Irreversible Processes and Entropy — Question 6

Question 6

Smaller temperature steps reduce entropy generation during heating

A body of constant heat capacity $C=100$ J/K starts at 300 K and finishes at 600 K. Compare direct contact with a 600 K reservoir to two stages: first equilibration with a 450 K reservoir, then with a 600 K reservoir. Neglect work and all other heat capacities.

1. Find the body entropy change and total entropy generation for direct heating.

2. Find entropy generation for the two-stage protocol and its reduction. Explain the limit of infinitely many reservoirs with infinitesimal temperature differences.

Question 6 - Solution

Smaller temperature steps reduce entropy generation during heating

(a) Method and calculation

$$\Delta S_b=C\ln2=69.3147\,\mathrm{J/K},\quad Q=30000\,\mathrm J.$$$$S_{\mathrm{gen,direct}}=C\ln2-30000/600=100\ln2-50\approx19.3147\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$S_{\mathrm{gen,two}}=100\ln2-\frac{15000}{450}-\frac{15000}{600}\approx10.9814\,\mathrm{J/K}.$$$$\text{reduction}=8.33333\,\mathrm{J/K},\quad\lim S_{\mathrm{gen}}=C\ln2-\int_{300}^{600}\frac{C\,dT}{T}=0.$$The body entropy change is unchanged; reservoir entropy loss approaches its magnitude as heat transfer becomes reversible.

Physical check

A reversible limit removes entropy generation, not the entropy change of the heated body.