Entropy of a thermostat

A thermostat (or reservoir, or thermal source) is a body so large that it can give up or absorb heat without changing its own temperature: the atmosphere of a room, a lake, the sea. For a thermostat the entropy formula is the simplest of all.

Key formula — Entropy of a thermostat

$\ev{\Delta S_\text{res} = \frac{Q}{T_\text{res}}}$

where $T_\text{res}$ is the (constant) temperature of the thermostat and $Q$ the heat received by it. The sign convention is crucial: $Q$ is positive if the thermostat absorbs heat (its entropy increases), negative if it gives up heat (its entropy decreases).

Why is there no logarithm here? Precisely because the temperature does not change. The solid’s formula, $mc_s\ln(T_B/T_A)$, reduces to $Q/T$ for small temperature changes; but a thermostat keeps $T$ rigorously constant by definition, so the relation $\Delta S = Q/T$ is exact, not approximate.

Example — the atmosphere receiving heat

If a cup of coffee gives up $25\,080\;\text{J}$ to the air of a room at $T = 293\;\text{K}$, the entropy of the atmosphere increases by $\Delta S_\text{res} = \frac{25\,080}{293} \approx 85{,}6\;\text{J/K}$

This formula, together with those for the gas and the solid, lets us calculate the complete entropy balance of a process: we add up the entropy change of each component (the bodies that change temperature and the thermostats involved) to obtain $\Delta S_\text{tot}$, the quantity on which the second law passes its verdict.

Links

Topics: Entropy and the second law Concepts: Entropy · Temperature Skills: Entropy balance

Related exercises: Heat from hot to cold: proof · T ratio · Melting 1 kg of ice: ΔS