Question

A heat engine having thermal efficiency of 40% receives heat from a source at 600 K and rejects heat to a sink at 300 K. The second-law efficiency (in %) of this engine is ......... (answer in integer).

Hint:

The second-law efficiency is a measure of how close the actual performance of a heat engine is to the ideal Carnot efficiency. It is calculated as the ratio of the actual efficiency to the Carnot efficiency.

Answer 1

The thermal efficiency \( \eta \) of the heat engine is given as: \[ \eta = 0.40 \] The second-law efficiency \( \eta_{{II}} \) is the ratio of the actual efficiency to the maximum possible efficiency, which is the Carnot efficiency. The Carnot efficiency \( \eta_{{Carnot}} \) is given by: \[ \eta_{{Carnot}} = 1 - \frac{T_L}{T_H} \] where \( T_H = 600 \, {K} \) is the temperature of the heat source and \( T_L = 300 \, {K} \) is the temperature of the heat sink. Substituting the values: \[ \eta_{{Carnot}} = 1 - \frac{300}{600} = 1 - 0.5 = 0.5 \] Now, the second-law efficiency is: \[ \eta_{{II}} = \frac{\eta}{\eta_{{Carnot}}} = \frac{0.40}{0.50} = 0.80 \] Thus, the second-law efficiency of the engine is \( 80% \).