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If a reservoir is in thermal contact with a system, why is the entropy simply Q/T ? Shouldn't this equation only valid for reversible process? Why is it reversible?
If the heat flow from the reservoir to another body is due to a finite temperature difference the process is not reversible. But ##\Delta S = \int dQ_{rev}/T## so you just have to find a reversible process between the initial and final state and work out the integral. The reversible process is one in which the same heat flow occurs at an infinitessimal temperature difference (eg. Insert a Carnot engine between the reservoir and body). [note: the reversible process between the initial and final states for each component will necessarily differ].If a reservoir is in thermal contact with a system, why is the entropy simply Q/T ? Shouldn't this equation only valid for reversible process? Why is it reversible?
In thermodynamics, an ideal constant temperature reservoir is defined as an entity for which the entropy change is always ##Q/T_R##. As such, one assumes that any irreversibility (and all entropy generation) takes place within the system, and none takes place within the reservoir. This means that the reservoir is implicitly assumed to have an infinite thermal conductivity and an infinite heat capacity, such that there are (a) never any temperature gradients within the reservoir and (b) its temperature never deviates from ##T_R## Thus, it always presents the temperature ##T_R## at its interface with the system.If a reservoir is in thermal contact with a system, why is the entropy simply Q/T ? Shouldn't this equation only valid for reversible process? Why is it reversible?