Irreversible Processes and Entropy — Question 9

Question 9

Entropy has an arbitrary additive reference but measurable differences

A body has constant heat capacity $C=50.0$ J/K. One analyst assigns $S_A(T)=C\ln(T/300\,\mathrm K)+S_0$; another assigns $S_B(T)=C\ln(T/600\,\mathrm K)+S_1$. Both describe the same temperature-dependent entropy up to a reference offset.

1. Find the condition on $S_1-S_0$ that makes the two functions identical at every temperature.

2. Compute entropy change from 300 K to 450 K using each reference, and explain why the logarithm requires a dimensionless argument.

Question 9 - Solution

Entropy has an arbitrary additive reference but measurable differences

(a) Method and calculation

$$S_B-S_A=C\ln(300/600)+S_1-S_0.$$$$S_B=S_A\ \Rightarrow\ S_1-S_0=C\ln2\approx34.6574\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S_A=C\ln(450/300)=50\ln(3/2)\approx20.2733\,\mathrm{J/K}.$$$$\Delta S_B=C[\ln(450/600)-\ln(300/600)]=50\ln(3/2).$$An additive reference cancels in differences; a logarithm of a dimensionful temperature alone would change under a unit conversion.

Physical check

The reference freedom does not alter the sign of entropy generation.