Entropy in the Real World: Refrigerators — Question 4

Question 4

Reversible cooling of a finite body requires an integral of changing work cost

A body with constant heat capacity $C=200$ J/K is cooled from 300 K to 250 K by a reversible refrigerator rejecting heat to a 300 K reservoir. At body temperature $T$, use $dW=dQ_c(300/T-1)$ with $dQ_c=-C\,dT$.

1. Find total heat removed and minimum work.

2. Find heat rejected and compare the work with an estimate using the final-temperature coefficient for the entire cooling process.

Question 4 - Solution

Reversible cooling of a finite body requires an integral of changing work cost

(a) Method and calculation

$$Q_c=C(300-250)=10000\,\mathrm J.$$$$W_{\min}=C\int_{250}^{300}\left(\frac{300}{T}-1\right)dT=200[300\ln(6/5)-50]\approx939.293\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q_h=Q_c+W_{\min}\approx10939.3\,\mathrm J.$$$$K_{R,f}=250/(300-250)=5,\quad W_{\mathrm{endpoint\ estimate}}=10000/5=2000\,\mathrm J.$$The endpoint estimate is high because earlier cooling occurs at a smaller temperature lift and requires less work per joule.

Physical check

The reversible work cost tends to zero as body temperature approaches the warm reservoir temperature.