The Second Law of Thermodynamics — Question 7

Question 7

Entropy generation measures the work shortfall against a fixed environment

A cyclic device has zero net internal-energy and entropy changes. Take heat magnitudes $Q_h,Q_c>0$ for absorption from a hot reservoir and rejection to a cold one. Then $W=Q_h-Q_c$ and $S_{\mathrm{gen}}=Q_c/T_c-Q_h/T_h\ge0$. A device receives 1200 J from 600 K and rejects heat to a 300 K environment. It generates .500 J/K of entropy per cycle.

1. Calculate rejected heat and actual work.

2. Derive $W_{\mathrm{rev}}-W=T_cS_{\mathrm{gen}}$ for the same hot-reservoir input. Evaluate the shortfall.

Question 7 - Solution

Entropy generation measures the work shortfall against a fixed environment

(a) Method and calculation

$$Q_c/T_c=Q_h/T_h+S_{\mathrm{gen}}\ \Rightarrow\ Q_c=300(1200/600+.500)=750\,\mathrm J.$$$$W=1200-750=450\,\mathrm J.$$

(b) Interpretation and further calculation

$$W=Q_h(1-T_c/T_h)-T_cS_{\mathrm{gen}}.$$$$W_{\mathrm{rev}}=600\,\mathrm J,\quad W_{\mathrm{rev}}-W=300(.500)=150\,\mathrm J.$$The energy has not disappeared; more energy must be rejected with the additional entropy.

Physical check

The work-deficit identity requires comparison at the same inputs and environmental temperature.