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Carnot efficiency 3: Proving that it is the most efficient

Proving that a Carnot Engine is the most efficient engine. Created by Sal Khan.

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  • leaf green style avatar for user J Szopi
    There is a serious issue with this proof. If the Carnot engine was replaced by some other reversible engine of efficiency n (lower than that of a Carnot engine working between the same temperatures), then we could prove that no other engine can exceed this arbitrarily chosen efficiency n. As this must not be correct, this proof is either not correct (although I cannot see any logical error) or some property of engine C, that makes it a Carnot engine rather just any other reversible engine, was assumed without mentioning.

    A similar proof I have seen in 'Classical Thermodynamics' by Elwell and Pointon. The issue was not mentioned there, which made me curious and led here. I see this question was asked earlier by Decio Pinto, but he did not get a reply, so I thought I would restate it in my own words. This is a serious flaw and should be explained, otherwise it is not a proof but merely a tricky story, which made many people falsely believe they understood why perpetual motion was not possible.
    (13 votes)
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    • male robot hal style avatar for user Reinhard Grünwald
      Sal has just proved that ALL reversible ideal engines have THE SAME efficiency, namely the Carnot efficiency.
      He shows in the video that if you assume an engine with higher efficiency you run into a conflict with the 2nd law of thermodynamics.
      But just the same: If there was a process with lower efficiency, simply switch the roles of the two engines and you again could pump heat from the cold to the hot reservoir and thus this is also in conflict with the 2nd law.
      (17 votes)
  • blobby green style avatar for user stanelburger
    What happens if instead of using your friend's "Super Engine" in the final engine you used a Carnot Engine? Would the Second law of thermodynamics still be violated?
    (8 votes)
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    • leafers sapling style avatar for user Peter Collingridge
      No - in that case, the two engines would cancel each other out and there would be no net work and no net heat transfer. The two engines would do nothing.

      It is the equivalent of x being equal to 0 in the video since the efficiency of one Carnot engine = (1+0) times the efficiency of the other.
      (26 votes)
  • duskpin ultimate style avatar for user Chris Bays
    Why do the temperatures have to be 500K and 300K yielding a 40% efficency? Couldn't you have a system which had temperatures of 500K and 100K yielding 80% efficency? So then the greater the difference in temperatures the higher the efficency according to the math. What am I missing here?
    (4 votes)
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    • leaf green style avatar for user Jan Suchánek
      I think you are right...but Sal said it - it is the most efficient engine, which operates between two teperatures - this means, that you cant get more efficiency than 80% between your temperatures (500 and 100 K)and you cant get more efficiency than 40% when working between 500 and 300 K ....so he just chose those temperatures randomly
      (18 votes)
  • leaf green style avatar for user Decio Pinto
    Hi. I am agreed with the final conclusion, that a system like that would oppose the second law. But I do not understand why we have to call the first engine or the refrigerator by Carnot engine/refrigerator. Let's suppose we have another engine "Y" that performs a different cycle than Carnot one. It takes some heat Q_1 from the hot reservoir, does some work W and leaves Q_2 to the cold reservoir, as well does Carnot engine, but not necessarily the same quantities. We can also make it works as a refrigerator. In the same way as in the video, if we imagine another engine more efficient than Y and couple it with an Y refrigerator, I guess we will reach the same (bad) result. The fact is that I did not need mention Y is a Carnot engine to conclude it. Actually, it looks like I can not couple machines with different efficiencies (such an odd conclusion, I know!). I am sure I am making a mistake, but I can not figure it out. Sorry about some English error. Thanks in advance.
    (5 votes)
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    • male robot hal style avatar for user Reinhard Grünwald
      Actually the title of the video is misleading. Sal did not prove that the Carnot engine is the most efficient one. He proved that it violates the 2nd law of thermodynamics if there were two ideal reversible engines with different efficiencies.
      In other words ALL ideal reversible engines have THE SAME efficiency, namely the Carnot efficiency. That is what is proved here!
      (9 votes)
  • blobby green style avatar for user sandeepdarknights
    Why does the reverse Carnot Engine take in (Q-W)(1+x) ? Why not Q1 - W(1+x)?
    (6 votes)
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    • leafers ultimate style avatar for user chulman444
      Because it's carnot engine, not the S engine.

      At tge end of the last video, he explained about xQ1, xW, xQ2 that x is the same for all three. Q2 is just Q1-W from equation (W=Q1-Q2) so xQ2=x(Q1-W). When x is (1+x) then the term will be (1+x)(Q-W).

      In this video, i think he's trying to say that such cannot happen where Q1, (1+x)W, Q2, where Q2 is ,from W=Q1-Q2, Q2=Q1-W, where Q1=xQ1 and W=(1+x)W, so Q2=Q1-(1+x)W.
      (0 votes)
  • blobby green style avatar for user congcongxiong09
    Nice video!
    What's the motivation for using the clever argument of combining the super engine with the reverse carnot to derive a contradiction? It all seems kind of magical :)
    (3 votes)
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  • leafers tree style avatar for user Dimitri Khoo
    Is it possible for you to relate this to a PV diagram if this has something to do with it?
    (3 votes)
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  • aqualine ultimate style avatar for user Sarthak
    Can we have a reversible engine less efficient than a Carnot engine with same source and sink?
    (3 votes)
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  • blobby green style avatar for user Sherman.DES
    Why did you choose the inputted Heat in the Reverse Carnot engine to also be multiplied by (1+x)? This video might have made sense if you left it as the pure reverse of the Super Engine (Q2 = Q1-W(1+x)), but of course the point would not have been made.
    (2 votes)
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  • purple pi purple style avatar for user hyoumans
    since the Carnot engine is at equilibrium it does not add entropy to the universe?
    (2 votes)
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Video transcript

Let's say we have two reservoirs. Let's say up here, I have my hot reservoir. It's at th for hot. And I have my engine. It's going to be a Carnot engine, because we'll learn that no engine is better, at least from an efficiency point of view. We have to be careful when we say "better." We have our Carnot engine, and it operates on heat differentials. So it takes some heat from our hot heat source. It takes some heat there. Let's call that q1. And it does some work. Work is a good thing, so I'll make that in green. It does some work, and then the surplus energy, the surplus heat, q2, then goes to our cold reservoir. I'll do that right there. t cold. Now, I made multiple insinuations in the previous video that this is the most efficient engine that can be created between two reservoirs, th and tc. Now, you come along and say, no, no, no, no, no. I know of a friend, he has invented a new engine that is more efficient than this engine between these same two reservoirs. And you go, and you proceed to draw this same type of diagram for your friend's engine. You say, look-- let me make it clear-- this is the same reservoir, these same reservoirs we're dealing with. Actually, I should probably draw this line all the way, because I'm going to have to do multiple engines here. So the same reservoir we're dealing with, right? This is all the hot reservoir, th th, and this is all the cold reservoir. I need space for multiple engines that we're going to deal with. So your friend has an engine. We'll call it the super engine. And your friend's claim-- and I'll show you why your friend's claim cannot be true, if you believe the second law of thermodynamics-- so your friend's claim, they have the super engine. And they claim that look, I can actually take in q1. I can take in that same heat from this heat source up here. But I can produce more work than your Carnot engine. I could produce 1 plus x. I don't want to get too algebraic. But let's say, you produce w. I produce w times 1 plus x of work. Where x is a positive number. So he's saying, look, x is greater than 0. Whatever that number he might feed you is. And then the rest of the energy that's left over is what? It's q1 minus this. So it's q1 minus w times 1 plus x. And just to be clear, q2 right here, I could rewrite that as q1 minus w. Fair enough. So you look at that. You come to me with this, and I say, no, no, no, no. This cannot be true. Because if this were, then we would solve literally all of the world's energy problems. And I'm about to show you why we could solve all of the world's energy problems, and we would have a perpetual motion machine, and be able to defy all sorts of things if we had this. Now, this is my Carnot engine. But I could devise a reverse Carnot engine, right? Let me make a reverse Carnot engine. So my reverse Carnot engine would look like this. And it's going to do the same thing, but in reverse. So instead of producing q1 minus w here and putting it into tc, it could take in q1 minus w from our cold source-- so could take that in-- or even better, let's scale it up a little bit. Let's say it takes in q1 minus w times 1 plus x. So I've just made a slightly larger reversed Carnot engine. Now, if I take in that much-- in order to do this in reverse, I'm going to have to take in, I'm going to have to scale up this Carnot engine and reverse everything. So instead of producing work, I'm now going to need work to go on the other direction, and I've scaled it up by 1 plus x. So I'm going to need the amount of work here times 1 plus x. And then I'm going to push q1, but I've scaled it up. I'm going to push in q1 times 1 plus x into my hot heat source. And once again, this isn't defying the laws of thermodynamics. I'm taking up some work. There's work that needs to be done in order to do this. But all of a sudden, you come to me and say, look, this is an awesome deal. You have this nice engine that works this way. My friend has a super engine. Let's just couple them together. Let's take the work that he produces right here-- he produces w times 1 plus x, and that just happens to be the amount of work that you need to operate your engine. So you just feed that into there. So what's the net effect of these two engines? So let me do another, scroll a little bit more. Actually, that might be the best way to do it. So let me make sure that we understand that these are the same heat sinks or heat sources that we're using the whole time. So that's my hot source, my cold source is down here. So if I add our two engines together-- so if I have a, you know, let me call it a-- I'm going to pick a new color. These colors are getting monotonous. Nope. I wanted to do the rectangle tool. There you go. All right. So I combine these two engines together. Essentially I just put a big box around them. They're both operating between these two heat sources. These two reservoirs. So I call this the, you know, your super engine plus my reverse Carnot engine. So what's happening now? What's the net heat that's being taken in or put out of here? So we have q1-- we have-- let me see. We have q1 minus w 1 plus x, but in this direction, we have q1-- so in this direction, we could rewrite this. I want to make sure you're clear on the algebra. This could be rewritten as what? As q1 times 1 plus x times, or minus, w times 1 place x. Right? Now, if you compare these terms, this is the same as this term. This term is bigger than this term. Right? This term is clearly bigger, because we're multiplying it by something larger than 1. It's bigger than this storm. So if we combine these two, the upward movement, or the amount of heat I'm taking up from my reverse Carnot, is going to be greater than the amount of heat being put in by your friend's super engine. And we can actually calculate the amount. We can just take this amount minus that amount, and that's the net upward movement. So the net upward movement from our cold reservoir is what? It's this value minus this value. So minus q1 minus w 1 plus x. If we take a minus, we're going to subtract it. So it's a minus and a plus. These cancel out. This minus cancels out with-- so this first term could be rewritten as q1 plus q1x. Right? We could rewrite it that way. So this cancels out with that. And so the net upward movement when we combine the two engines is q1 times x. Fair enough. Now what about the work transfer? Well, whatever work this guy produces is exactly the amount of work that I need. So no outside work has to be done on the system. It just works. This guy produces work, this guy uses the work. Now what's in that heat transferred up to our hot reservoir? What's the amount of heat? Well, it's the difference between these two. And this is clearly a larger number than this one, so the upward movement dominates. So what's this minus that? So this can be rewritten as q1 plus q1x, right? I just distributed the q1. We're going to subtract that out. Minus q1. You're left with q1x. So the net movement, when we combine the two engines, is q1x. So what's happening here? I have no external energy or work has to be expended into this system. And it's just taking heat from a cold body, and it's moving it to a warm body. And it does this indefinitely. It'll do this as much as I want to. I can just build a bigger one. It'll do it on even a larger and larger scale. So if you think about it, I could heat my house with ice by just making the ice colder. I could create steam from things that arbitrarily cold. This goes against the second law of thermodynamics. The net entropy in this world is going down. Because what's happening here? This is just a straight up transfer of q1x from a cold body to a hot body. So what's the net change in entropy here? The change in entropy of the universe. Well, the hot body is gaining some heat, so it's q1x over the temperature of the hot body, and then the cold body is losing the same amount. So it's minus q1x over the cold body. Now, this is a bigger number than this is, right? Because the denominator is smaller. This is a cold body. Its temperature in Kelvin will be a smaller number. So this is going to be less than 0, which the second law of thermodynamics tells us cannot be. The entropy cannot shrink in the universe. This whole thing is an independent system, and the entropy is shrinking. And we can make the entropy shrink arbitrarily if we just scale up our x's enough. So this is why the Carnot engine is the most efficient engine possible. Because if anyone claimed to have a more efficient engine, you could couple it with a reverse Carnot engine, and then create this perpetual reverse-- I guess you could call it a perpetual refrigeration machine that just out of the blue creates anti-entropy from anywhere, and it would be this perpetual energy source that creates energy out of nothing. And so this is just something that cannot be done in our world, especially if you believe the second law of thermodynamics. So the most efficient engine is the Carnot engine, where its efficiency is described as 1 minus the temperature of the cold body divided by the hot body. So if I have two temperature reservoirs, let's say that my hot one is at 500 Kelvin, and my cold one is at 300 Kelvin, and I have some engine that takes heat from there, and transfers it there, and does some work. The most efficient engine, if I were to remove all the friction in the engine, the highest efficiency I could get would be 1 minus 300 Kelvin over 500 Kelvin, which is 1 minus 3/5, which is 2/5, which is equal to 0.4, which is equal to 40%. So if someone tells you that they made an engine that operates between a reservoir that's 500 Kelvin and 300 Kelvin, and they say, oh, I've achieved 41% percent efficiency. I've really polished the thing well. You know that they are lying. So anyway, hopefully you found that reasonably interesting, and I'll see you in the next video.