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.

# The Carnot Cycle - Online Tutor, Practice Problems & Exam Prep

### The Carnot Cycle and Maximum Theoretical Efficiency

#### Video transcript

A theoretical heat engine in space could operate between the Sun's 5500°C surface and the –270.3°C temperature of intergalactic space. What would be its maximum theoretical efficiency?

A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?

### Lifting Mass with a Heat Engine

#### Video transcript

Hey, guys. So I've got a great example here for you. We're going to work this one out together. So we have this engine that's operating and it's extracting some heats and all this stuff, but we're asked to find out how many cycles it's going to take to lift a 10-kilogram mass. So I want to draw out what's going on here. So imagine that I have some kind of a box that's on the ground like this. It's 10 kilograms and we want to lift it up by, you know, 5 meters. So this height here is 5. Well, we remember back from energy conservation that if you want to lift a mass by some height, you have to do some work. So for example, the work that's required to lift this mass is equal to mgh, and so therefore it's just 10 times 9.8 times the 5 meters. So in other words, it takes 490 joules in order to lift this mass by 5 meters. So what does this have to do with the heat engine? Well, what's going on here is I have a Carnot engine that's operating between these temperatures and these heats, and that's the thing that's actually supplying the energy to lift this mass. So I'm going to draw out my sort of heat diagram or my my energy flow diagram. So I've got my hot reservoir that's connected to my engine, and this is connected to my cold reservoir. It's TC. So what's happening here is this engine, this is my Carnot engine here, is actually the thing that is supplying the work in order to lift this mass. So you can imagine that this Carnot engine, like, you know, is just like a box with a pulley or something. Maybe it's like a little pulley, and basically it's sort of cranking this this box to lift this mass by 5 meters. So that's basically what's going on here. The thing that's supplying the work here is my Carnot engine. And so remember, in a Carnot cycle or in any heat engine cycle, you have a heat that flows in. You have some work that's done by the engine, and then you have some heat that flows out to the cold reservoir. So what happens here is we have to figure out how many cycles it takes for this engine, this work that's produced by this engine, to lift this mass here. So basically what we're trying to find is the number of cycles times the work that's done in each cycle is going to equal the work that I need to lift this mass. So basically, so this is going to be number of cycles, like this, number of cycles times the work that I do per cycle is going to equal the work that is required to lift this mass here. And ultimately, what I'm trying to find is what is n? What is the number of cycles here? We know how much energy or how much work is required to lift, which is just the 490 that we calculated over here. So really all we have to do is figure out, well, how much work is produced by the engine each cycle? So basically, that's really what we're trying to figure out in this part of the problem. So how do we calculate the work that's done by the engine? Well now we're just going to stick to our heat engine equations. We have a couple of them that deal with work and heats. So let's just try to use our sort of basic work equation. So w=qh-qc. So do we have the qh? Well, in this problem we're told one of the heats. These are the 2 temperatures, 182 and 0, but we're told that it extracts 25 joules of energy from the hot reservoir. So that's my qh. This is going to be 25. And while I'm at it, I'm just going to fill in the rest of the variables here. My th, this hot reservoir here is 182. So if you add 273 to it, it's just going to be 455 Kelvin. If you do the same thing, this is just going to be tc=0. So that's just going to be 273 kelvin. Alright? So we don't have well, we do have what the qh is. Unfortunately, we don't have what qc is. What is the heat that gets discarded out to the cold reservoir? And we can't find it using this equation because we don't know w or qc. So it turns out we can't use this equation over here because we have multiple unknowns. And so we're going to have to use a different equation. The only other equation that has efficient or the work in it is going to be the efficiency equation. So we can't use this, but it turns out we're going to have to use this. So our efficiency equation, remember this is a Carnot engine, is equal to the work done divided by the heat that's taken in from the hot reservoir. Right? It's how much you get out versus how much you paid into this heat engine. However, because this is a Carnot engine, we also have one more equation. We have that this is 1-tcth. If you look through the variables here, what's going on by the way, this w here is the work done by the engine. That's not to confuse you by Wlift. Those are not the same thing. Okay? So what's going on here is that if you look through your variables, we have qh, we have tc, and th here. So what I can do is I can rearrange this equation. I can say the work that's done by the engine each cycle is equal to qh∗Ε. And because I don't have what the efficiency is, but I can solve it by using this part of the equation. In other words, it's qh∗1-tcth. This equation here gives me the work that's done each cycle. So basically, all I have to do is just plug some numbers in. So I've got the work done by the engine is going to equal to qh,whichis25∗1-tcth. Tc here is 273, and th is 455. So when you work this out, what you're going to get here is that the work done each cycle is 10 joules. So if you come up here, and this kinda makes sense here because you have 25 joules and you have a work that's done, it's 10. It should be a little bit less than the heat done, than the heat transferred from the hot reservoir. So what happens here is now that we figured out the work that's done by the engine each cycle, we can now just plug this into our last equation over here. That's the 10 joules. So all we have to do is we just do n equals the the work that's required to lift this mass here divided by the work that's done by the engine. So in other words, it's 490 joules, but each, each each cycle, the engine does 10 joules of work. So in other words, what you're going to need here is you're going to need 49 cycles of this heat engine running in order to lift this mass by 5 by 5 meters. Alright. So, hopefully, all that stuff made sense. Let me know if you guys have any questions in the comments. Alright. That's it for this one.

Your friend claims they have a design for a reversible heat engine that can operate between the freezing and boiling temperatures of water that has an efficiency of 30%. Is this possible?

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