The Second Law of Thermodynamics — Question 9

Question 9

Reversible work extraction makes equal finite bodies meet below the arithmetic mean

Two finite bodies of equal constant heat capacity $C=200$ J/K start at 600 K and 300 K. A cyclic reversible engine operates between them until their temperatures become equal. The composite is insulated except for work output.

1. Use total entropy conservation to find the final temperature and maximum work.

2. Compare with direct thermal contact, which produces no work. Relate the two final temperatures to their different energy balances.

Question 9 - Solution

Reversible work extraction makes equal finite bodies meet below the arithmetic mean

(a) Method and calculation

$$0=C\ln(T_f/600)+C\ln(T_f/300)\ \Rightarrow\ T_f=\sqrt{600(300)}\approx424.264\,\mathrm K.$$$$W_{\max}=C(600+300-2T_f)\approx10294.4\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{contact}}=0\ \Rightarrow\ T_{f,\mathrm{contact}}=(600+300)/2=450\,\mathrm K.$$$$S_{\mathrm{gen,contact}}=C\ln(450^2/[600(300)])=200\ln(9/8)>0.$$The reversible engine exports energy as work, so the remaining thermal energy and final temperature are lower.

Physical check

The geometric mean arises from entropy conservation; the arithmetic mean arises from no work output.