 Welcome back. We should be very clear about the importance of the Carnot theorem and the spatial status that it provides to a reversible 2 T heat engine. What we should realize that given T1 and T2, a reversible heat engine has the highest efficiency. Carnot theorem does not say what this efficiency will be. It does not even say whether it will be 1 or less than 1. All that it says that you have a reversible heat engine, compare it with any other engine working between the same temperatures T1 and T2, the reversible engine will not have an efficiency higher than the other engine. This characteristic depends only on the property of the reversible engine being reversible. It does not depend on how that engine is built, what are the materials used, whether it is wood, whether it is steel, whether it is cheese or whether it is moon rocks or martian rocks if you feel like. If there are any fluids used, it does not depend on whether the fluid is water, whether the fluid is oil, whether the fluid is mercury or any other fluid you can imagine. It does not depend on the processes or details of the processes that are involved in it. All that is required is each and every process involved in running that reversible engine must be a reversible process because the engine has to be a reversible engine. Now let us quickly prove a corollary. We ask ourselves this question that we have two fixed temperatures and we have said two reversible engines working between those two temperatures T1 and T2. How do the two efficiencies compare? We will soon demonstrate that if you have two heat engines working between two fixed temperatures T1 and T2, they should have the same efficiency and the proof is straight forward. Let us have our two reservoirs shown here as one at T1 and one at T2. T1 is fixed, T2 is some other fixed value. Obviously T1 must be greater than T2 and we have a reversible engine 1 and a reversible engine 2 working between the same two temperature levels. I am not even going to show the interaction because that detail is not necessary to prove the corollary. Let me say that the efficiency of this reversible heat engine is eta r1 and the efficiency of this reversible heat engine is eta r2. How do we prove that eta r1 must be equal to eta r2? It is simple. We will use the Carnot theorem. If we forget that re1 is reversible and just remain conscious of the fact that re2 is reversible, then by Carnot theorem, eta r1 must be less than or at most equal to eta r2 because this is any engine. I am comparing its efficiency with a reversible engine working between the same T1 and T2. So, this is one relation we must have. The other relation we must have is as follows. Now let us forget the fact that re2 is a reversible engine. Let us say it is an engine. A reversible engine is also an engine whereas be conscious of the fact that re1 is a reversible engine. So, the efficiency of any engine in this case eta r2 must be less than or at most equal to that of a reversible engine re1 working between the same two temperatures. Now both these relations have to be true. So, if this and this has to be true together, then naturally this implies that eta r1 must equal eta r2. Now what is the implication of this? The implication of this is as follows. We have shown that given T1, T2 and any two reversible heat engines re1 and re2 eta r1 must equal eta r2. And instead of 2, if you take three reversible heat engines, you should be able to prove that all three should have the same efficiency. Now that means given T1 and T2, any reversible heat engine working between T1 and T2 or fixed values of T1 and T2 must have the same efficiency. If you change T1 and T2, the efficiency may change. But if you have fixed T1 and T2, there is a fixed efficiency for any reversible heat engine working between them. It does not matter what type of engine it is. It does not matter about the details of the processes, materials of construction, fluids used, nothing. Only thing is that any engine must be reversible. Then you have the same efficiency. Turning this around, we can say that the efficiency of a reversible heat engine depends only on the temperature between which it works. In mathematical symbolism, if eta r is the efficiency of a reversible heat engine, it is a function only of T1 and T2 and of nothing else. It cannot depend on anything else other than T1 and T2 because if T1 and T2 are fixed, our eta r is fixed. If one of the T1 and T2 changes, well eta r might as well be different. In a mathematical sense, this is the symbolism that is used that eta r can only be a function of T1 and T2. This very important result will now be used in some very curious and important ways. Thank you.