Irreversible Processes and Entropy — Question 5

Question 5

The entropy decrease of a cooling object is balanced by its surroundings

A solid with constant heat capacity 200 J/K cools from 360 K to a large 300 K reservoir. No work occurs. Use $\Delta S_{\mathrm{solid}}=C\ln(T_f/T_i)$ and reservoir entropy gain $Q/T_r$.

1. Find heat delivered to the reservoir and entropy changes of solid, reservoir and composite.

2. If only the solid is chosen as the system, does its negative entropy change violate the entropy rule? Write the composite balance that resolves the apparent conflict.

Question 5 - Solution

The entropy decrease of a cooling object is balanced by its surroundings

(a) Method and calculation

$$Q_r=C(360-300)=12000\,\mathrm J,\quad\Delta S_s=200\ln(5/6)\approx-36.4643\,\mathrm{J/K}.$$$$\Delta S_r=12000/300=40.0\,\mathrm{J/K},\quad\Delta S_{\mathrm{total}}\approx3.53569\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S_s+\Delta S_r=S_{\mathrm{generated}}\ge0.$$$$\Delta S_s=-36.4643\,\mathrm{J/K}<0\quad\text{is allowed for a nonisolated subsystem}.$$The solid exports entropy with heat. The prohibition concerns decreasing total entropy of the isolated composite, not every chosen component.

Physical check

Always state the system boundary before applying an entropy inequality.