 All right, so we have two expressions that tell us how to calculate the rate at which the entropy is changing as we increase or decrease the temperature of something at either constant volume or at constant pressure. So these are going to be useful. We've had equations before. We've calculated the change in entropy upon heating of a gas, an ideal gas in particular, but these equations apply to any substance whatsoever. It doesn't have to be an ideal gas. So to see how those work, let's try an example. Let's say we have a substance. So instead of a gas, I'll choose a solid object. I'll say we have a penny. If I pull a penny out of my pocket, pennies are made of copper, or at least they were. Let's say I have an old penny in 1980, penny that I pull out of my pocket. Modern pennies are made of zinc with a copper coating. But before the early 1980s, they were made almost entirely of copper. So I'll say I've got a pure copper penny. Penny's way turns out 3.11 grams. So let's say I pull a copper penny out of my pocket. The initial temperature is 20 Celsius or 293 Kelvin. And I'm going to hold it in my hand until it heats up to body temperature. So at that point, it's reached 37 Celsius or 310 Kelvin. So the question is, how much does the entropy of that penny change when I just warm it up to body temperature from roughly room temperature? So knowing how much the entropy changes when I change the temperature, that sounds like a prescription for using one of these expressions. And I'm doing this at constant pressure. Penny's volume won't change much, but I didn't make any special efforts to keep the volume exactly constant. So I'm doing it at a constant one-atmosphere pressure. So this is the expression I would need. ds dt is molar constant pressure heat capacity over temperature. So I could rearrange that expression and say ds is equal to Cp over T dt with or without the bars on top to indicate molar quantities. If I integrate both sides of that equation, the total change in the entropy or the molar entropy is the integral of the molar heat capacity over temperature as I change the temperature from some initial temperature T1 to some final temperature T2. So the thing on the right-hand side is what I need to evaluate for my particular penny. So I need to know what is the heat capacity, the constant pressure heat capacity of a copper penny? Well, copper is a monatomic metal. So we know what the heat capacities of monatomic metals are, at least according to the Equal Partition Theorem or according to DuLong and Battit. So the constant pressure heat capacity of a solid-like copper is 3R, or should be close to 3R. So we can use that as our estimate of the heat capacity. So this would, 3R is just a constant. And since the heat capacity is constant, I can pull it out of the integral. So I would have the heat capacity, I'm sorry, the entropy. Being 3R times the integral of 1 over T dt, integrated from T1 to T2. Integral of 1 over T is natural log of the temperature. And again, if I evaluate that from T1 to T2, that'll give me 3 times R times the natural log of that ratio, T2 over T1. So that's the change in entropy, the molar change in entropy. Let's go ahead and evaluate that. Actually, I don't think I pre-calculated the answer to that one. So I'll instead will say, I don't want the entropy per mole. I want the entropy of this single actual penny that weighs 3.11 grams. So the total change in entropy is going to be the number of moles times the change in entropy per mole. So now, so that would be 3 times N times R natural log of T2 over T1. Now we have all the numbers, and we need to plug into that expression 3 times the number of moles. 3.11 grams is, if I divide by the molar mass of copper, that'll give me the number of moles. Grams will cancel, leaving me with just moles on top. Multiply by the gas constant, 8.314 joules per mole Kelvin. Also multiply by the natural log of this ratio of the temperatures. Final temperature is 310 in Kelvin. Initial temperature is 293. So 310 Kelvin divided by 293 Kelvin. Inside of the natural log will be unitless. Units cancel here. Units cancel. Grams cancel. These moles will cancel these moles. And when I'm done, final answer will have units of joules per Kelvin, as it should for an extensive entropy. And if we use our calculator to find the answer to that, we get a particular answer, 0.0689 joules per Kelvin. So that would be the change in the entropy of a penny as it's brought from roughly room temperature up to 37 Celsius. So maybe that's not a tremendously satisfying answer, because we don't have a very good intuitive grasp of what a joule per Kelvin is, or what a larger small quantity of entropy is. But at least we have a method now of calculating the change in entropy for some object that's heating up or cooling down at either constant volume or at constant pressure. In this case, specifically, when I was able to assume that the heat capacity was constant. Often, however, the heat capacity is not going to be constant. So the more general result, if I back up to this point in the calculation, if I don't know that the heat capacity is constant, because in fact it usually won't be, I would need to use one of these two expressions if I'm doing it at constant volume, if I'm doing it at constant pressure. These heat capacities might, in general, depend on the temperature. In fact, the heat capacity does depend on temperature in the general case. It's only in cases like this where we can assume the heat capacity is approximately constant that we can pull it out of the integral. In that case, for example, we've seen previously an example of what's often used as the temperature dependence, a model that can predict the temperature dependence or predict the temperature dependence of the heat capacity, this Schoemate equation. Just an empirical equation. If you know these coefficients, A, B, C, D, and E, you can evaluate what the heat capacity is at any range of temperature where this model is appropriate. So if you have a model like this that describes how the heat capacity is changing with temperature, you can take that heat capacity and plug it in to this expression, and it's not nearly as simple as a case where the heat capacity is constant, but if you take this expression and divide it by temperature, it still leaves you with something you could take the integral of and evaluate. That's a somewhat more involved calculation, so I won't try to do an example right here, but these same expressions give you the ability to calculate the entropy even when the heat capacity is something that's changing with respect to temperature because it's inside this integral with respect to temperature. So we have this procedure for evaluating the entropy, in particular, how much the entropy is changing. I can tell you the entropy of this penny changed by .0689 Joules per Kelvin as I warmed it up. What I can't tell you right now is how much entropy it had at the beginning or even how much entropy it had at the end. All I know is the entropy changed by this amount. Turns out we can also say something about the actual amount of entropy that an object has, so we'll consider that next.