Introduction to Heat

Concept: What is Heat?

13m
Video Transcript

Hey guys , now we really want to talk about heat and focus on developing a definition a physical definition for heat we already saw that heat is an energy and it's closely related to changes in temperature when substances that are in thermal contact exchange heat. So now we really want to talk about what exactly heat, alright let's get to it. Now remember two objects in thermal contact at different temperatures will reach thermal equilibrium via an exchange of heat. So we have some object that's a hot objects right its temperature is high some object that's a cold object that hot object is going to throw heat into the cold object whatever that he is we haven't defined it yet until their temperatures balance out the temperature of the hot object is going to drop the temperature of the cold object is going to rise and then eventually at thermal equilibrium their temperatures are going to equal lets say this is object one and that's subject two. Now thermal equilibrium is reached when as I just said the temperature is the same and this is achieved as we've talked about before by exchanging heat. Since temperature as we naively defined it depends on kinetic energy, then heat must be an energy quantity this is all a review this is all something that we've seen before alright now we want another naive definition we want to naively define heat this time remember what a naive definition is, it's a sort of basic definition that we're intending to build upon so our later definitions when we get into so much more advanced thermodynamics might not look exactly like this. Heat which is given by the letter Q is non-mechanical energy that is transferred from one substance to another to change temperatures. Mechanical energy is these. These are those macroscopic energies that you see the macroscopic kinetic energy of a substance that's moving as a whole. Potential energy between substances and external source's of a force those are mechanical energies and heat is a non mechanical energy, now as we've talked about later on heat will always flow from the hotter substance to the colder substance we've always said that, right just look at the diagram that I drew above look back to when we first talked about temperature we've always said that heat is going to flow from the hotter object or the substance to the colder substance this is actually a requirement of the second law of thermodynamics which is something that we'll cover later on. Heat doesn't have much meaning on its own it doesn't make any sense to think about how much heat a substance has right I never talked about how much heat object one has the hotter object or how much heat object two has the colder object at no time did I ever talk about how much heat an object carries heat only has relevance when we talk about exchanging heat, much like work is a change in energy no object has an amount of work but when an objects mechanical energy is changing work is being done work is that change in energy heat is absolutely always always going to be considered an amount of energy transferred from one substance to another, to another sometimes means what we would call the environment just the outside so if you have a hot objects. It might radiate heat outwards in order to drop its temperature and that he might not look like it's moving from one substance to another right it might look like its just going off into the abyss but it is that other substance is the environment it's that air around hot object, so there will always be an energy transfer from one substance to another that's the only way that heat makes sense. Now let's imagine an experimental apparatus as shown above me what we have is some sort of weight that's under the influence of gravity and it's free to fall downwards. As it falls it pulls this rope right it puts a tension in the rope which pulls the rope which spins this pulley and that pulley is connected to a rod through the water filled cylinder and since the pulley is spinning the rod is going to spin as well attached to this rod are these fans that spin in the water. So as the weight falls spinning fan. The fans spin inside the water because there's that friction because water has viscosity and it takes energy to move through the water, the water heats up it excepts some energy from those fans and it increases its temperature this is an experimental fact this is an experiment that one of the most famous physicists Joule did and showed that a falling block which has mechanical energy can actually exchange heat with a fluid like water increasing that water's temperature, some amount of heat must have been added to the water in order for the temperature of the water to rise. One calorie of heat that is added, calories the unit for heat is added when the block loses 4.186 joules of potential energy so if this block drops some height such that Delta U is negative. I need more room if it drops some height such that Delta U is -4.186 joules, then the spinning fan blades gave one calorie of heat to the water. This is known as the mechanical equivalence of heat that you can convert mechanical work an exchange of mechanical energy into heat technically the unit of heat is calorie but it has an equivalent to the unit of mechanical energy the joule one calorie is 4.16 joules. The common calorie that you see on foods which is where you guys have all heard the term calorie from has a capital C this is used primarily in the US to denote energy content of foods this common calorie with a capital C is actually 1000 calories as I defined it here or one kilocalorie. In Europe Company sorry countries that use calories to define energy continent food use kilocalories some countries don't use them and they use kilojoules to actually define the energy continent food which is pretty weird like I've eaten German candy and you look ar the back and it says that the amount of energy it has per serving is in kilojoules that's pretty weird. Now let's do an example to sort of round this out and it's going to be centered around food because we're talking about calories here. A can of soda has around 40 grams of carbs, each grams of carbohydrate is about 4 calories as we use them in the U.S. How many joules of energy does a can of soda contain in its carbohydrates alone. So it is also going to contain other things that have calories 1 gram of protein is roughly 4 U.S calories right 1 gram of fat is roughly 9 US calories. So the energy contained in the carbs alone are 40 grams of carbs and what we want is we want to convert that into some amount of joules. So we're going to need some conversion factors here and this is how you want to set up any of these what are called dimensional analysis problems where you're converting from one thing to another, we want to go from carbs to joules. Well we've never learned anything before about carbs we learned one thing about joules so the odds are the information that we need is going to be in the problem and we're told a gram of carbohydrate is about 4 calories so 1 gram of carbs is 4 calories. Now that's very very important because these are calories with a capital C. We just learned mechanical equivalents that one lower case c calorie is 4.186 joules so it turns out that we can't do this in two steps because we have the conversion 1 calorie is 4.186 joules but we're not in little c calorie we're in capital C calorie so let me move this back and let me add a third conversion factor. First we need to say one capital C calorie is 1000 lowercase calories and just to keep track or grams of carbs change into capital C calories our capital C calories change in the lower C calories and finally we have our mechanical equivalence which says one lower case Cal is 4.186 joules and if you plug this conversion into calculator you get 6609 sorry 669,760 joules man that took a lot to get out or 6.7 times 10 to the 5 joules however you want to write the answer. This is about mechanical equivalence a calorie is a unit of heat a joule is a unit of mechanical energy but they have an equivalence between the two so you can use the units interchangeably. Alright guys that wraps up our introduction into what exactly heat is . Thanks for watching.

Concept: A Microscopic View of Heat and Temperature

8m
Video Transcript

Hey guys, in this video we want to talk about a microscopic view of heat and how it relates to temperature. So far we've been talking everything mostly macroscopically, just looking at oh some substance has a temperature, when you put it into thermal contact with another substance, they exchange some amount of heat and the temperatures change but we haven't really looked at what this means about the particles moving inside of those substances since we know that temperature depends upon that internal kinetic energy of the particles within the substance. Since heat changes the temperature, it has to relate to changes in those particles themselves. Alright let's get to it. Remember heat transfer between two substances will change the temperatures of those substances. Most commonly, I had mentioned all the way back in the beginning when we talk about temperature that there are instances where this isn't true but that's for later on. For now let's just consider that as being true. The question is how exactly is heat exchange related to a change in temperature? And we want to analyse this on the microscopic level by looking at the actual particles inside of substances. Remember guys, temperature is a measure of what? It's a measure of the kinetic energy of the particles in a substance not the kinetic energy of the substance itself but that kinetic energy of the particles within the substance. He must change this quantity, this kinetic energy of the particles, in order to change the temperature right and therefore heat must be an energy quantity. This is a conclusion that we've arrived at several times but I keep arriving to the conclusion in several different ways because every time we sort of reaffirm exactly what heat is. Now let's consider two substances which I've drawn below. We have substance one and substance two. Substance one is hotter than substance two so T1 greater than T2. That means what about the particles moving in substance one versus the particles moving in substance two? Well the temperature of one is higher than the temperature two, the average kinetic energy of the particles in one is higher than the average kinetic energy in the particles of two. So particles in substance one are moving faster on average than those in substance two. Assuming they have the same mass. Now if thermal contact is allowed, then collisions occur between particles in the different substances. So we have these particles here that are just moving freely and every now and then one reaches this boundary between the two substances and we have these particles just moving randomly in the other substances and another one will reach the boundary randomly. Every now and then, two will reach the boundary at the same time and there will be a collision. Energy is exchanged between those collisions in the substances. That energy exchange leads to a change in the kinetic energy of the particles. This is not the kinetic energy of the substances but of the particles. A change in that kinetic energy leads to a change in temperature. So what heat is is at the microscopic level heat is this energy transfer in the collisions. The exchange of energy during the collisions between the particles in two substances is the transfer of heat. That is how heat is transferred when two substances are put in conductive contact with one another if heat is allowed to conduct between the two substances. This is how heat exchange directly determines temperature changes. Just as I said heat exchange changes the average kinetic energy of the particles which changes the average sorry which changes the temperature of the substance and heat will continue to be exchanged until thermal equilibrium is reached or T1 equals T2. At that point, when you have particles colliding, so maybe this one gives energy to this one then they collide here. This one gives energy to this one, they collide over here. This one gives energy to this one. Initially because substance one is hotter, the energy that these collisions are giving the substance two particles is higher but when they reach equilibrium, now it starts to balance out and then basically whatever energy substance one transfers over substance two, substance two transfers it back in other collisions. At equilibrium, heat is still exist changed but it's exchanged equally exactly what I said. However much heat substance one gives to substance two via these collisions, substance two gives back to substance one. So that the temperatures remain the same. When calculating heat exchange, Q. So Q we had originally said was heat when we defined heat but heat doesn't make any sense outside of talking about exchanges so Q really is a heat exchange or heat transfer or you can just say heat and understand that heat is always exchanged. So by convention heat is positive when entering a substance and heat is negative when leaving a substance or heat transfer is positive when entering and heat transfer is negative when leaving. Typically in physics, when we are analyzing and talking about energy of things whenever the energy of that thing, increases we say it was a positive change in energy. Whenever the energy of that thing decreases, we say it was a negative change in that energy. So this is the same sign convention for work when work is positive the kinetic energy of an object increased, right this is mechanical work sorry this is mechanical energy so this is macroscopic and whenever work is negative the change in kinetic energy for that object decreased. Alright? Okay guys, this wraps up our discussion on the microscopic view of heat and how it relates to temperature and more importantly temperature changes. Thanks for watching guys.

Example: Elastic Collisions at Boundaries Between Identical Substances

16m
Video Transcript

Hey guys, let's do an example. At the boundary between two identical substances, a particle of mass M in substance 1 moving at a speed perpendicular to the boundary of a V perpendicular, that's what this little symbol means, encounters a particle in substance 2 moving at a perpendicular speed of twice V perpendicular. If they collide elastically at the boundary, how much heat is transferred? And in which direction does the heat move, to substance 1 or to substance 2? So first we're talking about elastic collisions. So let's refresh ourselves on the the equations for elastic collisions and then we'll get to what's actually happening. So equation one is going to be momentum conservation which is in M1V1 plus M2V2 equals M1V1prime where the prime speeds are the final speeds post collision plus M2V2prime right. V prime, post collision. And in the second equation is a special equation that we use specifically for elastic collisions which says that V1 minus V2 equals V2 prime minus V1 prime. Once again the prime speeds are speeds post collision. This second equation is due to the fact that kinetic energy is conserved in elastic collisions but it's difficult to straight up use kinetic energy conservation because that equation is quadratic in speed right. It's proportional to speed squared whereas momentum conservation is linear in speed so the math gets really complicated just trust me you don't want to use it. Now let's look at the initial picture here before the collision. So we have our boundary, we have our particle let's say this is substance 1, this is substance 2. We have our particle in substance 1 moving with a speed V perpendicular to the boundary. It doesn't matter what angle it's moving at because there's no momentum exchange in the Y direction, there's only momentum exchange in sorry in the parallel direction there's only momentum exchange in the perpendicular direction so we can look at this exactly as if they were moving straight at each other perpendicularly and this has some mass M. Now in substance 2, this guy's moving at 2V perpendicular towards the boundary but what's its mass? We said that these were two identical substances, which means that they both have the same mass. So let's writes the initial sorry let's write the left hand side of equation one and equation two, the initial condition for equation one a equation two. So this is the left hand side of equation one, M1V1 plus M2V2 is MV perpendicular, now let me just define this as the positive direction so I can say that substance 1 is moving at positive V perpendicular and substance 2 is moving at -2V perpendicular. So this whole thing is -3MV perpendicular. And the left hand side of equation two is going to be V1 minus the V2. Now remember these are not speeds, these are still velocities so the sign still matters. V1 is positive V perpendicular and V2 is -2V perpendicular. So this equals positive 3V perpendicular and sorry this is just -MV perpendicular I cannot subtract. This is 1 minus 2 which is negative -1. Those are our initial that's our initial set up. Those are the left hand sides of our two elastic collision equations which are the initial conditions. Now let's look at the final. And I'll draw just for the sake of drawing it but we don't know anything about the velocities post collision so this is M and this is M and we don't know V1 final and we don't know V2 final. We know nothing about them. Let me minimize myself. Now the right hand side of the equation is an unknown. So we're going to leave this blank. Don't know anything post collision. So we cannot do anything with the right hand side we have to leave it as it's shown here just in variables. I mean technically if you want for equation one you can say that M1 and M2 are just M but that's all you can do. Now what we have to do is we have to equate the right hand side with the left hand side which is the whole point of these conservation problems. So what we have is for equation one, we have sorry not 3, -MV perpendicular. So this is left hand side equals right hand side. This is what we're doing. The actual conservation part. This equals M1 which is just MV1 final plus M2 which is just MV2 final. And I'm going to switch colors here, blue no longer means initial and green no longer means final we can draw a just a giant line across this if you want. Blue now means equation one, green means equation two, just keep this nice and separate. Equation two, the left hand side was 3V perpendicular and the right hand side was V2 final minus V1 final. Now we have two equations that we need to use substitution to solve them. So I'm going to say V2 final is just 3V perpendicular plus V1 final. I can just take this and add it over to that side. Now I can substitute this into here. So this is going to equal MV1 final plus 3V perpendicular plus V1 final. And now my first equation is entirely in terms of V1 final sorry I'm missing an M right here so this is MV1 final plus 3MV perpendicular plus and MV1 prime. All I did was I distributed the M. And now I can group the MV1 finals. So this is 2MV1 final plus 3MV perpendicular. So now let me write these equations again. Equation one, the left hand side never changed, the right hand side was simplified by using equation two, so this is -MV perpendicular equals 2MV1 prime plus 3MV perpendicular and before I do anything the equation two is our final form of equation two, V2 prime equals 3V perpendicular plus V1 prime. Now equation one is in a single variable so I can solve it. First what I'm going to do is I'm going to subtract that over to that side and I'm gonna get -1 minus 3 more so that's -4MV perpendicular equals 2MV perpendicular. I probably could have done this a while ago but I just didn't notice the M's can cancel or I could have done it as soon as I wrote down equation one and so all I have to do is divide the 2 over and V perpendicular is clearly negative sorry is V1 prime. This 2MV1 prime is this 2MV1 prime, that stayed the same that didn't change I just messed up. So V1 prime is -2V perpendicular. All I have to do is take this 2 and divide it over to that side. 4 divided by 2 is 2. So we officially have 1 result, now I can take that and I can plug it in for V1 prime and find V2 prime so this is going to be 3V perpendicular minus 2V perpendicular which is just V perpendicular so let me write it over here. V2 prime is just V perpendicular. So look at that, what was the original velocity of V2? It was negative V2 perpendicular. What was the original velocity V1? It was V perpendicular. These guys have switched velocities, so let me redraw the initial and the final. So once again, here's the initial. We've got our boundary, we have M, we have some V prime, we have M, we have some 2V prime and here is our final. We have our boundary right this is substance 1 and this is substance 2. Substance 1, substance 2. Now particle 1 is leaving to the left right because we said that to the right was positive. But V1 prime is negative so it's going to the left at 2V perpendicular. And particle two is going to the right because it's positive at V perpendicular. So how much energy was transferred from one substance to the other? Well let's look at the original energy. The initial kinetic energy of one was one half MV perpendicular squared, the initial kinetic energy of two was one half M2V perpendicular squared which is four halves MV perpendicular squared. Now let's look at the final kinetic energy. The final kinetic energy of particle one is one half M2V perpendicular squared which equals four halves MV perpendicular squared and the final kinetic energy of particle two is one half MV perpendicular squared. So how much was transferred? Well clearly, this guy gained three halves. This guy gained three halves MV perpendicular squared that's clear, it went from one half to four halves so it gained three halves. And this guy lost three halves. So you can see that clearly the magnitude of the heat transfer was just three halves MV perpendicular squared. That's how much heat was transferred and which direction did it go? It went from 2 to 1. Because the particle in substance 2 lost kinetic energy right this by the way, this arrow's backwards that's why I was I paused for a second because I confused myself. That arrow is backwards. Going from the initial state to the final state, substance 2 lost three halves MV perpendicular squared. Going from the initial state to the final state, substance 1 gained three halves MV perpendicular square. So the heat went from 2 to 1. So which substance was hotter? If that was another question which substance was at a higher temperature clearly substance 2 was at a higher temperature because it transferred heat to substance 1. Alright guys, that wraps up this really sort of in-depth look at exactly the amount of heat transfer for two substances that were not in thermal equilibrium. Thanks for watching guys.