The following heat engine produces power of 100,000 kW. The heat engine operates between 800 K and 300 K. It has a thermal efficiency equal to 50% of that of the Carnot engine for the same temperature. The rate at which heat is absorbed from the hot reservoir is____________________?

Question

Accepted Answer

320,000 kW. First, find the Carnot efficiency: $\eta_C = 1 - (300/800) = 0.625$. The engine's efficiency is $0.5 * 0.625 = 0.3125$. Then, use $	ext{Power} = Q_H \cdot \eta$, so $100,000 = Q_H \cdot 0.3125$, giving $Q_H = 320,000$ kW.