Hey, guys. So in previous videos, we've talked about the second law of thermodynamics, and there's a statement that said that no heat engine could ever have an efficiency of 100%. But some problems are going to ask you to calculate something called the maximum theoretical efficiency between reservoirs. And to answer that, we're going to look at something called the Carnot cycle or the Carnot engine. So this is something your professors and textbooks may make a big deal out of. What I'm going to show you in this video is that it's really just a special type of cycle, and we're going to see some equations that are really similar to ones that we've seen before. So let's check this out here. So basically, the Carnot cycle, some guy, Carnot, back in the 1800s, he sat down, played around with a bunch of these diagrams, and figured out that you could design an ideal reversible cycle. And if you do this, it has the maximum possible efficiency. So, basically, what he was like what he said was, well, if you can't get 100%, how high can you actually go? Now, before I give you the equation here, I want to talk about this word reversible for just a second because you might see it get thrown around a lot. And, really, anything, an engine or a cycle is reversible, basically if it happens infinitely slowly, really, really, really slow, and there's also no friction or anything like that that takes away energy from the system. That's what reversible means. So let's take a look at this Carnot cycle here. It really just has 4 steps, 2 isothermal and 2 adiabatic. So the first one from a to b is an isothermal expansion. So you're going to ride this isotherm down like this. And basically, this is the step here where it's absorbing heat from the hot reservoir. So here is the step where it's taking in heat from the hot reservoir. Then what happens is that there's an adiabatic expansion. So remember that's going to be steeper than an isotherm and it's going to look like this from b to c. And what happens in this adiabatic is that there is no heat transfer. Right? That's the definition. Then basically, from the next for the next two steps, it just gets reversed. So here now there is an isothermal compression from c to d. It goes the opposite way like this. And in this step, what happens is that heat is flowing out to the cold reservoir. And then what happens is, from d to a, there is an adiabatic compression, and then the cycle repeats itself. So 2 adiabatic, 2 isothermal. Alright. That's really what the Carnot cycle is, and then the work that is done is just going to be the area that's enclosed inside of this diagram. You don't really need to know that, but that's just the work. Alright. So what you might need to know though is where the heat gets transferred in the cycle here. Now we just said that the steps 2 and 4 are adiabatic, and what that means is that there's no heat transfer. So what that means is that it only gets transferred during steps 1 and 3, the 2 isothermal ones. This is where you absorb heat and then release it. Now your textbooks are going to have some pretty lengthy derivations. They're going to do some equations with molar specific heats and adiabatics and all this and that. And basically, what you're going to see here is that the maximum possible efficiency is \(1 - \frac{T_c}{T_h}\). Notice how it looks very similar to the equation over here that we've been working with so far. It's just instead of q's we're using t's. But that's basically what the maximum possible efficiency is. It depends only on the temperatures of the reservoirs themselves. That's what he found out. So the other equation they might need to know here is that the heat released and absorbed are basically the ratio of \( \frac{Q_c}{Q_h} \) is equal to the absolute value of \( \frac{T_c}{T_h} \). So notice how all these are going to be positive numbers when you do the absolute values, but that's really all that you need to know. You just need to know the maximum possible efficiency equation and then this one over here. That's really all you need. Let's go ahead and take a look at our example. So we have a Carnot engine that is operating between 520 and 300. So we have \(T_h\) here is 520. \(T_c\) here is 300. So what happens is the engine is going to take in some amount of heat from the hot reservoir. That's going to be \(Q_h\). So this is going to be 6.45 kilojoules. And then what happens is we want to calculate in part a what is the maximum theoretical efficiency. So what happens is \(e_{Carnot}\), this is going to be the maximum theoretical efficiency if this is a reversible engine. It's just going to be \(1 - \frac{T_c}{T_h}\). So this is going to be \(1 - \frac{300}{520}\), and you're going to get a theoretical maximum of 42%. So what this means here is that even if you could design the perfect engine that ran in this cycle here, the maximum efficiency that you could ever hope to even get out of this would be 42%, and that's the answer to part a. So part b now asks how much waste heat does the engine expel each cycle? We know \(Q_h\). We don't know what the work is, but now we want to calculate what is \(Q_c\). How do we calculate that? Well, remember that we have one equation now, that relates \(Q_c\) with all of the other variables. So, basically, what happens is if we want \(Q_c\), we're going to use \(\frac{Q_c}{Q_h}\), absolute value is going to equal \(\frac{T_c}{T_h}\). Right? The ratio of these heats is just for the ratio of the hot and cold reservoirs. So all you do here is \(Q_c\) is just going to be equal to \(Q_h \times \frac{T_c}{T_h}\). We're just going to use all positive numbers here. So they're going to be 6.45, and then this is going to be \( \frac{300}{520} \). When you work this out, what you're going to get is 3.72 kilojoules. So this is going to be 3.72 kilojoules. And that's the answer. Alright. So, finally now let's take a look at part c here. How much mechanical work does the engine produce? So, really what we want to calculate is this \(W\) here. We have a couple of different ways to calculate this. Remember that \(W\) for the engine is always just, the difference in \(Q_h - Q_c\). So we can actually just subtract these 2 right here. We You can also just get it from the efficiency equation. There's a bunch of different ways, but this is probably going to be the most straightforward. So \(W\) is just going to be the heat that gets absorbed, 6.45. This is going to be kilojoules minus 3.72 kilojoules. And then your answer is going to be 2.73 kilojoules. Alright? So, basically, if you had a Carnot cycle, this is as much work as you could possibly hope to extract per cycle. So that's it for this one, guys. Let me know if you have any questions.
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The Carnot Cycle
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