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Of course, I know the fact that the entropy of an isolated system never decreases. Neverthless what makes me confused about the entropy(or change of entropy) of an isolated system is the explanation used to tell the unavailable heat/work in a irreversible process.

With a closed system and its surrounding imagine that the system expanding isothermally(at T) then for

1) reversible process

Surely the total change of entropy is zero. Here I say the process is from a state A to a state B and the heat system absorbs is q.

2) irreversible process (from state A to B)

With very large surrounding compared to the system, the temperature of the surrounding can be assumed to be constant and same as that of system. Because the entropy is a state function the entropy change of the system is still q/T. It is obvious that q and q'(heat absorbed by system in irreversible process) are different. The problem is that a lot of books and lecture notes explain that the entropy change of the surrounding is -q'/T.

By the definition of entropy I cannot handle this statement because -q' is the heat extracted from the surrounding "irreversibly". If q' can be treated as a heat transferred reversibly why not use this quantity for the system? In the view of 2nd law of thermodynamics the difference between the magnitudes of entropy change of the system and the surrounding should exist in irreversible process but if we can say that -q'/T is the change of entropy of the surrounding we can also say that the change of entropy of the system is q'/T. As a result the total entropy change will be zero, which doesn't obey the law of thermodynamics.

Books/lecture notes examples for this problem:

Introduction to the thermodynamics of materials, D. R. Gaskell, 5th ed., p.44. & A note for lost work

In addition if the entropy of the system is a state function can't we regard the entropy of the surrounding as a state function? The entropy changes of the system are same for both cases, reversible and irreversible processes because the first and final states are unchanged. In this situation I think the surrounding also have the same first and final states for both reversible and irreversible processes. In this way the total entropy change will be zero since q/T - q/T = 0. If reversible this result is not strange, but state function doesn't depend on paths but states...

In short, my troubles are

  1. Many books/notes argue that q-q' is the heat unavailable or consumption of free energy, and I think this statement is plausible but the reason that we can take q' is not permissible thing. q' is not the reversibly transferred heat. It has irreversible property and thus we can't use q' for calculating entropy.

  2. Entropy is a state function and this is true for both entropies of systems and surroudings. If we take an irreversible path between states(A to B) the total entropy change would be zero for we can always draw a reversible path between the states(also A to B). However this violates the law of thermodynamics.

share|improve this question
I don't quite understand which sentence contains your question but if your real question is what the entropy of something else isn't viewed as the state function of the original object, why would it? It belongs to someone else. Moreover, if the something else is a heat bath, it should really be infinite and its entropy is therefore also infinite and the operations that involve its subtractions are ill-defined. Also, your $q,q'$ were not really defined. Otherwise a transfer of heat between two equal temperatures is reversible but "equilibrization" of temperature is not, what's the problem? –  LuboŇ° Motl Mar 27 '13 at 6:38

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