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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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• 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. • 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.
• 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? •  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.
• 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? • 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
• 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. • Why does the reverse Carnot Engine take in (Q-W)(1+x) ? Why not Q1 - W(1+x)? • 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.
• 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 :) • Is it possible for you to relate this to a PV diagram if this has something to do with it? • Can we have a reversible engine less efficient than a Carnot engine with same source and sink?   