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Entropy change does not depend on the heat that is actually absorbed,but it depends on the heat absorbed reversibly.

How?

Since the surroundings emit irreversible heat then how the system which is also a part of universe absorb heat reversibly? I can't get the concept.

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To determine the change in entropy of a system that has experienced an irreversible process between an initial and final thermodynamic equilibrium state, you must first use the 1st law of thermodynamics to establish the final equilibrium state. After that, you can totally and entirely forget about the irreversible process. It is of no further use. Instead, you need to devise a reversible process between the initial and final states. You can even apply different reversible processes to the system and to the surroundings, as long as each of them passes through a continuous sequence of thermodynamic equilibrium states between the two end points. There are an infinite number of reversible processes for the system and the surroundings that can take them each between their initial and final states; each of these paths give exactly the same entropy change, as determined by the integral of dq/T. The reversible paths do not need to bear any resemblance whatsoever to the real irreversible path, as long as they start and end at the same initial and final states. We need to follow a reversible path to determine the entropy change since this is the only way we know of to determine $\Delta S$.

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  • $\begingroup$ Maybe it would be useful to add that although heat is a path dependent variable, entropy is a state function. $\endgroup$ – Kinformationist Apr 6 '17 at 16:33
  • $\begingroup$ Chester Miller does 1st law is applicable to only reversible process? $\endgroup$ – M.Naeem Ahmad Apr 6 '17 at 16:37
  • $\begingroup$ The 1st law is applicable to all processes, both reversible and irreversible. Have you never applied the 1st law to an irreversible process? $\endgroup$ – Chet Miller Apr 6 '17 at 16:47

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