Change in Entropy — Question 8

Question 8

A nonuniform temperature field has an entropy below its uniform equilibrium

A thin insulated rod has uniform heat capacity per length $C/L$, with total $C=100$ J/K. Initially $T(x)=300+300x/L$ K for $0\le x\le L$. Ignore expansion, work and temperature dependence of capacity. Internal conduction brings it to uniform temperature.

1. Find final temperature and express entropy change as a spatial integral.

2. Evaluate the integral using $\int\ln T\,dT=T\ln T-T$. Explain why replacing the initial field by its mean would miss entropy generation.

Question 8 - Solution

A nonuniform temperature field has an entropy below its uniform equilibrium

(a) Method and calculation

$$T_f=\frac1L\int_0^LT(x)dx=450\,\mathrm K.$$$$\Delta S=\frac CL\int_0^L\ln\frac{450}{T(x)}dx=\frac C{300}\int_{300}^{600}\ln\frac{450}{T}\,dT.$$

(b) Interpretation and further calculation

$$\Delta S=C\left[1+\ln(450/300)-2\ln2\right]\approx1.91707\,\mathrm{J/K}.$$$$S_{\mathrm{gen}}=\Delta S>0.$$Averaging temperature preserves energy for constant capacity, but entropy depends on the spatial average of the logarithm, not the logarithm of the average.

Physical check

The logarithm is concave, so a uniform field maximizes entropy at fixed total thermal energy.