."},{"Start":"03:36.785 ","End":"03:41.675","Text":"Now, we saw that in the question that we only have 1 gram of sucrose."},{"Start":"03:41.675 ","End":"03:46.365","Text":"That\u0027s 16.5 kilojoules per gram."},{"Start":"03:46.365 ","End":"03:47.730","Text":"If we had 2 grams,"},{"Start":"03:47.730 ","End":"03:49.560","Text":"that would be twice as much."},{"Start":"03:49.560 ","End":"03:55.609","Text":"Now, we have to work out what the heat of reaction is per mole."},{"Start":"03:55.609 ","End":"03:57.425","Text":"In order to do this,"},{"Start":"03:57.425 ","End":"04:01.040","Text":"we need to know the molar mass of sucrose and we can easily"},{"Start":"04:01.040 ","End":"04:05.735","Text":"work out that\u0027s 342.3 grams per mole."},{"Start":"04:05.735 ","End":"04:10.370","Text":"Now, we can calculate the heat of reaction per mole."},{"Start":"04:10.370 ","End":"04:13.595","Text":"Here\u0027s the heat of reaction per gram,"},{"Start":"04:13.595 ","End":"04:20.810","Text":"and we multiply it by the molar mass which is 342.3 grams per mole."},{"Start":"04:20.810 ","End":"04:22.370","Text":"What about the units grams,"},{"Start":"04:22.370 ","End":"04:25.550","Text":"the power minus 1 times grams, just 1."},{"Start":"04:25.550 ","End":"04:28.465","Text":"We\u0027re left with kilojoules per mole."},{"Start":"04:28.465 ","End":"04:37.190","Text":"When we work out the numbers we get minus 5.65 times 10^3 kilojoules per mole."},{"Start":"04:37.190 ","End":"04:40.325","Text":"We can see from the negative sign,"},{"Start":"04:40.325 ","End":"04:44.520","Text":"that this is an exothermic reaction."},{"Start":"04:48.460 ","End":"04:53.090","Text":"It produces heat, so it\u0027s exothermic."},{"Start":"04:53.090 ","End":"05:01.215","Text":"Another thing we need to note that this was all done in a steel bomb of fixed volumes."},{"Start":"05:01.215 ","End":"05:05.930","Text":"This was an exothermic reaction and we have constant volume."},{"Start":"05:05.930 ","End":"05:09.800","Text":"Later on, we\u0027ll see the difference between working out the heat"},{"Start":"05:09.800 ","End":"05:13.880","Text":"of reaction at constant volume and constant pressure."},{"Start":"05:13.880 ","End":"05:19.520","Text":"Now, bomb calorimeters are not something you\u0027ll find very easily around labs."},{"Start":"05:19.520 ","End":"05:26.375","Text":"They usually confined to government laboratories where they work out standards."},{"Start":"05:26.375 ","End":"05:33.005","Text":"What you\u0027re more likely to see in your chemistry lab is a coffee cup calorimeter."},{"Start":"05:33.005 ","End":"05:38.075","Text":"It\u0027s made of Styrofoam coffee cups."},{"Start":"05:38.075 ","End":"05:40.055","Text":"Usually 2 of them,"},{"Start":"05:40.055 ","End":"05:45.500","Text":"1 inside the other to make sure that heat isn\u0027t lost to the environment."},{"Start":"05:45.500 ","End":"05:56.075","Text":"Usually, there is a cork lid on it also to make sure that heat doesn\u0027t get lost."},{"Start":"05:56.075 ","End":"05:59.930","Text":"Then inside through this cork lid,"},{"Start":"05:59.930 ","End":"06:04.535","Text":"usually have a thermometer and a stirrer."},{"Start":"06:04.535 ","End":"06:09.470","Text":"Lots of assumptions are made here."},{"Start":"06:09.470 ","End":"06:13.850","Text":"Usually, it\u0027s just assumed that the temperature of the water,"},{"Start":"06:13.850 ","End":"06:18.145","Text":"that only the water rises in temperature there\u0027s water."},{"Start":"06:18.145 ","End":"06:24.475","Text":"This calorimeter is only used for aqueous solutions."},{"Start":"06:24.475 ","End":"06:28.580","Text":"No gas is released as in a combustion gas and"},{"Start":"06:28.580 ","End":"06:33.270","Text":"we can assume that it\u0027s at constant pressure."},{"Start":"06:36.050 ","End":"06:38.990","Text":"As we\u0027re only dealing with aqueous solutions,"},{"Start":"06:38.990 ","End":"06:45.690","Text":"it\u0027s probably nearly constant volume."},{"Start":"06:48.520 ","End":"06:55.670","Text":"This is what you\u0027re likely to meet in your chemistry lab rather than a bomb calorimeter."},{"Start":"06:55.670 ","End":"07:02.220","Text":"In this video, we described how to use calorimeters to measure heats of reaction."}],"ID":34068},{"Watched":false,"Name":"Exercise 6","Duration":"4m 46s","ChapterTopicVideoID":31833,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34069},{"Watched":false,"Name":"Exercise 7","Duration":"9m 36s","ChapterTopicVideoID":31834,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34070},{"Watched":false,"Name":"Exercise 8","Duration":"3m 39s","ChapterTopicVideoID":31835,"CourseChapterTopicPlaylistID":81299,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34071}],"Thumbnail":null,"ID":81299},{"Name":"Work","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Pressure-Volume Work","Duration":"6m 13s","ChapterTopicVideoID":18670,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.720","Text":"The previous video, we talked about energy and work."},{"Start":"00:03.720 ","End":"00:10.110","Text":"In this video, we\u0027ve discuss a particular kind of work called pressure volume work."},{"Start":"00:10.110 ","End":"00:12.240","Text":"There are many types of work."},{"Start":"00:12.240 ","End":"00:14.865","Text":"But in thermochemistry and thermodynamics,"},{"Start":"00:14.865 ","End":"00:18.270","Text":"the most common type is pressure volume work."},{"Start":"00:18.270 ","End":"00:20.775","Text":"Let\u0027s try to explain what it means."},{"Start":"00:20.775 ","End":"00:26.925","Text":"Supposing we have a cylinder of area, the base is A."},{"Start":"00:26.925 ","End":"00:30.945","Text":"In this cylinder, there is a gas."},{"Start":"00:30.945 ","End":"00:38.940","Text":"We have a piston that somebody can move freely with a pressure on the piston."},{"Start":"00:38.940 ","End":"00:43.695","Text":"Now, the height of the gas we\u0027ll write as h_i."},{"Start":"00:43.695 ","End":"00:46.005","Text":"The volume V_i,"},{"Start":"00:46.005 ","End":"00:50.415","Text":"which is just the height, times the area."},{"Start":"00:50.415 ","End":"00:53.805","Text":"Supposing we hit the gas."},{"Start":"00:53.805 ","End":"01:02.825","Text":"Now, the piston rises because the volume of the gas changes."},{"Start":"01:02.825 ","End":"01:12.530","Text":"The new volume is V_f and the new height is h_f."},{"Start":"01:12.530 ","End":"01:14.915","Text":"V_f=h_f times A."},{"Start":"01:14.915 ","End":"01:19.670","Text":"The difference in height between h_f and h_i,"},{"Start":"01:19.670 ","End":"01:28.595","Text":"we\u0027ll call Delta h. If we draw it here,"},{"Start":"01:28.595 ","End":"01:35.165","Text":"this is Delta h. The gas has done work."},{"Start":"01:35.165 ","End":"01:39.080","Text":"It\u0027s moved this piston further up,"},{"Start":"01:39.080 ","End":"01:41.600","Text":"so it\u0027s done work on the piston."},{"Start":"01:41.600 ","End":"01:44.660","Text":"We could write that the change in volume is"},{"Start":"01:44.660 ","End":"01:48.050","Text":"equal to the final volume minus initial volume,"},{"Start":"01:48.050 ","End":"01:52.715","Text":"which is the area times the difference in height."},{"Start":"01:52.715 ","End":"01:54.605","Text":"That\u0027s our first equation."},{"Start":"01:54.605 ","End":"01:59.795","Text":"Now we said before the previous videos that work is equal to force."},{"Start":"01:59.795 ","End":"02:03.890","Text":"The force acting times the distance moved."},{"Start":"02:03.890 ","End":"02:11.310","Text":"In this case, the distance moved is Delta h. We went from h_i to h_f."},{"Start":"02:11.310 ","End":"02:13.060","Text":"It\u0027s Delta h,"},{"Start":"02:13.060 ","End":"02:17.660","Text":"so w is equal to F times Delta h. When we talked about gases,"},{"Start":"02:17.660 ","End":"02:22.145","Text":"we defined the pressure as a force divided by the area."},{"Start":"02:22.145 ","End":"02:25.760","Text":"We can multiply both sides by A and write that"},{"Start":"02:25.760 ","End":"02:29.895","Text":"the force is equal to the pressure times the area."},{"Start":"02:29.895 ","End":"02:32.285","Text":"Well, now we could write, again,"},{"Start":"02:32.285 ","End":"02:38.645","Text":"work is equal to force times Delta h. Instead of writing the force,"},{"Start":"02:38.645 ","End":"02:41.165","Text":"we could write P times A."},{"Start":"02:41.165 ","End":"02:45.510","Text":"Instead of force, we\u0027re writing P times A."},{"Start":"02:45.510 ","End":"02:52.620","Text":"We have P times A times Delta h. That\u0027s work is equal to P times A times"},{"Start":"02:52.620 ","End":"03:00.199","Text":"Delta h. We can take A times Delta h as difference in the volume,"},{"Start":"03:00.199 ","End":"03:06.755","Text":"so now we have P times Delta V. This is what we call pressure volume work,"},{"Start":"03:06.755 ","End":"03:10.100","Text":"your pressure times the change in volume."},{"Start":"03:10.100 ","End":"03:11.720","Text":"Now if we look at this,"},{"Start":"03:11.720 ","End":"03:17.120","Text":"we see that Delta V is greater than 0 because our volume increased."},{"Start":"03:17.120 ","End":"03:20.930","Text":"The sides, there are lots of conventions."},{"Start":"03:20.930 ","End":"03:23.960","Text":"We decide what side something has to have,"},{"Start":"03:23.960 ","End":"03:27.695","Text":"and that everything is treated in a consistent way."},{"Start":"03:27.695 ","End":"03:34.475","Text":"By convention, w is supposed to be negative when the system does work."},{"Start":"03:34.475 ","End":"03:37.245","Text":"We need something negative."},{"Start":"03:37.245 ","End":"03:39.100","Text":"In order to make it negative,"},{"Start":"03:39.100 ","End":"03:44.185","Text":"we write a minus sign in front of P times Delta V."},{"Start":"03:44.185 ","End":"03:50.215","Text":"The external here is just written because this is a force acting from the outside."},{"Start":"03:50.215 ","End":"03:52.600","Text":"It\u0027s external force."},{"Start":"03:52.600 ","End":"03:55.675","Text":"Now our work is negative."},{"Start":"03:55.675 ","End":"03:57.805","Text":"Now, what about the units?"},{"Start":"03:57.805 ","End":"03:59.080","Text":"In the previous video,"},{"Start":"03:59.080 ","End":"04:02.320","Text":"we showed that the units of work are joules or"},{"Start":"04:02.320 ","End":"04:06.175","Text":"kilograms times meter squared per second squared."},{"Start":"04:06.175 ","End":"04:10.195","Text":"But suppose you were given the pressure in bars or atmospheres,"},{"Start":"04:10.195 ","End":"04:12.290","Text":"and the volume in liters,"},{"Start":"04:12.290 ","End":"04:16.220","Text":"so then we have pressure times volume."},{"Start":"04:16.220 ","End":"04:19.085","Text":"We have bars times liters,"},{"Start":"04:19.085 ","End":"04:20.705","Text":"or atmospheres times liter."},{"Start":"04:20.705 ","End":"04:23.899","Text":"We need to convert that into joules."},{"Start":"04:23.899 ","End":"04:28.550","Text":"If we go back and recall what we learned about gases,"},{"Start":"04:28.550 ","End":"04:33.685","Text":"we learned there that 1 bar times liter is 100 joules,"},{"Start":"04:33.685 ","End":"04:39.570","Text":"and 1 atmosphere times a liter is 101.325 joules."},{"Start":"04:39.570 ","End":"04:41.700","Text":"They\u0027re very close."},{"Start":"04:41.700 ","End":"04:44.015","Text":"Let\u0027s solve a little problem."},{"Start":"04:44.015 ","End":"04:46.610","Text":"If the pressure on the piston is 1 atmosphere,"},{"Start":"04:46.610 ","End":"04:48.425","Text":"it\u0027s just atmospheric pressure,"},{"Start":"04:48.425 ","End":"04:54.755","Text":"and the volume changes by 10 milliliters when the gas is heated,"},{"Start":"04:54.755 ","End":"04:58.430","Text":"what is the work done by the gas?"},{"Start":"04:58.430 ","End":"05:00.230","Text":"Well, here\u0027s our expression."},{"Start":"05:00.230 ","End":"05:04.040","Text":"Work is equal to minus P external,"},{"Start":"05:04.040 ","End":"05:07.865","Text":"that\u0027s pressure exerted times Delta V,"},{"Start":"05:07.865 ","End":"05:16.265","Text":"and here\u0027s minus 1 atmosphere Delta V is 10^-2 liters,"},{"Start":"05:16.265 ","End":"05:19.520","Text":"because we have here 10 milliliters."},{"Start":"05:19.520 ","End":"05:26.285","Text":"10 milliliters is 10 times 10^-3, which is liters."},{"Start":"05:26.285 ","End":"05:30.940","Text":"That\u0027s 10^-2. Here,"},{"Start":"05:30.940 ","End":"05:34.700","Text":"1 atmosphere times 10^-2 liters,"},{"Start":"05:34.700 ","End":"05:39.710","Text":"which we could write as -10^-2 atmosphere liters."},{"Start":"05:39.710 ","End":"05:44.215","Text":"Now we want to convert atmosphere liter into joules."},{"Start":"05:44.215 ","End":"05:46.575","Text":"Here\u0027s our conversion factor,"},{"Start":"05:46.575 ","End":"05:51.750","Text":"101.325 joules is the same as 1 atmosphere times a liter."},{"Start":"05:51.750 ","End":"06:01.155","Text":"The atmosphere liter cancels and we multiply and we get -1.01 joules."},{"Start":"06:01.155 ","End":"06:07.745","Text":"The work done by the system is -1.01 joules."},{"Start":"06:07.745 ","End":"06:12.900","Text":"In this video, we learned about pressure volume work."}],"ID":19394},{"Watched":false,"Name":"Exercise 1","Duration":"2m 29s","ChapterTopicVideoID":31839,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34072},{"Watched":false,"Name":"Exercise 2","Duration":"3m 20s","ChapterTopicVideoID":31840,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34073},{"Watched":false,"Name":"Exercise 3","Duration":"4m 7s","ChapterTopicVideoID":31836,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34074},{"Watched":false,"Name":"Exercise 4","Duration":"1m 56s","ChapterTopicVideoID":31837,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34075},{"Watched":false,"Name":"Exercise 5","Duration":"2m 28s","ChapterTopicVideoID":31838,"CourseChapterTopicPlaylistID":81300,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34076}],"Thumbnail":null,"ID":81300},{"Name":"First Law of Thermodynamics","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"First Law of Thermodynamics","Duration":"5m 25s","ChapterTopicVideoID":18674,"CourseChapterTopicPlaylistID":81301,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.419","Text":"In previous videos, we learned that heat and work are forms of energy."},{"Start":"00:06.419 ","End":"00:09.480","Text":"In this video, we will discuss the effect of heat"},{"Start":"00:09.480 ","End":"00:12.900","Text":"and work on the internal energy of a system."},{"Start":"00:12.900 ","End":"00:15.015","Text":"What\u0027s the internal energy?"},{"Start":"00:15.015 ","End":"00:18.810","Text":"The internal energy, which we\u0027re going to write as U,"},{"Start":"00:18.810 ","End":"00:21.420","Text":"of an atomic or molecular system includes"},{"Start":"00:21.420 ","End":"00:24.870","Text":"all forms of energy: translational, vibrational,"},{"Start":"00:24.870 ","End":"00:30.150","Text":"rotational, the energy stored in the bonds and intermolecular interactions,"},{"Start":"00:30.150 ","End":"00:33.375","Text":"and even the energy of the electrons in the atoms,"},{"Start":"00:33.375 ","End":"00:35.570","Text":"and it can also be other things."},{"Start":"00:35.570 ","End":"00:39.380","Text":"Fortunately, only changes in the internal energy,"},{"Start":"00:39.380 ","End":"00:43.820","Text":"which we\u0027re going to call Delta U are as a result of heat and work,"},{"Start":"00:43.820 ","End":"00:46.250","Text":"are of interest in thermodynamics."},{"Start":"00:46.250 ","End":"00:48.310","Text":"We will only need Delta U,"},{"Start":"00:48.310 ","End":"00:52.355","Text":"we don\u0027t need to be concerned with all the other forms of energy."},{"Start":"00:52.355 ","End":"00:54.260","Text":"Now we\u0027re going to write the first law of"},{"Start":"00:54.260 ","End":"00:56.720","Text":"thermodynamics and we\u0027re going to write it according to"},{"Start":"00:56.720 ","End":"01:03.020","Text":"the convention set by the International Union of Pure and Applied Chemistry, IUPAC."},{"Start":"01:03.020 ","End":"01:07.310","Text":"Delta U, the change in internal energy is equal to q,"},{"Start":"01:07.310 ","End":"01:10.685","Text":"the heat, plus w, work."},{"Start":"01:10.685 ","End":"01:15.500","Text":"The internal energy changes when the system gains or loses heat,"},{"Start":"01:15.500 ","End":"01:19.310","Text":"or does work, or has work done on it."},{"Start":"01:19.310 ","End":"01:26.575","Text":"I\u0027m just going to note that some textbooks write the law as Delta U equal to q minus w,"},{"Start":"01:26.575 ","End":"01:28.745","Text":"so you should take note of that."},{"Start":"01:28.745 ","End":"01:32.480","Text":"Now we have to choose the signs of heat and work that are"},{"Start":"01:32.480 ","End":"01:36.985","Text":"consistent with how we\u0027ve written the first law of thermodynamics."},{"Start":"01:36.985 ","End":"01:38.990","Text":"This is very important."},{"Start":"01:38.990 ","End":"01:43.370","Text":"The signs have to be consistent with q plus"},{"Start":"01:43.370 ","End":"01:49.130","Text":"w. Let\u0027s assume first of all that q and w are positive,"},{"Start":"01:49.130 ","End":"01:50.630","Text":"q is greater than 0,"},{"Start":"01:50.630 ","End":"01:52.505","Text":"w is greater than 0."},{"Start":"01:52.505 ","End":"01:54.305","Text":"They\u0027re both positive,"},{"Start":"01:54.305 ","End":"01:57.160","Text":"and Delta U will also be positive."},{"Start":"01:57.160 ","End":"02:01.670","Text":"So q being positive means that the system absorbs"},{"Start":"02:01.670 ","End":"02:07.910","Text":"heat so that it increases Delta U so that Delta U will be positive."},{"Start":"02:07.910 ","End":"02:10.145","Text":"If w is positive,"},{"Start":"02:10.145 ","End":"02:13.220","Text":"that means that work is done on"},{"Start":"02:13.220 ","End":"02:18.770","Text":"the system because that will also increase the internal energy."},{"Start":"02:18.770 ","End":"02:21.810","Text":"We can draw a little diagram."},{"Start":"02:22.360 ","End":"02:25.160","Text":"Here\u0027s our system."},{"Start":"02:25.160 ","End":"02:28.325","Text":"If the system absorbs heat that way,"},{"Start":"02:28.325 ","End":"02:31.160","Text":"q is going into the system."},{"Start":"02:31.160 ","End":"02:34.265","Text":"When work is done on the system,"},{"Start":"02:34.265 ","End":"02:36.560","Text":"w is going into the system,"},{"Start":"02:36.560 ","End":"02:40.190","Text":"then q and w will be positive."},{"Start":"02:40.190 ","End":"02:44.860","Text":"Now, if q is negative and w is negative,"},{"Start":"02:44.860 ","End":"02:48.210","Text":"Delta U will also be negative."},{"Start":"02:48.210 ","End":"02:55.145","Text":"q being negative means the system loses heat so that Delta U decreases."},{"Start":"02:55.145 ","End":"02:57.570","Text":"Delta U will be negative."},{"Start":"02:57.570 ","End":"03:02.815","Text":"w less than 0 means that the work is done by the system."},{"Start":"03:02.815 ","End":"03:06.310","Text":"Once again, Delta U will be negative."},{"Start":"03:06.310 ","End":"03:07.870","Text":"We can draw a little picture."},{"Start":"03:07.870 ","End":"03:10.010","Text":"Here\u0027s our system."},{"Start":"03:10.040 ","End":"03:14.985","Text":"q going out the way losing heat,"},{"Start":"03:14.985 ","End":"03:17.760","Text":"w going out the way,"},{"Start":"03:17.760 ","End":"03:25.390","Text":"work being done by the system on the surroundings and then we have both of them negative."},{"Start":"03:25.390 ","End":"03:27.215","Text":"Let\u0027s take an example."},{"Start":"03:27.215 ","End":"03:34.325","Text":"If a gas loses 30 joules in heat and has 330 joules of work done on it,"},{"Start":"03:34.325 ","End":"03:36.250","Text":"what is Delta U?"},{"Start":"03:36.250 ","End":"03:40.610","Text":"We know that Delta U is q plus w. We have to choose"},{"Start":"03:40.610 ","End":"03:45.845","Text":"the correct signs for q and w. If it loses heat,"},{"Start":"03:45.845 ","End":"03:49.730","Text":"the sign of q will be negative minus 30 joules."},{"Start":"03:49.730 ","End":"03:51.485","Text":"If work is done on it,"},{"Start":"03:51.485 ","End":"03:56.925","Text":"the sign of the work will be positive plus 330 joules."},{"Start":"03:56.925 ","End":"04:00.710","Text":"The answer will be 330 minus 30,"},{"Start":"04:00.710 ","End":"04:03.500","Text":"which of course is 300 joules."},{"Start":"04:03.500 ","End":"04:08.585","Text":"It\u0027s very important to use the correct signs in order to get the correct answer."},{"Start":"04:08.585 ","End":"04:12.095","Text":"Now a few other points about the first law of thermodynamics."},{"Start":"04:12.095 ","End":"04:17.870","Text":"First of all, remember we divided the universe into a system and its surroundings."},{"Start":"04:17.870 ","End":"04:20.195","Text":"The system is everything we\u0027re interested in,"},{"Start":"04:20.195 ","End":"04:22.895","Text":"everything else is called surroundings."},{"Start":"04:22.895 ","End":"04:27.020","Text":"The first law is a form of the law of conservation of energy."},{"Start":"04:27.020 ","End":"04:30.185","Text":"The energy of the universe is constant,"},{"Start":"04:30.185 ","End":"04:34.100","Text":"doesn\u0027t change so Delta U of universe is 0."},{"Start":"04:34.100 ","End":"04:37.970","Text":"We said Delta U of the universe consisted"},{"Start":"04:37.970 ","End":"04:41.645","Text":"of Delta U of the system plus Delta U of surroundings."},{"Start":"04:41.645 ","End":"04:44.090","Text":"Now we have, if the sum is 0,"},{"Start":"04:44.090 ","End":"04:49.150","Text":"Delta U of the system will be equal to minus Delta U of the surroundings."},{"Start":"04:49.150 ","End":"04:51.395","Text":"Now if no work is done,"},{"Start":"04:51.395 ","End":"04:57.005","Text":"this will reduce to q of the system being equal to minus q of the surroundings."},{"Start":"04:57.005 ","End":"05:01.280","Text":"We used that in the video on heat."},{"Start":"05:01.280 ","End":"05:04.160","Text":"Supposing we have an isolated system."},{"Start":"05:04.160 ","End":"05:05.995","Text":"In an isolated system,"},{"Start":"05:05.995 ","End":"05:10.740","Text":"Delta U isolated system is 0 because no heat"},{"Start":"05:10.740 ","End":"05:16.010","Text":"can come in or out and no work could be done on the system or by the system."},{"Start":"05:16.010 ","End":"05:17.600","Text":"For an isolated system,"},{"Start":"05:17.600 ","End":"05:20.405","Text":"Delta U is 0."},{"Start":"05:20.405 ","End":"05:25.470","Text":"In this video, we discussed the first law of thermodynamics."}],"ID":19398},{"Watched":false,"Name":"State Functions and Path-Dependent Functions","Duration":"4m 55s","ChapterTopicVideoID":19122,"CourseChapterTopicPlaylistID":81301,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"In the previous video,"},{"Start":"00:02.145 ","End":"00:05.325","Text":"we studied the first law of thermodynamics."},{"Start":"00:05.325 ","End":"00:11.640","Text":"In this video, we will discuss state functions and path dependent functions."},{"Start":"00:11.640 ","End":"00:16.800","Text":"Let\u0027s begin by defining a thermodynamic state of a system."},{"Start":"00:16.800 ","End":"00:21.480","Text":"The thermodynamic state of a system refers to the temperature,"},{"Start":"00:21.480 ","End":"00:25.835","Text":"pressure, and quantity of substances present."},{"Start":"00:25.835 ","End":"00:28.910","Text":"It\u0027s the ID of the system."},{"Start":"00:28.910 ","End":"00:31.130","Text":"Now, what\u0027s a state function?"},{"Start":"00:31.130 ","End":"00:35.090","Text":"A state function only depends on these parameters."},{"Start":"00:35.090 ","End":"00:37.910","Text":"The parameters define the state,"},{"Start":"00:37.910 ","End":"00:40.924","Text":"and not how the state was reached."},{"Start":"00:40.924 ","End":"00:43.025","Text":"Let\u0027s take some examples."},{"Start":"00:43.025 ","End":"00:45.515","Text":"Simple examples are mass,"},{"Start":"00:45.515 ","End":"00:49.910","Text":"pressure, density, temperature, and volume."},{"Start":"00:49.910 ","End":"00:53.210","Text":"It doesn\u0027t matter how we reached any of these,"},{"Start":"00:53.210 ","End":"00:55.750","Text":"they are simply state functions."},{"Start":"00:55.750 ","End":"00:59.930","Text":"Now, what\u0027s very important for the first law of thermodynamics"},{"Start":"00:59.930 ","End":"01:04.925","Text":"is that the internal energy U of a system is a state function."},{"Start":"01:04.925 ","End":"01:06.845","Text":"Of course, Delta U,"},{"Start":"01:06.845 ","End":"01:09.740","Text":"which is U final minus U initial,"},{"Start":"01:09.740 ","End":"01:14.255","Text":"U_f minus U_i is a state function."},{"Start":"01:14.255 ","End":"01:18.065","Text":"It doesn\u0027t matter how we go from the initial to final state,"},{"Start":"01:18.065 ","End":"01:21.095","Text":"Delta U will have the same value."},{"Start":"01:21.095 ","End":"01:26.780","Text":"Now, we saw that U_f minus U_i is equal to Delta U."},{"Start":"01:26.780 ","End":"01:28.490","Text":"Let us draw a little diagram."},{"Start":"01:28.490 ","End":"01:32.355","Text":"Here is i, here\u0027s f,"},{"Start":"01:32.355 ","End":"01:36.695","Text":"the initial and the final state and go from one to the other,"},{"Start":"01:36.695 ","End":"01:38.840","Text":"we have Delta U,"},{"Start":"01:38.840 ","End":"01:42.155","Text":"the difference in internal energy."},{"Start":"01:42.155 ","End":"01:47.940","Text":"Now supposing we now are in state f,"},{"Start":"01:47.940 ","End":"01:49.230","Text":"we got to state f,"},{"Start":"01:49.230 ","End":"01:51.825","Text":"we want to go back to state i,"},{"Start":"01:51.825 ","End":"01:55.535","Text":"we want to go from the final to the initial state."},{"Start":"01:55.535 ","End":"01:58.860","Text":"What will be the internal energy?"},{"Start":"01:58.860 ","End":"02:02.190","Text":"It\u0027ll be U_i, which is now the final state,"},{"Start":"02:02.190 ","End":"02:06.825","Text":"minus U_f, which is minus Delta U."},{"Start":"02:06.825 ","End":"02:10.795","Text":"In going from i to f and f to i,"},{"Start":"02:10.795 ","End":"02:16.445","Text":"the total change in internal energy is Delta U minus Delta U,"},{"Start":"02:16.445 ","End":"02:18.760","Text":"so it is precisely 0."},{"Start":"02:18.760 ","End":"02:20.060","Text":"It doesn\u0027t matter."},{"Start":"02:20.060 ","End":"02:24.770","Text":"We\u0027ll come back to the exact same value of the internal energy."},{"Start":"02:24.770 ","End":"02:26.780","Text":"Here I\u0027ve written it out mathematically,"},{"Start":"02:26.780 ","End":"02:34.865","Text":"U_f minus U_i is Delta U. U_i minus U_f is simply minus delta U."},{"Start":"02:34.865 ","End":"02:36.215","Text":"In the next video,"},{"Start":"02:36.215 ","End":"02:43.295","Text":"we\u0027ll see the enthalpy Delta H is also a state function."},{"Start":"02:43.295 ","End":"02:45.770","Text":"That\u0027s something that\u0027s very important."},{"Start":"02:45.770 ","End":"02:52.325","Text":"Now let\u0027s take an analogy so we totally understand the meaning of a state function."},{"Start":"02:52.325 ","End":"02:57.395","Text":"Supposing we have a mountain whose height is 1 kilometer above sea level,"},{"Start":"02:57.395 ","End":"03:06.745","Text":"here\u0027s our mountain, and its height is Delta H equal to 1 kilometer."},{"Start":"03:06.745 ","End":"03:11.130","Text":"Now, supposing we want to climb the mountain,"},{"Start":"03:11.130 ","End":"03:13.550","Text":"now, we can climb it several ways."},{"Start":"03:13.550 ","End":"03:15.575","Text":"One way is straight up."},{"Start":"03:15.575 ","End":"03:18.170","Text":"That\u0027s if you\u0027re very, very energetic."},{"Start":"03:18.170 ","End":"03:23.240","Text":"Another way for the lazier or the less athletic,"},{"Start":"03:23.240 ","End":"03:25.820","Text":"is to go on a sneak path."},{"Start":"03:25.820 ","End":"03:29.255","Text":"Now, it doesn\u0027t matter which path you take."},{"Start":"03:29.255 ","End":"03:32.270","Text":"The height of the mountain won\u0027t change."},{"Start":"03:32.270 ","End":"03:36.410","Text":"The height will still be Delta H, 1 kilometer."},{"Start":"03:36.410 ","End":"03:41.300","Text":"However, the amount of work you had to do to get up to"},{"Start":"03:41.300 ","End":"03:47.750","Text":"the top is different because work depends on the distance you travel."},{"Start":"03:47.750 ","End":"03:55.265","Text":"The direct path will involve less work than the sneak path."},{"Start":"03:55.265 ","End":"04:02.690","Text":"From all this, we can conclude that work is a path dependent function."},{"Start":"04:02.690 ","End":"04:05.540","Text":"Here we have a path dependent function."},{"Start":"04:05.540 ","End":"04:08.300","Text":"A path dependent function depends on the path"},{"Start":"04:08.300 ","End":"04:12.385","Text":"taken to go from the initial state to the final state."},{"Start":"04:12.385 ","End":"04:18.890","Text":"Really anything that isn\u0027t a state function is a path dependent function."},{"Start":"04:18.890 ","End":"04:25.160","Text":"Examples we\u0027ve already seen that work is a path dependent function."},{"Start":"04:25.160 ","End":"04:26.720","Text":"What about heat?"},{"Start":"04:26.720 ","End":"04:33.290","Text":"Now, we know the first law of thermodynamics is Delta U is equal to q plus"},{"Start":"04:33.290 ","End":"04:41.400","Text":"w. We can write that q is equal to Delta U minus w,"},{"Start":"04:41.400 ","End":"04:49.220","Text":"so q depends on W. Q must be also a path dependent function."},{"Start":"04:49.220 ","End":"04:55.560","Text":"In this video, we defined state and path dependent functions."}],"ID":19787},{"Watched":false,"Name":"Exercise 1","Duration":"5m 11s","ChapterTopicVideoID":31841,"CourseChapterTopicPlaylistID":81301,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34077},{"Watched":false,"Name":"Exercise 2","Duration":"5m 13s","ChapterTopicVideoID":31842,"CourseChapterTopicPlaylistID":81301,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34078}],"Thumbnail":null,"ID":81301},{"Name":"Enthalpy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Definition of Enthalpy","Duration":"4m 30s","ChapterTopicVideoID":19309,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"In a previous video,"},{"Start":"00:01.650 ","End":"00:04.440","Text":"we talked about the first law of thermodynamics."},{"Start":"00:04.440 ","End":"00:07.560","Text":"In this video, we\u0027ll use the first law to define"},{"Start":"00:07.560 ","End":"00:11.130","Text":"a very important concept called enthalpy."},{"Start":"00:11.130 ","End":"00:13.470","Text":"Let\u0027s consider a chemical reaction."},{"Start":"00:13.470 ","End":"00:17.130","Text":"We have reactants which we\u0027re going to call the initial state that"},{"Start":"00:17.130 ","End":"00:21.570","Text":"react to form products and we\u0027re going to call the products the final state."},{"Start":"00:21.570 ","End":"00:24.240","Text":"Now, let\u0027s write the first law of thermodynamics,"},{"Start":"00:24.240 ","End":"00:28.545","Text":"Delta U, which is U final minus U initial,"},{"Start":"00:28.545 ","End":"00:35.240","Text":"the change in the internal energy is equal to q plus w. Since it\u0027s a reaction,"},{"Start":"00:35.240 ","End":"00:38.974","Text":"we\u0027re going to write q_rxn, q reaction."},{"Start":"00:38.974 ","End":"00:42.680","Text":"Previous video, we learned about pressure volume work."},{"Start":"00:42.680 ","End":"00:45.770","Text":"We learned that when there\u0027s pressure volume work,"},{"Start":"00:45.770 ","End":"00:52.310","Text":"w is equal to minus P Delta V. Now we have Delta U is"},{"Start":"00:52.310 ","End":"00:58.790","Text":"equal to q reaction minus P Delta V. Let\u0027s first consider constant volume,"},{"Start":"00:58.790 ","End":"01:01.130","Text":"a reaction occurring at constant volume."},{"Start":"01:01.130 ","End":"01:05.075","Text":"We met this when we discussed the bomb calorimeter."},{"Start":"01:05.075 ","End":"01:06.980","Text":"Its volume is constant."},{"Start":"01:06.980 ","End":"01:09.335","Text":"Delta V is equal to 0."},{"Start":"01:09.335 ","End":"01:12.155","Text":"Now, if Delta V is equal to 0,"},{"Start":"01:12.155 ","End":"01:14.480","Text":"P Delta V is 0,"},{"Start":"01:14.480 ","End":"01:18.095","Text":"and Delta U is just equal to q reaction,"},{"Start":"01:18.095 ","End":"01:19.640","Text":"so at constant volume,"},{"Start":"01:19.640 ","End":"01:22.055","Text":"Delta U is equal to q reaction."},{"Start":"01:22.055 ","End":"01:26.285","Text":"To stress this constant force we can call this qv."},{"Start":"01:26.285 ","End":"01:30.125","Text":"Now let us consider constant pressure."},{"Start":"01:30.125 ","End":"01:33.740","Text":"We have Delta U is equal to q,"},{"Start":"01:33.740 ","End":"01:36.725","Text":"and we\u0027re going to write P because it\u0027s constant pressure,"},{"Start":"01:36.725 ","End":"01:40.820","Text":"minus P Delta V minus P Delta V is the work."},{"Start":"01:40.820 ","End":"01:42.600","Text":"Now, instead of writing Delta U,"},{"Start":"01:42.600 ","End":"01:46.865","Text":"we can write qv heat at constant volume."},{"Start":"01:46.865 ","End":"01:56.480","Text":"That\u0027s now equal to qp minus P Delta V. We can rearrange this equation and write it as qp"},{"Start":"01:56.480 ","End":"02:02.270","Text":"equal to qv plus P Delta V. All we\u0027ve done is"},{"Start":"02:02.270 ","End":"02:07.610","Text":"to take the minus P Delta V over to the other side of the equation and change the order."},{"Start":"02:07.610 ","End":"02:12.870","Text":"We\u0027ll write qp is equal to qv plus P Delta V. Now,"},{"Start":"02:12.870 ","End":"02:16.500","Text":"remember qv is equal to Delta U."},{"Start":"02:16.500 ","End":"02:23.450","Text":"We can write qp is equal to Delta U plus P Delta V. Now,"},{"Start":"02:23.450 ","End":"02:25.100","Text":"when we look at this expression,"},{"Start":"02:25.100 ","End":"02:28.534","Text":"we can see that Delta U is a state function,"},{"Start":"02:28.534 ","End":"02:31.195","Text":"P is a state function,"},{"Start":"02:31.195 ","End":"02:34.050","Text":"Delta V is a state function."},{"Start":"02:34.050 ","End":"02:37.235","Text":"All the right-hand side is a state function,"},{"Start":"02:37.235 ","End":"02:40.445","Text":"so qp must be a state function."},{"Start":"02:40.445 ","End":"02:43.955","Text":"Now we define something called enthalpy."},{"Start":"02:43.955 ","End":"02:46.820","Text":"Let\u0027s first look at the change in enthalpy."},{"Start":"02:46.820 ","End":"02:51.205","Text":"The change in enthalpy is the same as qp,"},{"Start":"02:51.205 ","End":"02:55.042","Text":"and it\u0027s written Delta U plus P Delta V,"},{"Start":"02:55.042 ","End":"03:01.085","Text":"so we\u0027re redefining q at constant pressure as Delta H."},{"Start":"03:01.085 ","End":"03:07.639","Text":"This expression is so important that it gets its own name and its own symbol,"},{"Start":"03:07.639 ","End":"03:08.930","Text":"it\u0027s called the enthalpy."},{"Start":"03:08.930 ","End":"03:11.300","Text":"This is the change in enthalpy."},{"Start":"03:11.300 ","End":"03:16.055","Text":"Delta H is equal to Delta U plus P Delta V,"},{"Start":"03:16.055 ","End":"03:20.165","Text":"is a very important equation and you should remember it."},{"Start":"03:20.165 ","End":"03:27.140","Text":"The other equation that\u0027s very important is that Delta U is equal to qv."},{"Start":"03:27.140 ","End":"03:30.020","Text":"That\u0027s another very important equation."},{"Start":"03:30.020 ","End":"03:32.720","Text":"We have these 2 important equations."},{"Start":"03:32.720 ","End":"03:37.075","Text":"Now, we could have defined the enthalpy in a more general way."},{"Start":"03:37.075 ","End":"03:40.310","Text":"I\u0027m just noting this though it\u0027s not essential."},{"Start":"03:40.310 ","End":"03:47.735","Text":"H, the enthalpy, not the change in the enthalpy itself is defined as U plus PV."},{"Start":"03:47.735 ","End":"03:50.544","Text":"If we have Delta H,"},{"Start":"03:50.544 ","End":"03:57.825","Text":"it\u0027s Delta U plus Delta PV."},{"Start":"03:57.825 ","End":"04:00.260","Text":"If the pressure doesn\u0027t change,"},{"Start":"04:00.260 ","End":"04:06.270","Text":"we can take it out and then we get Delta U plus"},{"Start":"04:06.270 ","End":"04:14.075","Text":"P Delta V. This is another way of defining the change in enthalpy."},{"Start":"04:14.075 ","End":"04:21.875","Text":"An important thing is that Delta H is equal to Delta U plus P Delta V. In this video,"},{"Start":"04:21.875 ","End":"04:24.200","Text":"we defined the enthalpy."},{"Start":"04:24.200 ","End":"04:25.505","Text":"In the next video,"},{"Start":"04:25.505 ","End":"04:30.390","Text":"we\u0027ll discuss the enthalpy change in a chemical reaction."}],"ID":34079},{"Watched":false,"Name":"Enthalpy Change in a Chemical Reaction","Duration":"6m 32s","ChapterTopicVideoID":18675,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.265","Text":"In the previous video,"},{"Start":"00:02.265 ","End":"00:05.250","Text":"we introduced the concept of enthalpy."},{"Start":"00:05.250 ","End":"00:08.190","Text":"In this video, we will learn about the changes in"},{"Start":"00:08.190 ","End":"00:12.270","Text":"internal energy and enthalpy in a chemical reaction."},{"Start":"00:12.270 ","End":"00:18.540","Text":"Let\u0027s begin by recalling what we learned already about internal energy and enthalpy."},{"Start":"00:18.540 ","End":"00:21.810","Text":"The heat of reaction at constant volume is"},{"Start":"00:21.810 ","End":"00:25.710","Text":"equal to the change in internal energy Delta U."},{"Start":"00:25.710 ","End":"00:27.075","Text":"On the other hand,"},{"Start":"00:27.075 ","End":"00:32.895","Text":"the heat of reaction at constant pressure is the same as the change in enthalpy."},{"Start":"00:32.895 ","End":"00:34.985","Text":"That\u0027s equal to Delta H,"},{"Start":"00:34.985 ","End":"00:40.550","Text":"which is equal to Delta U plus P Delta V. Another way of writing"},{"Start":"00:40.550 ","End":"00:47.025","Text":"this is to replace Delta H by q_p and Delta U by q_v."},{"Start":"00:47.025 ","End":"00:52.610","Text":"We get q_p the heat of reaction at constant pressure equal to the heat of reaction at"},{"Start":"00:52.610 ","End":"01:00.575","Text":"constant volume plus P Delta V. In order to know the difference between q_p and q_v,"},{"Start":"01:00.575 ","End":"01:04.265","Text":"we need to calculate P Delta V. Now,"},{"Start":"01:04.265 ","End":"01:09.410","Text":"most reactions are performed constant pressure rather than constant volume,"},{"Start":"01:09.410 ","End":"01:15.055","Text":"so that Delta H is more commonly used for heat of reaction than Delta U."},{"Start":"01:15.055 ","End":"01:19.729","Text":"The tables and books are all related to Delta H."},{"Start":"01:19.729 ","End":"01:24.875","Text":"Let\u0027s revisit the pressure volume work in a chemical reaction."},{"Start":"01:24.875 ","End":"01:26.600","Text":"First thing to realize is that"},{"Start":"01:26.600 ","End":"01:30.860","Text":"pressure volume work is only relevant for reactions that involve"},{"Start":"01:30.860 ","End":"01:37.910","Text":"gases as either the reactants or products or both reactants and products."},{"Start":"01:37.910 ","End":"01:44.210","Text":"We don\u0027t need to really consider the changes in volume of liquids and solids."},{"Start":"01:44.210 ","End":"01:49.720","Text":"P Delta V=P*V_f,"},{"Start":"01:49.720 ","End":"01:53.850","Text":"that\u0027s the volume of the products - V_i,"},{"Start":"01:53.850 ","End":"01:56.835","Text":"that\u0027s the volume of the reactants."},{"Start":"01:56.835 ","End":"02:03.930","Text":"Now, we know from the gas law, PV=nRT."},{"Start":"02:03.930 ","End":"02:11.970","Text":"If we write PV_f=n_fRT and the same for initial."},{"Start":"02:11.970 ","End":"02:15.210","Text":"PV_f-PV_i="},{"Start":"02:15.210 ","End":"02:23.590","Text":"n_fRT-n_iRT. We\u0027re left with n_f-n_i,"},{"Start":"02:23.590 ","End":"02:28.115","Text":"which is the change in the number of moles."},{"Start":"02:28.115 ","End":"02:33.500","Text":"n_f-n_i will write as Delta n. Now we"},{"Start":"02:33.500 ","End":"02:39.135","Text":"have P Delta V=Delta nRT."},{"Start":"02:39.135 ","End":"02:43.940","Text":"We can write that Delta H change in enthalpy is equal to Delta U,"},{"Start":"02:43.940 ","End":"02:48.585","Text":"the change in internal energy plus Delta nRT."},{"Start":"02:48.585 ","End":"02:50.555","Text":"Let\u0027s take an example."},{"Start":"02:50.555 ","End":"02:55.579","Text":"Delta H for the combustion of 1 mole of propane gas is -"},{"Start":"02:55.579 ","End":"03:05.470","Text":"2,219.9 kilojoules at 25 degrees Celsius and a pressure of 1 atmosphere."},{"Start":"03:05.470 ","End":"03:07.990","Text":"Calculate Delta U?"},{"Start":"03:07.990 ","End":"03:12.240","Text":"The first thing to do is directly equation for the reaction."},{"Start":"03:12.850 ","End":"03:15.340","Text":"C_3H_8, that\u0027s propane."},{"Start":"03:15.340 ","End":"03:20.000","Text":"1 mole of propane is a gas plus 5 moles of oxygen,"},{"Start":"03:20.000 ","End":"03:21.410","Text":"which is also a gas,"},{"Start":"03:21.410 ","End":"03:23.899","Text":"gives us 3 moles of carbon dioxide,"},{"Start":"03:23.899 ","End":"03:25.355","Text":"which is also a gas,"},{"Start":"03:25.355 ","End":"03:27.080","Text":"and 4 moles of water,"},{"Start":"03:27.080 ","End":"03:28.490","Text":"which is a liquid."},{"Start":"03:28.490 ","End":"03:35.780","Text":"The number of moles of gas in the reactants is 1 for propane plus 5 for oxygen,"},{"Start":"03:35.780 ","End":"03:40.965","Text":"giving you a total of 6. n_i is 6 and in the products,"},{"Start":"03:40.965 ","End":"03:43.815","Text":"we have 3 moles of carbon dioxide."},{"Start":"03:43.815 ","End":"03:46.140","Text":"n_f is equal to 3."},{"Start":"03:46.140 ","End":"03:52.500","Text":"Delta n, which is n_f-n_i is -3."},{"Start":"03:52.500 ","End":"03:57.830","Text":"Let\u0027s recall that Delta H=Delta U+Delta nRT."},{"Start":"03:57.830 ","End":"04:02.375","Text":"But here we\u0027re given Delta H and asked to find Delta U."},{"Start":"04:02.375 ","End":"04:08.130","Text":"We can rewrite this as Delta U=H-Delta"},{"Start":"04:08.130 ","End":"04:14.570","Text":"nRT over done is to subtract Delta nRT from both sides of the equation."},{"Start":"04:14.570 ","End":"04:16.985","Text":"Now we can put in the numbers,"},{"Start":"04:16.985 ","End":"04:22.810","Text":"Delta H we\u0027re told is -2,219.9 kilojoules."},{"Start":"04:22.810 ","End":"04:25.578","Text":"Now we need minus Delta nRT."},{"Start":"04:25.578 ","End":"04:28.680","Text":"Delta n is minus 3."},{"Start":"04:28.680 ","End":"04:31.620","Text":"Minus Delta n is plus 3."},{"Start":"04:31.620 ","End":"04:35.155","Text":"Of course we\u0027re talking about moles, so 3 moles."},{"Start":"04:35.155 ","End":"04:38.510","Text":"Now we need to put in the value of R,"},{"Start":"04:38.510 ","End":"04:40.670","Text":"anytime we talk about energy,"},{"Start":"04:40.670 ","End":"04:47.530","Text":"we need the value 8.3145 joules."},{"Start":"04:47.530 ","End":"04:53.825","Text":"But we want everything in kilojoules because Delta H has given us an kilojoules."},{"Start":"04:53.825 ","End":"04:58.530","Text":"We can write this as kilojoules, it\u0027s 8.3145*10^-3"},{"Start":"05:09.470 ","End":"05:14.759","Text":"kilojoules per mole per Kelvin."},{"Start":"05:14.759 ","End":"05:16.565","Text":"We need the temperature now,"},{"Start":"05:16.565 ","End":"05:23.365","Text":"298.15 Kelvin, that\u0027s 25 degrees Celsius."},{"Start":"05:23.365 ","End":"05:34.175","Text":"We have minus 2,219.9 kilojoules and when we multiply out the numbers, we get 7.4369."},{"Start":"05:34.175 ","End":"05:35.975","Text":"What about the units?"},{"Start":"05:35.975 ","End":"05:39.720","Text":"Moles times mole to the power -1 is 1."},{"Start":"05:39.720 ","End":"05:43.350","Text":"K^-1*K is 1."},{"Start":"05:43.350 ","End":"05:45.755","Text":"We\u0027re left just with kilojoules,"},{"Start":"05:45.755 ","End":"05:49.940","Text":"and it turns out to be 7.4369 kilojoules."},{"Start":"05:49.940 ","End":"05:52.340","Text":"When we do the arithmetic,"},{"Start":"05:52.340 ","End":"05:58.780","Text":"we get -2,212.5 kilojoules,"},{"Start":"05:58.780 ","End":"06:03.055","Text":"so that\u0027s the value of Delta U."},{"Start":"06:03.055 ","End":"06:11.500","Text":"Now let\u0027s compare that with the value of Delta H here is -2,219.9 kilojoules."},{"Start":"06:11.500 ","End":"06:16.760","Text":"It\u0027s really very little difference between the two and that\u0027s quite usual."},{"Start":"06:16.760 ","End":"06:20.179","Text":"Delta H and Delta U are very similar."},{"Start":"06:20.179 ","End":"06:26.180","Text":"We write it, note that Delta H and delta you are very close in numerical volume."},{"Start":"06:26.180 ","End":"06:32.520","Text":"In this video, we talked about the changes in enthalpy and internal energy in a reaction."}],"ID":34080},{"Watched":false,"Name":"Exercise 1","Duration":"5m 16s","ChapterTopicVideoID":31831,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34081},{"Watched":false,"Name":"Exercise 2","Duration":"2m 52s","ChapterTopicVideoID":31832,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34082},{"Watched":false,"Name":"Enthalpy of Vaporization and Fusion","Duration":"4m 47s","ChapterTopicVideoID":19310,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"In the previous video,"},{"Start":"00:01.680 ","End":"00:06.360","Text":"we learned about the enthalpy change that occurs in a chemical reaction."},{"Start":"00:06.360 ","End":"00:12.240","Text":"In this video, we will see that phase changes also involve a change in the enthalpy."},{"Start":"00:12.240 ","End":"00:18.570","Text":"Phase changes occur at constant temperature and involve a change in enthalpy."},{"Start":"00:18.570 ","End":"00:22.410","Text":"To remind you what a phase change is something like going"},{"Start":"00:22.410 ","End":"00:26.550","Text":"from a solid to a liquid or a liquid to a gas."},{"Start":"00:26.550 ","End":"00:29.775","Text":"Let\u0027s first consider enthalpy of vaporization."},{"Start":"00:29.775 ","End":"00:34.425","Text":"The enthalpy of vaporization is the heat required to vaporize the liquid."},{"Start":"00:34.425 ","End":"00:37.160","Text":"In other words, to turn a liquid into a gas."},{"Start":"00:37.160 ","End":"00:40.055","Text":"It\u0027s an endothermic process."},{"Start":"00:40.055 ","End":"00:41.750","Text":"Let\u0027s take an example."},{"Start":"00:41.750 ","End":"00:43.700","Text":"The vaporization of water,"},{"Start":"00:43.700 ","End":"00:50.810","Text":"the amount of heat required to change a mole of water into a mole of gas is"},{"Start":"00:50.810 ","End":"00:59.239","Text":"the Delta H vaporization equal to 44.0 kilojoules at 298 Kelvin."},{"Start":"00:59.239 ","End":"01:01.760","Text":"Now let\u0027s consider the enthalpy of fusion."},{"Start":"01:01.760 ","End":"01:05.660","Text":"The enthalpy of fusion is the heat required to melt a solid,"},{"Start":"01:05.660 ","End":"01:08.330","Text":"in other words, to turn a solid into a liquid."},{"Start":"01:08.330 ","End":"01:11.315","Text":"It\u0027s also an endothermic process."},{"Start":"01:11.315 ","End":"01:13.640","Text":"For example, the melting of ice."},{"Start":"01:13.640 ","End":"01:19.655","Text":"In order to melt 1 mole of ice into 1 mole of liquid water,"},{"Start":"01:19.655 ","End":"01:25.735","Text":"we require an enthalpy change of 6.01 kilojoules."},{"Start":"01:25.735 ","End":"01:28.580","Text":"This is 0 degrees Celsius,"},{"Start":"01:28.580 ","End":"01:33.815","Text":"273.15 Kelvin, 0 degrees Celsius."},{"Start":"01:33.815 ","End":"01:35.465","Text":"Let\u0027s take an example."},{"Start":"01:35.465 ","End":"01:39.440","Text":"How much heat is required to melt 10.0 grams of"},{"Start":"01:39.440 ","End":"01:44.005","Text":"ice and heat the water formed to 60 degrees Celsius?"},{"Start":"01:44.005 ","End":"01:46.370","Text":"The 2 points involved here,"},{"Start":"01:46.370 ","End":"01:48.665","Text":"first of all, we have to melt the ice,"},{"Start":"01:48.665 ","End":"01:53.825","Text":"and then we have to heat the water from 0-60 degrees Celsius."},{"Start":"01:53.825 ","End":"02:00.020","Text":"The temperature stays constant all the time the ice is melting and then rises."},{"Start":"02:00.020 ","End":"02:04.700","Text":"This is the ice and this is water."},{"Start":"02:04.700 ","End":"02:14.345","Text":"Now, the information we were given above was about the molar enthalpy of ice."},{"Start":"02:14.345 ","End":"02:17.915","Text":"We need to know the number of moles of ice."},{"Start":"02:17.915 ","End":"02:24.140","Text":"We have number of moles of ice and each tool is equal to 10.0 grams"},{"Start":"02:24.140 ","End":"02:31.550","Text":"multiplied by our conversion factor of 1 mole is equivalent to 18.0 grams."},{"Start":"02:31.550 ","End":"02:34.220","Text":"That\u0027s the molar mass of water."},{"Start":"02:34.220 ","End":"02:36.185","Text":"Of course, the grams cancel."},{"Start":"02:36.185 ","End":"02:37.520","Text":"When we do the arithmetic,"},{"Start":"02:37.520 ","End":"02:41.125","Text":"we get 0.056 moles."},{"Start":"02:41.125 ","End":"02:45.740","Text":"Now the amount of heat required for this process is first of all,"},{"Start":"02:45.740 ","End":"02:47.555","Text":"the number of moles of water,"},{"Start":"02:47.555 ","End":"02:49.295","Text":"number moles of ice,"},{"Start":"02:49.295 ","End":"02:53.870","Text":"times the enthalpy of fusion of ice."},{"Start":"02:53.870 ","End":"02:58.400","Text":"Then to heat the water from 0-60 degrees"},{"Start":"02:58.400 ","End":"03:03.500","Text":"Celsius is the mass of the water times the specific heat of water,"},{"Start":"03:03.500 ","End":"03:06.110","Text":"times the temperature difference."},{"Start":"03:06.110 ","End":"03:13.445","Text":"So that\u0027s 0.556 moles times 6.01 kilojoules per mole."},{"Start":"03:13.445 ","End":"03:16.090","Text":"That\u0027s the first part, that\u0027s to melt the ice."},{"Start":"03:16.090 ","End":"03:21.635","Text":"The moles cancels with moles or moles times more power minus 1 is 1."},{"Start":"03:21.635 ","End":"03:25.460","Text":"The second part is the mass of water,"},{"Start":"03:25.460 ","End":"03:30.245","Text":"which is 10.0 grams times the specific heat of water,"},{"Start":"03:30.245 ","End":"03:38.990","Text":"which is 4.18 times 10^minus 3 kilojoules per gram per degrees Celsius."},{"Start":"03:38.990 ","End":"03:41.615","Text":"We\u0027ve changed the joules to kilojoules"},{"Start":"03:41.615 ","End":"03:45.605","Text":"because the first part of the problem is in kilojoules."},{"Start":"03:45.605 ","End":"03:48.950","Text":"We want the second part also to be in kilojoules."},{"Start":"03:48.950 ","End":"03:52.730","Text":"Then we multiply by the temperature difference at"},{"Start":"03:52.730 ","End":"03:57.080","Text":"60 degrees Celsius minus 0 degrees Celsius,"},{"Start":"03:57.080 ","End":"04:00.595","Text":"60 minus 0 degrees Celsius."},{"Start":"04:00.595 ","End":"04:02.485","Text":"Now let\u0027s look at the units here."},{"Start":"04:02.485 ","End":"04:05.830","Text":"We have grams times grams power minus 1,"},{"Start":"04:05.830 ","End":"04:10.240","Text":"that\u0027s 1 degrees Celsius to power minus 1 times degrees Celsius."},{"Start":"04:10.240 ","End":"04:11.620","Text":"That\u0027s also 1."},{"Start":"04:11.620 ","End":"04:13.450","Text":"We\u0027re left here with kilojoules,"},{"Start":"04:13.450 ","End":"04:15.250","Text":"just as in the first part."},{"Start":"04:15.250 ","End":"04:16.710","Text":"When you do the arithmetic,"},{"Start":"04:16.710 ","End":"04:21.715","Text":"we get 3.34 kilojoules required to melt the ice,"},{"Start":"04:21.715 ","End":"04:26.280","Text":"plus 2.51 kilojoules required to heat to the water."},{"Start":"04:26.280 ","End":"04:27.660","Text":"When we add these 2,"},{"Start":"04:27.660 ","End":"04:30.690","Text":"we get 5.85 kilojoules."},{"Start":"04:30.690 ","End":"04:35.140","Text":"That\u0027s the amount of heat required to melt"},{"Start":"04:35.140 ","End":"04:41.055","Text":"the ice and then heat the water from 0 degrees to 60 degrees Celsius."},{"Start":"04:41.055 ","End":"04:43.490","Text":"In this video, we learned about the change in"},{"Start":"04:43.490 ","End":"04:47.789","Text":"enthalpy that accompanies a change in phase."}],"ID":34083},{"Watched":false,"Name":"Exercise 3","Duration":"5m 59s","ChapterTopicVideoID":22988,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.879","Text":"Hi. We\u0027re going to solve the following exercise."},{"Start":"00:02.879 ","End":"00:04.830","Text":"What mass of ice can be melted with"},{"Start":"00:04.830 ","End":"00:07.920","Text":"the same amount of heat required to raise the temperature"},{"Start":"00:07.920 ","End":"00:13.335","Text":"of 5.4 moles of liquid water by 23-degrees Celsius?"},{"Start":"00:13.335 ","End":"00:17.700","Text":"We know that the amount of heat which is used to"},{"Start":"00:17.700 ","End":"00:22.770","Text":"raise the water equals the amount of heat of the ice."},{"Start":"00:22.770 ","End":"00:31.110","Text":"The amount of heat of water equals m times s times Delta T. M is the mass,"},{"Start":"00:31.110 ","End":"00:32.600","Text":"s is the specific heat of water,"},{"Start":"00:32.600 ","End":"00:36.200","Text":"and Delta T is the difference in the temperature."},{"Start":"00:36.200 ","End":"00:38.210","Text":"We\u0027re given the number of moles of water,"},{"Start":"00:38.210 ","End":"00:40.355","Text":"so we\u0027re going to calculate the mass of water."},{"Start":"00:40.355 ","End":"00:41.630","Text":"Recall that n,"},{"Start":"00:41.630 ","End":"00:45.080","Text":"the number of moles equals the mass divided by the molar mass."},{"Start":"00:45.080 ","End":"00:47.705","Text":"In our case, we need the mass."},{"Start":"00:47.705 ","End":"00:50.570","Text":"We\u0027re going to multiply both sides by the molar mass."},{"Start":"00:50.570 ","End":"00:55.445","Text":"The mass equals the number of moles times the molar mass."},{"Start":"00:55.445 ","End":"00:58.280","Text":"We\u0027re going to calculate the molar mass of water."},{"Start":"00:58.280 ","End":"01:00.080","Text":"The molar mass of water,"},{"Start":"01:00.080 ","End":"01:01.640","Text":"which is H_2O,"},{"Start":"01:01.640 ","End":"01:05.030","Text":"equals 2 times the molar mass of hydrogen,"},{"Start":"01:05.030 ","End":"01:06.815","Text":"since we have 2 hydrogens,"},{"Start":"01:06.815 ","End":"01:11.850","Text":"plus the molar mass of oxygen."},{"Start":"01:11.850 ","End":"01:19.610","Text":"This equals 2 times 1.01 grams per mole plus 16 grams per mole,"},{"Start":"01:19.610 ","End":"01:22.410","Text":"which is the molar mass of oxygen."},{"Start":"01:22.420 ","End":"01:28.055","Text":"This gives us 18.02 grams per mole."},{"Start":"01:28.055 ","End":"01:30.080","Text":"We have the molar mass of water,"},{"Start":"01:30.080 ","End":"01:32.105","Text":"which is 18.02 grams per mole,"},{"Start":"01:32.105 ","End":"01:35.885","Text":"times the number of moles of water and that will give us the mass of water."},{"Start":"01:35.885 ","End":"01:38.900","Text":"Again, the amount of heat of water equals the mass of"},{"Start":"01:38.900 ","End":"01:43.620","Text":"water times the specific heat of water times the difference in temperature."},{"Start":"01:44.390 ","End":"01:48.920","Text":"As we said, mass equals number of moles"},{"Start":"01:48.920 ","End":"01:57.135","Text":"times molar mass of water,"},{"Start":"01:57.135 ","End":"01:59.530","Text":"times s, which is the specific heat of water,"},{"Start":"01:59.530 ","End":"02:03.820","Text":"which equals 4.18 joules per gram degrees Celsius."},{"Start":"02:03.820 ","End":"02:05.890","Text":"This is times the difference in temperature,"},{"Start":"02:05.890 ","End":"02:08.960","Text":"which equals 23-degrees Celsius."},{"Start":"02:10.080 ","End":"02:14.140","Text":"We can already mark off our degrees Celsius"},{"Start":"02:14.140 ","End":"02:17.710","Text":"because they cancel out and we\u0027re just going to continue right here."},{"Start":"02:17.710 ","End":"02:19.930","Text":"This equals the number of moles of water,"},{"Start":"02:19.930 ","End":"02:27.160","Text":"which has given it equals 5.4 moles times the molar mass of water,"},{"Start":"02:27.160 ","End":"02:28.450","Text":"which is what we calculated,"},{"Start":"02:28.450 ","End":"02:32.780","Text":"equals 18.02 grams per mole."},{"Start":"02:33.300 ","End":"02:36.400","Text":"You can already see that the moles cancel out."},{"Start":"02:36.400 ","End":"02:46.130","Text":"Times 4.18 joule per grams times 23."},{"Start":"02:46.830 ","End":"02:55.050","Text":"This equals 9,355.19."},{"Start":"02:55.050 ","End":"02:57.130","Text":"We can see also the grams cancel out,"},{"Start":"02:57.130 ","End":"02:59.470","Text":"so we\u0027re left with joules,"},{"Start":"02:59.470 ","End":"03:04.915","Text":"so this equals 9,355.19 joules."},{"Start":"03:04.915 ","End":"03:08.180","Text":"Now we\u0027re going to look at the amount of heat of the ice."},{"Start":"03:10.830 ","End":"03:14.140","Text":"The amount of heat of the ice equals the number of moles of"},{"Start":"03:14.140 ","End":"03:20.030","Text":"the ice times the heat of fusion of the ice."},{"Start":"03:22.250 ","End":"03:27.185","Text":"Again, the number of moles equals the mass divided by the molar mass."},{"Start":"03:27.185 ","End":"03:28.670","Text":"Instead of writing number of moles,"},{"Start":"03:28.670 ","End":"03:30.380","Text":"we\u0027re just going to go ahead and write the mass of"},{"Start":"03:30.380 ","End":"03:35.760","Text":"the ice divided by the molar mass of the ice."},{"Start":"03:36.550 ","End":"03:40.460","Text":"This is times the heat of fusion of the ice."},{"Start":"03:40.460 ","End":"03:46.260","Text":"The heat of fusion of ice equals 6.01 kilojoules per mole."},{"Start":"03:48.520 ","End":"03:54.240","Text":"This equals the mass of the ice times 6.01"},{"Start":"03:54.240 ","End":"03:59.420","Text":"kilojoules per mole divided by the molar mass of the ice."},{"Start":"03:59.420 ","End":"04:02.045","Text":"The molar mass of the ice is the same as the molar mass of water."},{"Start":"04:02.045 ","End":"04:05.400","Text":"It equals 18.02 grams per mole."},{"Start":"04:08.480 ","End":"04:18.315","Text":"This equals 0.34 kilojoules per gram times the mass of the ice."},{"Start":"04:18.315 ","End":"04:20.260","Text":"In our regarding the units,"},{"Start":"04:20.260 ","End":"04:28.070","Text":"we can see that we have kilojoules per mole divided by grams per mole."},{"Start":"04:29.940 ","End":"04:34.840","Text":"We multiply the numerator times the denominator in both cases and this equals"},{"Start":"04:34.840 ","End":"04:40.420","Text":"kilojoules times mole divided by mole times grams."},{"Start":"04:40.420 ","End":"04:45.670","Text":"Moles cancel out and we\u0027re left with kilojoules per gram."},{"Start":"04:45.670 ","End":"04:50.890","Text":"Again, the amount of heat of the water equals the amount of heat of the ice."},{"Start":"04:54.530 ","End":"05:03.570","Text":"The amount of heat of the water equals 9,355.19 joules,"},{"Start":"05:03.570 ","End":"05:07.130","Text":"and this equals the amount of heat of the ice,"},{"Start":"05:07.130 ","End":"05:15.850","Text":"which is 0.34 kilojoules per gram divided by the mass of the ice."},{"Start":"05:15.850 ","End":"05:19.535","Text":"Since we have joules on one side and kilojoules on the other,"},{"Start":"05:19.535 ","End":"05:22.700","Text":"just going to multiply by a conversion factor."},{"Start":"05:22.700 ","End":"05:24.845","Text":"In every 1 kilojoule,"},{"Start":"05:24.845 ","End":"05:27.190","Text":"we have 1,000 joules."},{"Start":"05:27.190 ","End":"05:33.850","Text":"The joules cancel out and 9.36 kilojoules"},{"Start":"05:33.850 ","End":"05:41.435","Text":"equals 0.34 kilojoules per gram times the mass of the ice."},{"Start":"05:41.435 ","End":"05:51.450","Text":"Kilojoules cancel out and the mass of ice equals 27.53 grams."},{"Start":"05:51.450 ","End":"05:56.060","Text":"The mass of the ice, which we calculated, equals 27.53 grams."},{"Start":"05:56.060 ","End":"05:57.590","Text":"That is our final answer."},{"Start":"05:57.590 ","End":"06:00.060","Text":"Thank you very much for watching."}],"ID":34084},{"Watched":false,"Name":"Standard States and Standard Enthalpy Changes","Duration":"1m 20s","ChapterTopicVideoID":19313,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"In the previous videos,"},{"Start":"00:01.680 ","End":"00:04.664","Text":"we talked about enthalpy change in a reaction."},{"Start":"00:04.664 ","End":"00:07.890","Text":"In order to know the exact value of the enthalpy change,"},{"Start":"00:07.890 ","End":"00:11.940","Text":"we need to know the precise initial and final states."},{"Start":"00:11.940 ","End":"00:15.495","Text":"We\u0027re going to talk about the standard enthalpy of reaction."},{"Start":"00:15.495 ","End":"00:20.670","Text":"These are the enthalpies that are listed in books and in tables."},{"Start":"00:20.670 ","End":"00:22.830","Text":"The standard enthalpy of reaction,"},{"Start":"00:22.830 ","End":"00:26.535","Text":"which we write as Delta H with a little 0 at the top,"},{"Start":"00:26.535 ","End":"00:32.965","Text":"is the enthalpy change when both the reactants and products are in their standard states."},{"Start":"00:32.965 ","End":"00:36.185","Text":"We need to know what the standard state is."},{"Start":"00:36.185 ","End":"00:38.690","Text":"Standard states, first of all,"},{"Start":"00:38.690 ","End":"00:40.895","Text":"for a liquid or solid,"},{"Start":"00:40.895 ","End":"00:48.065","Text":"is a pure element or a pure compound at a pressure of 1 bar and specify temperature."},{"Start":"00:48.065 ","End":"00:52.685","Text":"The specify temperature\u0027s usually 25 degrees Celsius."},{"Start":"00:52.685 ","End":"00:54.215","Text":"What brought a gas?"},{"Start":"00:54.215 ","End":"00:58.340","Text":"If it\u0027s a pure gas behavior as an ideal gas, again,"},{"Start":"00:58.340 ","End":"01:01.985","Text":"it\u0027s a pressure of 1 bar and specify temperature,"},{"Start":"01:01.985 ","End":"01:05.555","Text":"usually 25 degrees Celsius."},{"Start":"01:05.555 ","End":"01:12.035","Text":"In this video, we talked about the standard enthalpy of reaction Delta H_0."},{"Start":"01:12.035 ","End":"01:20.250","Text":"This is the enthalpy that\u0027s usually listed in tables and we\u0027ll beat it in future videos."}],"ID":34085},{"Watched":false,"Name":"Hess\u0027s Law","Duration":"6m 50s","ChapterTopicVideoID":19123,"CourseChapterTopicPlaylistID":81302,"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":"In a previous video,"},{"Start":"00:01.785 ","End":"00:06.465","Text":"we learned about the enthalpy change that accompanies a chemical reaction."},{"Start":"00:06.465 ","End":"00:12.240","Text":"In this video, we will discuss Hess\u0027s law for the enthalpy change in a reaction."},{"Start":"00:12.240 ","End":"00:15.480","Text":"Let\u0027s recall what we learned about enthalpy change."},{"Start":"00:15.480 ","End":"00:18.480","Text":"The change in enthalpy in a chemical reaction is"},{"Start":"00:18.480 ","End":"00:22.530","Text":"the heat change at constant pressure so Delta H,"},{"Start":"00:22.530 ","End":"00:26.160","Text":"that\u0027s the change in enthalpy is equal to q_p,"},{"Start":"00:26.160 ","End":"00:29.055","Text":"where the p stands for constant pressure;"},{"Start":"00:29.055 ","End":"00:31.320","Text":"that\u0027s the heat at constant pressure."},{"Start":"00:31.320 ","End":"00:35.510","Text":"Now, we should recall that Delta H is a state function."},{"Start":"00:35.510 ","End":"00:39.305","Text":"That means it only depends on the initial and final states"},{"Start":"00:39.305 ","End":"00:43.730","Text":"but not on the path taken to go from the initial to final states."},{"Start":"00:43.730 ","End":"00:47.970","Text":"If we have a reaction, A to B or A to X,"},{"Start":"00:47.970 ","End":"00:49.140","Text":"then X to Y,"},{"Start":"00:49.140 ","End":"00:51.890","Text":"and then Y to B, it doesn\u0027t matter."},{"Start":"00:51.890 ","End":"00:55.565","Text":"Delta H will be the same for both of them."},{"Start":"00:55.565 ","End":"00:59.570","Text":"If we know Delta H for all the intermediate steps of A to B,"},{"Start":"00:59.570 ","End":"01:02.470","Text":"we can find it for the overall reaction."},{"Start":"01:02.470 ","End":"01:05.855","Text":"Here\u0027s Hess\u0027s law that helps us to do this."},{"Start":"01:05.855 ","End":"01:08.375","Text":"If a reaction occurs in steps,"},{"Start":"01:08.375 ","End":"01:11.300","Text":"the enthalpy change for the overall process is"},{"Start":"01:11.300 ","End":"01:14.780","Text":"the sum of the enthalpy changes for the individual steps."},{"Start":"01:14.780 ","End":"01:20.480","Text":"We can sum up the enthalpy change for A to X and for X to Y and for Y"},{"Start":"01:20.480 ","End":"01:26.165","Text":"to B and that will give us the overall Delta H for A to B."},{"Start":"01:26.165 ","End":"01:27.590","Text":"Now, in order to do this,"},{"Start":"01:27.590 ","End":"01:29.170","Text":"we need some rules."},{"Start":"01:29.170 ","End":"01:32.270","Text":"Delta H is an extensive property."},{"Start":"01:32.270 ","End":"01:34.700","Text":"That means it depends on the amount of substance and"},{"Start":"01:34.700 ","End":"01:38.060","Text":"so it depends on the number of moles that react."},{"Start":"01:38.060 ","End":"01:44.670","Text":"If we have a reaction A to B and Delta H has some number X joules here,"},{"Start":"01:44.670 ","End":"01:53.120","Text":"then if we multiply this equation by 2A to 2B the Delta H will be twice as much,"},{"Start":"01:53.120 ","End":"01:55.390","Text":"it\u0027ll be 2X Joules."},{"Start":"01:55.390 ","End":"02:01.730","Text":"The second rule is Delta H changes sign when the chemical reaction is reversed."},{"Start":"02:01.730 ","End":"02:03.635","Text":"If we have A to B,"},{"Start":"02:03.635 ","End":"02:07.510","Text":"we have Delta H equal to X joules,"},{"Start":"02:07.510 ","End":"02:09.420","Text":"and then we reverse that."},{"Start":"02:09.420 ","End":"02:11.220","Text":"We have B to A,"},{"Start":"02:11.220 ","End":"02:16.360","Text":"Delta H will be minus X joules."},{"Start":"02:16.360 ","End":"02:19.650","Text":"If we had an endothermic process,"},{"Start":"02:19.650 ","End":"02:23.435","Text":"A to B then B to A would be exothermic."},{"Start":"02:23.435 ","End":"02:25.955","Text":"If A to B was exothermic,"},{"Start":"02:25.955 ","End":"02:28.940","Text":"B to A would be endothermic."},{"Start":"02:28.940 ","End":"02:32.480","Text":"Now, when the individual steps sum up to the direct reaction,"},{"Start":"02:32.480 ","End":"02:35.795","Text":"we\u0027ve arranged all the individual steps correctly."},{"Start":"02:35.795 ","End":"02:41.540","Text":"We can apply Hess\u0027s law and sum up the enthalpy changes for the individual steps."},{"Start":"02:41.540 ","End":"02:42.890","Text":"Now let\u0027s do this,"},{"Start":"02:42.890 ","End":"02:46.375","Text":"let\u0027s have an example that will illustrate these points."},{"Start":"02:46.375 ","End":"02:50.359","Text":"Here\u0027s our example; use the following data to calculate"},{"Start":"02:50.359 ","End":"02:55.580","Text":"Delta H for the reaction 2S solid sulfur,"},{"Start":"02:55.580 ","End":"03:03.090","Text":"plus 3O_2, which is oxygen gas to give 2SO_3 gas,"},{"Start":"03:03.090 ","End":"03:05.970","Text":"that says sulfur trioxide as a gas."},{"Start":"03:05.970 ","End":"03:09.740","Text":"It should be Delta H^0 because we\u0027ll see that"},{"Start":"03:09.740 ","End":"03:14.495","Text":"the information about the individual steps is given as Delta H^0,"},{"Start":"03:14.495 ","End":"03:17.800","Text":"the standard enthalpy of reaction."},{"Start":"03:17.800 ","End":"03:19.860","Text":"Here\u0027s the information given to us."},{"Start":"03:19.860 ","End":"03:24.735","Text":"Solid Sulfur reacting with oxygen gas to give SO_2,"},{"Start":"03:24.735 ","End":"03:26.355","Text":"which is also a gas."},{"Start":"03:26.355 ","End":"03:32.780","Text":"Here, Delta H^0 is equal to minus 297 kilojoules."},{"Start":"03:32.780 ","End":"03:37.920","Text":"Also given, that 2 moles of SO_3 gas reacts to"},{"Start":"03:37.920 ","End":"03:43.245","Text":"give 2 moles of SO_2 gas and 1 mole of oxygen gas."},{"Start":"03:43.245 ","End":"03:48.645","Text":"The Delta H^0 for this process is 198 kilojoules."},{"Start":"03:48.645 ","End":"03:53.040","Text":"Now we see that the equation we want to get to,"},{"Start":"03:53.040 ","End":"03:57.255","Text":"the overall equation has 2 moles of sulfur."},{"Start":"03:57.255 ","End":"04:04.435","Text":"What we\u0027re given is 1 mole of sulfur so we need to multiply the first equation by 2."},{"Start":"04:04.435 ","End":"04:08.145","Text":"Then when we look at the equation we want to get to,"},{"Start":"04:08.145 ","End":"04:11.750","Text":"we see that SO_3 is a product,"},{"Start":"04:11.750 ","End":"04:14.495","Text":"whereas in the equation we have here,"},{"Start":"04:14.495 ","End":"04:17.105","Text":"SO_3 is a reactant."},{"Start":"04:17.105 ","End":"04:20.360","Text":"We will need to reverse this equation."},{"Start":"04:20.360 ","End":"04:26.105","Text":"Let\u0027s do that. The first thing we\u0027re going to do is multiply the equation by 2."},{"Start":"04:26.105 ","End":"04:29.330","Text":"We have 2 S plus 2 O_2,"},{"Start":"04:29.330 ","End":"04:32.830","Text":"giving us 2 moles of SO_2."},{"Start":"04:32.830 ","End":"04:37.140","Text":"Now, Delta H was minus 297 kilojoules."},{"Start":"04:37.140 ","End":"04:39.195","Text":"We have to multiply that by 2."},{"Start":"04:39.195 ","End":"04:41.790","Text":"We get minus 594."},{"Start":"04:41.790 ","End":"04:46.175","Text":"That\u0027s 2 times minus 297."},{"Start":"04:46.175 ","End":"04:49.300","Text":"Now we have to reverse the second equation."},{"Start":"04:49.300 ","End":"04:52.300","Text":"Instead of having 2 SO_3 as a reactant,"},{"Start":"04:52.300 ","End":"04:54.819","Text":"2 SO_3 is now a product."},{"Start":"04:54.819 ","End":"04:59.350","Text":"Instead of having 2 SO_2 plus O_2 as a product,"},{"Start":"04:59.350 ","End":"05:01.660","Text":"we need to add it as a reactant."},{"Start":"05:01.660 ","End":"05:04.390","Text":"We have 2 SO_2 plus O_2."},{"Start":"05:04.390 ","End":"05:09.050","Text":"Now Delta H^0 has reversed its sign."},{"Start":"05:09.050 ","End":"05:13.740","Text":"Delta H^0 equal to 198 kilojoules,"},{"Start":"05:13.740 ","End":"05:20.030","Text":"we have Delta H^0 equal to minus 198 kilojoules."},{"Start":"05:20.030 ","End":"05:22.145","Text":"These are 2 equations."},{"Start":"05:22.145 ","End":"05:26.990","Text":"Let\u0027s sum them up and see that we get to the equation we desire."},{"Start":"05:26.990 ","End":"05:28.675","Text":"So we have 2 S,"},{"Start":"05:28.675 ","End":"05:34.025","Text":"2S we have 2O_2 plus O_2."},{"Start":"05:34.025 ","End":"05:36.855","Text":"We have 3O_2, 3O_2."},{"Start":"05:36.855 ","End":"05:41.040","Text":"Now, the SO_2 cancels 2 SO_2 is"},{"Start":"05:41.040 ","End":"05:45.765","Text":"a product here and 2 SO_2 is a reactant here, so they cancel."},{"Start":"05:45.765 ","End":"05:50.085","Text":"The product that\u0027s left is 2 SO_3."},{"Start":"05:50.085 ","End":"05:52.510","Text":"So we have here 2 SO_3."},{"Start":"05:52.510 ","End":"05:54.575","Text":"If added up the equations,"},{"Start":"05:54.575 ","End":"05:56.450","Text":"we got to the final equation."},{"Start":"05:56.450 ","End":"06:00.500","Text":"That means when we add up the enthalpy changes,"},{"Start":"06:00.500 ","End":"06:03.925","Text":"we\u0027ll get to the great enthalpy change."},{"Start":"06:03.925 ","End":"06:09.083","Text":"Here we have minus 594 and we\u0027re going to add to that minus"},{"Start":"06:09.083 ","End":"06:14.505","Text":"198 and we get minus 792 kilojoules."},{"Start":"06:14.505 ","End":"06:22.040","Text":"That\u0027s the standard enthalpy of reaction for this reaction as it is written."},{"Start":"06:22.040 ","End":"06:26.795","Text":"If we want to compare it with what we had before where there was just 1 mole of sulfur,"},{"Start":"06:26.795 ","End":"06:30.800","Text":"we need to divide by 2 to get 1 mole of sulfur in"},{"Start":"06:30.800 ","End":"06:35.890","Text":"this respect and then we can divide Delta H^0 by 2."},{"Start":"06:35.890 ","End":"06:41.960","Text":"The Delta H is always for the equation as it is written,"},{"Start":"06:41.960 ","End":"06:44.090","Text":"that\u0027s a very important point."},{"Start":"06:44.090 ","End":"06:50.850","Text":"In this video, we used Hess\u0027s law to calculate the enthalpy of the overall reaction."}],"ID":34086},{"Watched":false,"Name":"Exercise 4","Duration":"4m 24s","ChapterTopicVideoID":31822,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34087},{"Watched":false,"Name":"Exercise 5","Duration":"6m 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37s","ChapterTopicVideoID":31826,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34091},{"Watched":false,"Name":"Standard Enthalpy of Formation","Duration":"5m 28s","ChapterTopicVideoID":19311,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.545","Text":"In a previous video,"},{"Start":"00:01.545 ","End":"00:04.500","Text":"we talked about the standard enthalpy of reaction."},{"Start":"00:04.500 ","End":"00:10.170","Text":"This video, we\u0027ll talk about the standard enthalpy of formation of a substance."},{"Start":"00:10.170 ","End":"00:12.810","Text":"What\u0027s a standard enthalpy of formation?"},{"Start":"00:12.810 ","End":"00:15.075","Text":"The standard enthalpy of formation,"},{"Start":"00:15.075 ","End":"00:16.935","Text":"which we\u0027re going to write it as delta H^0_f;"},{"Start":"00:16.935 ","End":"00:19.815","Text":"0 for standard,"},{"Start":"00:19.815 ","End":"00:21.000","Text":"f for formation,"},{"Start":"00:21.000 ","End":"00:26.730","Text":"of a substance is the enthalpy change that occurs when 1 mole of a substance in"},{"Start":"00:26.730 ","End":"00:29.640","Text":"its standard state is formed from"},{"Start":"00:29.640 ","End":"00:33.689","Text":"the reference forms of elements in their standard states."},{"Start":"00:33.689 ","End":"00:38.535","Text":"There\u0027s a few words here that we have to make sure we understand."},{"Start":"00:38.535 ","End":"00:40.785","Text":"The first one is standard,"},{"Start":"00:40.785 ","End":"00:43.690","Text":"and the second one is reference."},{"Start":"00:43.690 ","End":"00:45.590","Text":"What\u0027s a standard state?"},{"Start":"00:45.590 ","End":"00:47.030","Text":"We did define it before,"},{"Start":"00:47.030 ","End":"00:48.395","Text":"we\u0027ll see it once again."},{"Start":"00:48.395 ","End":"00:52.430","Text":"The standard state of an element or compound is its state at a pressure of"},{"Start":"00:52.430 ","End":"00:56.480","Text":"1 bar and temperature 25 degrees Celsius."},{"Start":"00:56.480 ","End":"00:59.540","Text":"What are the reference forms of the elements?"},{"Start":"00:59.540 ","End":"01:03.575","Text":"The reference form of an element is its most stable form"},{"Start":"01:03.575 ","End":"01:07.670","Text":"at a pressure of 1 bar and the temperature 25 degrees Celsius."},{"Start":"01:07.670 ","End":"01:10.250","Text":"In a few minutes, we\u0027ll give some examples."},{"Start":"01:10.250 ","End":"01:11.975","Text":"Now, why is this important?"},{"Start":"01:11.975 ","End":"01:14.570","Text":"It\u0027s important because we take the enthalpy of"},{"Start":"01:14.570 ","End":"01:18.200","Text":"the reference forms of the elements to be 0."},{"Start":"01:18.200 ","End":"01:20.510","Text":"Now, this is just for convenience,"},{"Start":"01:20.510 ","End":"01:26.480","Text":"and this is analogous to taking the sea level at 0 when measuring heights."},{"Start":"01:26.480 ","End":"01:29.130","Text":"Supposing we have 2 mountains."},{"Start":"01:29.130 ","End":"01:33.210","Text":"Now, the first mountain we take as 1,000"},{"Start":"01:33.210 ","End":"01:37.385","Text":"meters high and the second mountain as 500 meters high."},{"Start":"01:37.385 ","End":"01:38.930","Text":"What do we mean by this?"},{"Start":"01:38.930 ","End":"01:44.975","Text":"We mean that the height of the mountain is 1,000 meters above sea level."},{"Start":"01:44.975 ","End":"01:48.260","Text":"We take sea level as 0 of height."},{"Start":"01:48.260 ","End":"01:52.265","Text":"The second mountain is 500 meters above sea level."},{"Start":"01:52.265 ","End":"01:54.470","Text":"Again, sea level is 0."},{"Start":"01:54.470 ","End":"02:02.120","Text":"Now the difference in height between the 2 mountains is 1,000 minus 500 meters,"},{"Start":"02:02.120 ","End":"02:04.625","Text":"which is of course, 500 meters."},{"Start":"02:04.625 ","End":"02:10.580","Text":"Now this difference in height is independent of how we define the 0."},{"Start":"02:10.580 ","End":"02:14.780","Text":"We could take the 0 of height to be the bottom of the Dead Sea,"},{"Start":"02:14.780 ","End":"02:19.255","Text":"and the difference in height of the 2 mountains would just be the same."},{"Start":"02:19.255 ","End":"02:22.740","Text":"The 0 is chosen for convenience."},{"Start":"02:22.740 ","End":"02:27.065","Text":"In enthalpy, we\u0027re usually talking about distances,"},{"Start":"02:27.065 ","End":"02:30.695","Text":"differences between two different enthalpies."},{"Start":"02:30.695 ","End":"02:34.520","Text":"How we choose the 0 is really not very important."},{"Start":"02:34.520 ","End":"02:37.955","Text":"We just need subform that is convenient."},{"Start":"02:37.955 ","End":"02:40.160","Text":"Let\u0027s take some examples."},{"Start":"02:40.160 ","End":"02:43.970","Text":"Now, the reference form of sodium, hydrogen,"},{"Start":"02:43.970 ","End":"02:47.690","Text":"nitrogen, oxygen, where sodium is a solid, hydrogen,"},{"Start":"02:47.690 ","End":"02:50.005","Text":"nitrogen, oxygen are gases,"},{"Start":"02:50.005 ","End":"02:52.095","Text":"also fluorine is a gas,"},{"Start":"02:52.095 ","End":"02:55.505","Text":"these are all considered to be reference forms."},{"Start":"02:55.505 ","End":"03:00.920","Text":"Bromine liquid form is also the reference form of bromine,"},{"Start":"03:00.920 ","End":"03:04.235","Text":"not bromine gas but bromine liquid."},{"Start":"03:04.235 ","End":"03:09.515","Text":"Now, an interesting case is what is the reference form of carbon?"},{"Start":"03:09.515 ","End":"03:11.000","Text":"Carbon, as you may know,"},{"Start":"03:11.000 ","End":"03:13.010","Text":"has more than one allotrope."},{"Start":"03:13.010 ","End":"03:16.130","Text":"It has graphite and it has diamond."},{"Start":"03:16.130 ","End":"03:19.250","Text":"Nowadays it\u0027s known that has even more than that."},{"Start":"03:19.250 ","End":"03:24.190","Text":"Now, graphite is slightly more stable than diamond."},{"Start":"03:24.190 ","End":"03:28.035","Text":"Delta H^0 for the formation of"},{"Start":"03:28.035 ","End":"03:33.980","Text":"the transformation of graphite to diamond is 1.9 kilojoules,"},{"Start":"03:33.980 ","End":"03:39.300","Text":"at of course, 1 bar and 25 degrees Celsius."},{"Start":"03:39.300 ","End":"03:43.685","Text":"Under those conditions, graphite is more stable."},{"Start":"03:43.685 ","End":"03:47.090","Text":"We take graphite to be the reference form."},{"Start":"03:47.090 ","End":"03:52.550","Text":"Now, let\u0027s say a few words about the sign of the standard enthalpy of formation."},{"Start":"03:52.550 ","End":"03:55.625","Text":"What does it mean if it\u0027s positive or if it\u0027s negative?"},{"Start":"03:55.625 ","End":"03:59.690","Text":"Supposing delta H of formation is positive,"},{"Start":"03:59.690 ","End":"04:03.710","Text":"that means that the substance is less stable than the elements."},{"Start":"04:03.710 ","End":"04:05.350","Text":"Let\u0027s draw this."},{"Start":"04:05.350 ","End":"04:08.805","Text":"Here\u0027s the substance, that\u0027s its energy,"},{"Start":"04:08.805 ","End":"04:10.670","Text":"and here are the elements."},{"Start":"04:10.670 ","End":"04:14.180","Text":"If the difference between the two is positive,"},{"Start":"04:14.180 ","End":"04:17.190","Text":"this is a substance,"},{"Start":"04:17.500 ","End":"04:20.720","Text":"this is the elements."},{"Start":"04:20.720 ","End":"04:26.075","Text":"Delta H formation is positive."},{"Start":"04:26.075 ","End":"04:33.635","Text":"That means that the elements which have a lower energy is more stable than the substance."},{"Start":"04:33.635 ","End":"04:36.275","Text":"However, if the opposite is the case,"},{"Start":"04:36.275 ","End":"04:42.680","Text":"if delta H of formation is negative then the substance is more stable than the elements."},{"Start":"04:42.680 ","End":"04:44.405","Text":"Here\u0027s a substance."},{"Start":"04:44.405 ","End":"04:46.145","Text":"Here are the elements."},{"Start":"04:46.145 ","End":"04:52.740","Text":"The substance has a lower energy than the elements."},{"Start":"04:53.320 ","End":"05:00.095","Text":"Because this distance, the energy of the substance minus the energy of the elements,"},{"Start":"05:00.095 ","End":"05:02.600","Text":"or the enthalpy of the substance minus"},{"Start":"05:02.600 ","End":"05:06.410","Text":"the enthalpy of the elements is going to be negative."},{"Start":"05:06.410 ","End":"05:10.810","Text":"Delta H_f^0 will be negative."},{"Start":"05:10.810 ","End":"05:15.665","Text":"Now we know that when the delta H of formation is positive,"},{"Start":"05:15.665 ","End":"05:18.605","Text":"the substance is less stable than its elements."},{"Start":"05:18.605 ","End":"05:20.075","Text":"Whereas if it\u0027s negative,"},{"Start":"05:20.075 ","End":"05:23.390","Text":"the substance is more stable than its elements."},{"Start":"05:23.390 ","End":"05:28.860","Text":"In this video, we defined the standard enthalpy of formation of a substance."}],"ID":34092},{"Watched":false,"Name":"Standard Enthalpy of Reaction","Duration":"5m 23s","ChapterTopicVideoID":19312,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"In a previous video,"},{"Start":"00:01.800 ","End":"00:04.799","Text":"we defined the standard enthalpy of formation."},{"Start":"00:04.799 ","End":"00:10.335","Text":"In this video, we\u0027ll use it to calculate the standard enthalpy of reaction."},{"Start":"00:10.335 ","End":"00:14.520","Text":"Let\u0027s recall what we learned about the standard enthalpy of formation."},{"Start":"00:14.520 ","End":"00:18.724","Text":"The standard enthalpy of formation Delta H^0_f,"},{"Start":"00:18.724 ","End":"00:20.055","Text":"0 for standard,"},{"Start":"00:20.055 ","End":"00:25.280","Text":"f of formation of many substances are listed in tables,"},{"Start":"00:25.280 ","End":"00:30.695","Text":"and it can be used to calculate the enthalpy of reaction under standard conditions,"},{"Start":"00:30.695 ","End":"00:35.420","Text":"which are 1 bar and 25 degrees Celsius."},{"Start":"00:35.420 ","End":"00:37.425","Text":"We need to remember 2 things."},{"Start":"00:37.425 ","End":"00:42.485","Text":"First of all, Delta H of formation is defined for 1 mole."},{"Start":"00:42.485 ","End":"00:44.195","Text":"If there are n moles,"},{"Start":"00:44.195 ","End":"00:49.190","Text":"we need to multiply it by n. The second thing we need to recall is that"},{"Start":"00:49.190 ","End":"00:53.450","Text":"Delta H formation is 0 for an element in"},{"Start":"00:53.450 ","End":"00:57.980","Text":"its reference form and we\u0027ll see in the problem that we\u0027re going to solve,"},{"Start":"00:57.980 ","End":"01:02.180","Text":"that is 0 for oxygen gas."},{"Start":"01:02.180 ","End":"01:03.915","Text":"Let\u0027s solve an example."},{"Start":"01:03.915 ","End":"01:06.950","Text":"Consider the combustion of methane and calculate"},{"Start":"01:06.950 ","End":"01:10.760","Text":"the enthalpy of reaction using the enthalpies of formation."},{"Start":"01:10.760 ","End":"01:15.440","Text":"First thing to do is to write the equation for combustion of methane."},{"Start":"01:15.440 ","End":"01:18.080","Text":"Here it is CH_4, it\u0027s a gas,"},{"Start":"01:18.080 ","End":"01:21.895","Text":"that\u0027s methane, plus 2O_2 gas,"},{"Start":"01:21.895 ","End":"01:26.603","Text":"that\u0027s oxygen gas, gives us CO_2 carbon dioxide gas,"},{"Start":"01:26.603 ","End":"01:30.065","Text":"and 2 moles of water, which is a liquid."},{"Start":"01:30.065 ","End":"01:33.035","Text":"Here\u0027s the information we need to solve the problem."},{"Start":"01:33.035 ","End":"01:39.235","Text":"The Delta H of formation of methane is -74.5 kilojoules per mole."},{"Start":"01:39.235 ","End":"01:43.865","Text":"The Delta H of formation of oxygen gas,"},{"Start":"01:43.865 ","End":"01:46.370","Text":"which is its reference state, 0."},{"Start":"01:46.370 ","End":"01:53.405","Text":"Delta H of formation of water is -393.5 kilojoules per mole."},{"Start":"01:53.405 ","End":"02:02.800","Text":"Finally, the Delta H of formation of carbon dioxide gas is -285.8 kilojoules per mole."},{"Start":"02:02.800 ","End":"02:06.050","Text":"Now the Delta H of reaction under 0,"},{"Start":"02:06.050 ","End":"02:09.260","Text":"we\u0027re putting it because we\u0027re doing it in the standard state,"},{"Start":"02:09.260 ","End":"02:12.530","Text":"is Delta H^0 of the products."},{"Start":"02:12.530 ","End":"02:19.010","Text":"That\u0027s Delta H final minus Delta H^0 of the reactants."},{"Start":"02:19.010 ","End":"02:21.365","Text":"That\u0027s like Delta H initial."},{"Start":"02:21.365 ","End":"02:25.490","Text":"Let\u0027s first calculate the Delta H^0 of the products."},{"Start":"02:25.490 ","End":"02:27.975","Text":"Now, let\u0027s look at the products."},{"Start":"02:27.975 ","End":"02:33.686","Text":"The number of moles of CO_2 times the enthalpy of formation of CO_2,"},{"Start":"02:33.686 ","End":"02:38.550","Text":"plus the number of moles of water times the enthalpy of formation of water."},{"Start":"02:38.550 ","End":"02:39.935","Text":"We can work that out."},{"Start":"02:39.935 ","End":"02:42.145","Text":"There is only 1 mole of CO_2,"},{"Start":"02:42.145 ","End":"02:45.555","Text":"so it\u0027s -285.8 kilojoules,"},{"Start":"02:45.555 ","End":"02:48.930","Text":"but there are 2 moles of water."},{"Start":"02:48.930 ","End":"02:51.600","Text":"N_H_2O is 2."},{"Start":"02:51.600 ","End":"02:57.705","Text":"We have +2 times -393.5 kilojoules."},{"Start":"02:57.705 ","End":"02:59.070","Text":"We work all that out."},{"Start":"02:59.070 ","End":"03:03.570","Text":"It\u0027s -1,072.8 kilojoules."},{"Start":"03:03.570 ","End":"03:06.390","Text":"Now let\u0027s look at the reactants."},{"Start":"03:06.390 ","End":"03:13.025","Text":"Delta H^0 of the reactants is the number of moles of CH_4 times"},{"Start":"03:13.025 ","End":"03:16.790","Text":"the enthalpy of formation of CH_4 plus the number of moles"},{"Start":"03:16.790 ","End":"03:20.765","Text":"of oxygen times the enthalpy of formation of oxygen."},{"Start":"03:20.765 ","End":"03:24.410","Text":"Now for methane, we have only 1 mole."},{"Start":"03:24.410 ","End":"03:26.540","Text":"So n_CH_4 is 1."},{"Start":"03:26.540 ","End":"03:34.800","Text":"Delta H^0_f for methane we were given before is -74.5 kilojoules,"},{"Start":"03:34.800 ","End":"03:38.480","Text":"and oxygen is in its reference state,"},{"Start":"03:38.480 ","End":"03:40.445","Text":"as we\u0027ve said several times before,"},{"Start":"03:40.445 ","End":"03:43.925","Text":"so Delta H^0_f for oxygen 0,"},{"Start":"03:43.925 ","End":"03:45.965","Text":"so we don\u0027t include that."},{"Start":"03:45.965 ","End":"03:49.790","Text":"So here\u0027s our total for the reactants."},{"Start":"03:49.790 ","End":"03:53.720","Text":"Now we can work out Delta H for the reaction,"},{"Start":"03:53.720 ","End":"03:56.735","Text":"which is a standard enthalpy of reaction."},{"Start":"03:56.735 ","End":"03:58.385","Text":"We take first of all,"},{"Start":"03:58.385 ","End":"04:08.355","Text":"the products -1,072.8 kilojoules and subtract the Delta H for the reactants."},{"Start":"04:08.355 ","End":"04:11.939","Text":"The reactants is -74.5 kilojoules."},{"Start":"04:11.939 ","End":"04:17.280","Text":"Work that out, we get -998.3 kilojoules."},{"Start":"04:17.280 ","End":"04:20.675","Text":"Now, it\u0027s useful to write a general equation,"},{"Start":"04:20.675 ","End":"04:24.280","Text":"one that isn\u0027t specific to this particular problem."},{"Start":"04:24.280 ","End":"04:28.110","Text":"We can write that Delta H of the reaction in"},{"Start":"04:28.110 ","End":"04:32.420","Text":"its standard state that\u0027s 0 is equal to the sum,"},{"Start":"04:32.420 ","End":"04:35.255","Text":"going to the sum over all the products,"},{"Start":"04:35.255 ","End":"04:42.155","Text":"the number of moles of each product times its Delta H of formation."},{"Start":"04:42.155 ","End":"04:49.040","Text":"We\u0027re going to do Delta H formation of each product minus now we go to the reactants."},{"Start":"04:49.040 ","End":"04:51.410","Text":"We\u0027re going to sum over all the reactants."},{"Start":"04:51.410 ","End":"04:55.628","Text":"The sum over all the reactants times n_r,"},{"Start":"04:55.628 ","End":"04:59.090","Text":"that\u0027s the number of moles of a particular reactant,"},{"Start":"04:59.090 ","End":"05:04.160","Text":"times Delta H_f^0 for that reactant."},{"Start":"05:04.160 ","End":"05:09.140","Text":"This is a general equation which we can use in"},{"Start":"05:09.140 ","End":"05:15.005","Text":"every case and it\u0027s totally equivalent to what we did in solving the problem."},{"Start":"05:15.005 ","End":"05:18.680","Text":"In this video, we calculated the standard enthalpy"},{"Start":"05:18.680 ","End":"05:23.550","Text":"of reaction from the standard enthalpies of formation."}],"ID":34093},{"Watched":false,"Name":"Exercise 9","Duration":"3m 14s","ChapterTopicVideoID":31827,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34094},{"Watched":false,"Name":"Exercise 10","Duration":"4m 5s","ChapterTopicVideoID":31828,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34095},{"Watched":false,"Name":"Exercise 11","Duration":"4m 22s","ChapterTopicVideoID":31829,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34096},{"Watched":false,"Name":"Exercise 12","Duration":"2m 40s","ChapterTopicVideoID":31830,"CourseChapterTopicPlaylistID":81302,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":34097}],"Thumbnail":null,"ID":81302}]

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