I have seen many similar questions but have not found my answer. Why do we use the actual heat involved in the process to calculate entropy change of surrounding? The only answer I can think of is that we are treating the process as reversible with respect to the surroundings even if it is irreversible with respect to the system. But, this is an irreversible path for the system, and a reversible path will not involve the same heat, so now, the entropy of surrounding calculated must be different. This seems to contradict the idea that the entropy of surroundings must be a state function. So why is this reasoning incorrect?
Answer
When you evaluate the change in entropy of an entity (in this case the surroundings) between an initial and final (equilibrium state), you separate that entity from everything else, and then you only look at how that entity has changed between the two states, disregarding the actual process path. In the case of the surroundings (treated as an infinite reservoir), it had received a certain amount of heat between the initial and final states, and its internal energy and entropy increased as a result. In the formalism of thermodynamics, changes in an ideal infinite reservoir are always regarded as taking place reversibly, irrespective of other irreversibilities in the actual process. So if the actual process is irreversible, the irreversibility (entropy generation) is all considered to take place in the system, rather than in the surroundings (in the case of an ideal infinite reservoir).
You may also find my response in the following link pertinent to this question: Thermodynamics entropy and temperature
No comments:
Post a Comment