Irreversible Processes and Entropy — Question 7

Question 7

Steady heat conduction stores no entropy while continuously generating it

A rod conducts 900 J each second from a 450 K reservoir to a 300 K reservoir. It is in steady state with no energy or entropy accumulation. Ignore heat loss through its sides. Reservoirs remain at fixed temperatures.

1. Calculate entropy transfer rates into and out of the rod and its entropy generation rate.

2. Find total generated entropy in two minutes. Explain why a constant temperature profile does not mean a reversible process.

Question 7 - Solution

Steady heat conduction stores no entropy while continuously generating it

(a) Method and calculation

$$\dot S_{\mathrm{in}}=900/450=2.00\,\mathrm{J/(K\,s)},\quad\dot S_{\mathrm{out}}=900/300=3.00\,\mathrm{J/(K\,s)}.$$$$0=\dot S_{\mathrm{in}}-\dot S_{\mathrm{out}}+\dot S_{\mathrm{gen}}\ \Rightarrow\dot S_{\mathrm{gen}}=1.00\,\mathrm{J/(K\,s)}.$$

(b) Interpretation and further calculation

$$S_{\mathrm{gen}}(120\,\mathrm s)=120\,\mathrm{J/K}.$$$$\dot U=900-900=0,\quad\dot S=2-3+1=0.$$A steady process can create entropy as fast as it exports the excess. A finite temperature gradient sustains irreversibility.

Physical check

Entropy is not conserved in the rod, even though steady energy flow is conserved.