The Second Law of Thermodynamics — Question 8

Question 8

An irreversible path and a reversible return produce a negative Clausius integral

A closed system follows path A to B, absorbing 800 J through a 400 K boundary and rejecting 300 J through a 300 K boundary. Its entropy increases by 1.50 J/K. It returns from B to A reversibly, with whatever changing-temperature heat contacts are needed.

1. Find entropy transfer and entropy generation on the outward path.

2. Find $\int\delta Q_{\mathrm{rev}}/T$ on the return and $\oint\delta Q/T_b$ over the cycle. Verify the cyclic inequality.

Question 8 - Solution

An irreversible path and a reversible return produce a negative Clausius integral

(a) Method and calculation

$$\int_A^B\frac{\delta Q}{T_b}=800/400-300/300=1.00\,\mathrm{J/K}.$$$$S_{{\mathrm{gen}},AB}=\Delta S_{AB}-\int_A^B\delta Q/T_b=.500\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\int_B^A\frac{\delta Q_{\mathrm{rev}}}T=S_A-S_B=-1.50\,\mathrm{J/K}.$$$$\oint\frac{\delta Q}{T_b}=1.00-1.50=-.500\,\mathrm{J/K}=-S_{\mathrm{gen,cycle}}\le0.$$The return removes the system entropy increase but cannot undo the generation from the outward leg.

Physical check

The closed integral of $dS$ is zero; the closed integral of actual heat divided by boundary temperature need not be.