# I Question about entropy change in a reservoir

1. Jan 13, 2019

### Clara Chung

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?

2. Jan 13, 2019

### Andrew Mason

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].

In such case the change in entropy of the reservoir is always $\Delta S_{reservoir}= -|Q|/T_h$ since its temperature does not change. If the cooler body also has infinite heat capacity, then $\Delta S_{body} = |Q|/T_c$. If temperature of the body changes with heat flow it is a bit more complicated to work out the integral for the body.

AM

Last edited: Jan 13, 2019
3. Jan 13, 2019

### Staff: Mentor

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.