Entropy in the Real World: Engines — Question 6

Question 6

A cooling finite source cannot maintain its initial Carnot efficiency

A finite body of constant heat capacity $C=1000$ J/K starts at 600 K. A succession of reversible cyclic engines extracts work while cooling it to a fixed 300 K environment. At instantaneous source temperature $T$, the engine absorbs $dQ_h=-C\,dT$ and has efficiency $1-300/T$.

1. Integrate to find maximum work and total rejected heat.

2. Find overall work divided by total source heat loss. Compare with the estimate obtained by applying the initial efficiency to all heat.

Question 6 - Solution

A cooling finite source cannot maintain its initial Carnot efficiency

(a) Method and calculation

$$W_{\max}=C\int_{300}^{600}\left(1-\frac{300}{T}\right)dT=C[300-300\ln2]\approx92055.8\,\mathrm J.$$$$Q_h=C(600-300)=300000\,\mathrm J,\quad Q_c=C(300)\ln2\approx207944\,\mathrm J.$$

(b) Interpretation and further calculation

$$\eta_{\mathrm{overall}}=W_{\max}/Q_h=1-\ln2\approx.306853.$$$$W_{\mathrm{initial\ efficiency}}=.500(300000)=150000\,\mathrm J>W_{\max}.$$The source temperature falls continuously, reducing the available work per joule of heat.

Physical check

The available-work integral vanishes as the source approaches the environmental temperature.