Change in Entropy — Question 7

Question 7

Unequal thermal capacities require an energy-weighted final temperature

Two insulated-together bodies have constant heat capacities $C_A=100$ J/K and $C_B=300$ J/K. Initial temperatures are 600 K and 300 K. They do no work and finally equilibrate.

1. Find final temperature from energy conservation and total entropy generation.

2. Calculate the entropy changes separately and show why averaging the initial temperatures without heat-capacity weights gives an inconsistent energy balance.

Question 7 - Solution

Unequal thermal capacities require an energy-weighted final temperature

(a) Method and calculation

$$T_f=\frac{C_AT_A+C_BT_B}{C_A+C_B}=\frac{100(600)+300(300)}{400}=375\,\mathrm K.$$$$S_{\mathrm{gen}}=100\ln(375/600)+300\ln(375/300)\approx19.9427\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S_A=100\ln(5/8)\approx-47.0004\,\mathrm{J/K},\quad\Delta S_B=300\ln(5/4)\approx66.9431\,\mathrm{J/K}.$$$$\Delta U_{\mathrm{if\ }}T_f=450=100(450-600)+300(450-300)=30000\,\mathrm J\ne0.$$The unweighted temperature would require an unprovided energy input.

Physical check

The equilibrium temperature lies closer to the initial temperature of the body with greater heat capacity.