The Second Law of Thermodynamics — Question 3

Question 3

Three-reservoir operation requires the sum of all entropy transfers

A cyclic device absorbs 800 J from an 800 K reservoir and 600 J from a 400 K reservoir. It rejects heat $Q_0$ to a 300 K environment and produces net work. There are no other exchanges.

1. Find the minimum $Q_0$ and maximum work allowed by the second law.

2. Test a claim of 700 J work per cycle. Explain why combining the two heat inputs and using an arithmetic-mean source temperature is unjustified.

Question 3 - Solution

Three-reservoir operation requires the sum of all entropy transfers

(a) Method and calculation

$$S_{\mathrm{gen}}=Q_0/300-800/800-600/400=Q_0/300-2.50\ge0.$$$$Q_{0,\min}=750\,\mathrm J,\quad W_{\max}=800+600-750=650\,\mathrm J.$$

(b) Interpretation and further calculation

$$W=700\ \Rightarrow\ Q_0=700\,\mathrm J,\quad S_{\mathrm{gen}}=700/300-2.50=-1/6\,\mathrm{J/K}.$$$$S_{\mathrm{in}}=\sum_iQ_i/T_i\ne(Q_1+Q_2)/[(T_1+T_2)/2].$$Each heat input carries an entropy fixed by its own source temperature.

Physical check

Available work depends on both the energy and entropy supplied by the sources.