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Andrew Mason
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Why would Tb always be the same for both the system and the reservoir? What if there is no direct thermal contact? (eg they are connected via a Carnot device so that all heat flow occurs isothermally at the temperature of the system or reservoir). And how do we maintain the first law if dQsystem = -dQsurr? Are you assuming that no work is being done?Chestermiller said:It doesn't imply this at all. At the boundary between the system and surroundings, the rate of heat flow from the surroundings to the system is always minus the rate of heat flow from the system to the surroundings, and the temperature at the boundary is the same for both the system and the surroundings, say TB. (This assumes that there is only one location where heat is being exchanged between the system and the surroundings). This all implies that dQ/TB for the system is always equal to -dQ/TB for the surroundings.
This cannot be true if work is being done. For example, this is not true for a reversible Carnot engine operating between the system and the surroundings. Qsys ≠ -Qsurr and Tsys ≠ TBThe entropy change for the system is only equal to dQ/TB if the system is undergoing a reversible path, and the entropy change for the surroundings is only equal to -dQ/TB if the surroundings are undergoing a reversible path.
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