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I have this homework question:

"Show that any reversible engine operating between T1 and T2 is a carnot engine."

I think I have a solution, but it feels very hand-wavy. We know that any process that can be represented as a loop in the PV plane is reversible as the net entropy change will be zero. We must operate between two specifice temperatures, so the loop must comprise of two isotherms a T1 and T2. So the question is what curves join the isotherms. As a heat engine comprises of energy input at constant temperature, there will be no energy change between the isotherms. So the curves connecting the isotherms must be adiabatic curves. So we have a carnot cycle.

Is this sufficient? I don't know why, but I doubt it.

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up vote 1 down vote accepted

I suspect the expression "operating between T1 and T2" actually means "operating between heat reservoirs with temperatures T1 and T2". But even then I am not sure "any reversible engine operating between heat reservoirs with temperatures T1 and T2 is a Carnot engine." As far as I know, a Carnot engine is an engine "that operates on the reversible Carnot cycle" ( ), which cycle consists of two isothermal processes and two adiabatic processes. However, it seems that more complex reversible processes can exist that use the same isothermal processes (but maybe different parts of them) and more than two adiabatic processes (e.g., T1S1-T1S2-T2S2-T2S3-T1S3-T1S4-T2S4-T2S1-T1S1). so maybe the condition of the problem lacks some additional requirement.

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Another way to show this is through the non-existence of perpetual motion, the Feynman way. As he says, we first assume that perpetual motion and hence the creation of energy is not possible. Next, let engine $A$ be a reversible engine working between temperatures $T_1$ and $T_2$ where $T_1>T_2$ and let it absorb heat $Q_1$ frome the reservoir at $T_1$ and give out heat $Q_2$ at reservoir at temperature $T_2$ doing work $W=Q_1-Q_2$ in the process.

Consider another reversible engine $B$ working between the same temperature but having different efficiency. On taking heat $Q_1$ from the hotter reservoir, let us suppose it does work $W'>W$ and therefore gives out heat $Q'$ at the colder reservoir which is less than $Q_2$. After operating engine $B$ for one cycle, we could use $A$ in reverse to siphon out $Q_2$ from colder reservoir, submit heat $Q_1$ at the hotter reservoir with a work input of $W$. But since $W'>W$, therefore the net result of oprating $B$ followed by reverse $A$ will be to have drawn heat $Q_2-Q'$ from the colder reservoir and completely converted it into work $W'-W$, without any other change or entropy since the engines are reversible. But this contradicts the well know statement of second law of thermodynamics:-

Clausius staement-"Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."

This argument shows that working between any two temperatures, irrespective of the nature of the engine, a reversible engine shows the maximum efficiency and all reversible engines must show the same efficiency (Carnot principle) to not violate second law of thermodynamics. This idea, originally Carnot's proves that all reversible engines have the same efficiency as a Carnot engine. Your argument uses certain assumption about the working of a reversible engine which might not be universally true.

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Thanks for the detailed response. What exactly am I assuming that "might not be universally true"? – Kieran Cooney Sep 20 '13 at 3:37
@KieranCooney You are assuming that there are only four quasistaic processes in a reversible engine, but that is not true. You can design a reversible engine which performs several transitions, which may not be adiabatic and isothermal only. You may use a complex setup with several compartments and mechanical constraints which would ultimately complicate the PV loop of the system by including several processes unlike only 4 used in Carnot engine. But if it is reversible and operates between 2 reservoirs at different temperatures, it's efficiency must be = to Carnot efficiency. – Satwik Pasani Sep 20 '13 at 7:01

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