Irreversible Processes and Entropy — Question 10

Question 10

Thermal equilibrium maximizes entropy at fixed total energy

Two bodies of equal constant heat capacity $C$ are isolated together with total energy fixing $T_A+T_B=2T_0$. Write $T_A=T_0+\delta$ and $T_B=T_0-\delta$, where $|\delta|<T_0$. Their entropies have the form $C\ln T$ plus reference constants; only dimensionless ratios will be used.

1. Find entropy relative to the equal-temperature state and prove it is maximal at $\delta=0$.

2. For $C=300$ J/K, $T_0=300$ K and initial $\delta=30$ K, find entropy generated on equilibration and its leading small-$\delta$ approximation.

Question 10 - Solution

Thermal equilibrium maximizes entropy at fixed total energy

(a) Method and calculation

$$S(\delta)-S(0)=C\ln\frac{(T_0+\delta)(T_0-\delta)}{T_0^2}=C\ln[1-(\delta/T_0)^2]\le0.$$$$\frac{dS}{d\delta}=-\frac{2C\delta}{T_0^2-\delta^2}=0\ \Rightarrow\delta=0.$$

(b) Interpretation and further calculation

$$S_{\mathrm{gen}}=-300\ln(.990)\approx3.01510\,\mathrm{J/K}.$$$$-\ln(1-x)=x+O(x^2)\ \Rightarrow\ S_{\mathrm{gen}}\simeq C(\delta/T_0)^2=3.00\,\mathrm{J/K}.$$The leading entropy penalty is quadratic in a small temperature imbalance, consistent with a smooth maximum.

Physical check

The fixed-energy constraint is essential; entropy is not maximized by holding only the body temperatures arbitrary.