A heat engine absorbs 64 kcal of heat from a hot reservoir and exhausts 42 kcal to a cold reservoir each cycle.Its efficiency is

Question

Quizplus · Accepted Answer

The efficiency of a heat engine is given by $ \eta = \frac{Q_{hot} - Q_{cold}}{Q_{hot}} \times 100\% $, where $ Q_{hot} $ is the heat absorbed from the hot reservoir and $ Q_{cold} $ is the heat exhausted to the cold reservoir. Plugging in the given values, $ \eta = \frac{64 - 42}{64} \times 100\% = \frac{22}{64} \times 100\% = 34.375\% $, which rounds to 34%.

A Heat Engine Absorbs 64 Kcal of Heat from a Hot