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welcome to the next section in our unit on the first law of thermodynamics. In this video, we're going to talk about heat can cause a change in temperature specifically And now, because we're gonna be talking about heat specifically, we're gonna be thinking about situations like putting a pan on the hot stove where their work is going to be zero jewels. Now, this is convenient because we can establish a model. What says For some change in thermal energy, we get some proportional amount of change in temperature and the proportionality factor here we call the heat capacity and it is unique to the object that we are considering. So the units of heat capacity are jewels per Calvin. As in, we add a certain number of jewels in order to a chain. A certain change in Calvin. I'm looking over here. Then we know if we have plug in the first law, we get Q minus. W equals ch times Delta T. But we're assuming work a zero. So we have given some heat. We get some change in temperature according to our heat capacity notice here. If we had a negative heat, that is to say, If we're pulling energy out of the system, then we'd have a decrease in temperature. Now, Um, because it's inconvenient to major the heat capacity of every object in the world. We like to talk about specific heat and Mueller specific heat where instead, what we do is we explicitly pull out the mount of material that we have. So the mass or the number of moles that we have and then we can go in and calculate the specific heat or Mueller specific heat for a particular material so we can find the specific heat of lead or aluminum instead of that particular piece of aluminum. So this is convenient. You can see the equation looks very similar to what we had before. Where we have heat is equal to M. Times, C. S, Delta T and Q equals end time cm Delta T. It's important that whenever you look at a table of specific heats, you know what if there if you're looking at, uh, Mueller specific heats or standards specific heats in order to do this, just look at the units. A standard specific heat will be jewels per kilogram. Kelvin and a Mueller specific heat will be jewels per mole, Kelvin. Now, we don't just have to worry about specific heats. We also have to worry about phase changes. When we go from a solid to a liquid to a gas or from a gas to a liquid to a solid, it costs extra energy in the form of heat in order to either break or form the bonds. So we have what we call the heat of transformation. Looking at our plot here, you can see what happens is we're going to have an increasing amount of heat, but no change in temperature when we have these heats of transformation. So this is when, for example, here were either. If we're on our way up in temperature, we're going to boil our liquid so that it becomes a gas or we could be condensing our gas into a liquid. So this is going to be the heat of vaporization. So this is also some in some books called the Latent heat of vaporization and latent heat of fusion. Um, now l fusion, then is going to be the amount of energy that it's, or the amount of heat of transformation that is going to require in order to either melt or to freeze something. So we have to keep an on eye on whether or not we are going through a phase change. And if we do, then we need to take into account this amount of extra heat that it will take to cause that total change in phase. For example, if I were to ask you how much energy does it take to go from 50 degrees Celsius water to 150 degrees Celsius water? Well, you might know that it's gonna pack across the boiling point and so you need to take into account not just that we have heat capacities, but that we also have heats of transformation that we need to calculate. Now. The field of keeping track of that change in temperature is called calorie mitri. The idea of Kalorama tree is simply that we're going to put two objects together A and B here, where their initial temperatures are different from each other and then the total exchange of heat must add up to be zero jewels. Noticed that since B is hotter, it will lose heat to a and A will lose heat back to be, and the idea is that we're going to wait until they come to thermal equilibrium. So when they come to their common equilibrium, we know that Q A Plus Q B will be zero, which means we can write out this equation by plugging in our heat capacity are specific heat capacity equations. So once we have our specific heat written out, we can put in are changing temperatures, recognizing we want the same final temperature and then solve for the final temperature of the system. Now we only have to solve for one because both A and B will be at the same temperature notice here that we have to keep track of a lot of different numbers. The mass of a the massive B, the specific heat of a the specific heat of B don't get a mixed up. It's best to work with variables in this case so that you can put in your numbers at the end. Now it gets a little tricky with gas is because we know we have all these different processes that we've been talking about. Thankfully, the is a thermal process simply means that we're not gonna have any change in temperature, so that's a bit of a null result there. But it turns out that if we want Thio, consider a constant volume process or a constant pressure process. Then we need to consider different specific heats. So here we have Moeller specific heats that are for constant volume and constant pressure. Mueller specific heat is the preferred style when you're working with a gas as opposed to the standard specific heat. And if you were to look at the table in your textbook, where it lists all of these specific heat, the Mueller specific heats with constant volume in the Muellers specific heat with constant pressure. What you're going to find is that this relationship is going to hold true for an individual gas. That the Mueller specific heat, with respect to a constant pressure, is going to be equal to the Mueller specific heat with respect to a constant volume. Plus the idea the universal gas constant are, it's an interesting relationship here, and because of it we will find that the total thermal change in thermal energy for any ideal gas process. So this is any ideal gas process can be calculated as n times the specific Mueller specific heat for a constant volume. Times Delta T. So this is interesting, and it allows us to do quite a bit with it now. One other thing you should check is that in your table you will find that there is a difference in between the Mueller specific heats for a mon atomic gas and for a dye atomic gas. Most tables will actually list both, even if it is rare to find the mon atomic or die atomic form. The reason for that will become apparent in the next unit, but it's an important thing to notice right now. Now, in order to hint, there's one other process that we need to talk about, and we've already alluded to it. And that is Theodora Batic process. Remember, this is the process for which Q is equal to zero. But remember, that does not mean that we will not see a change in temperature because we still have work done on and by the system that could change the thermal energy. So in order to look at this, then we're going to end up with a work that is equal to this number because we have, uh changing thermal energy is going to be equal to this defining, then a constant gamma, which is the ratio of our to Mueller specific heats. We can actually come up with these two equations, which will help us solve for what's happening in an ad automatic process. Now, on Addie, a bat is not going to look like a nice a therm, though they look kind of similar on a PV diagram where an addi a bat will look like this. A nice awesome will look something more like this. So not exactly the same, though they will look similar on the plots.

University of North Carolina at Chapel Hill