A Carnot cycle engine operates between a low temperature reservoir at 20°C and a high temperature reservoir at 800°C. If the engine is required to output 20.0 kJ of work per cycle, how much heat must the high temperature reservoir transfer to the engine during each cycle?
A) 27.5 kJ
B) 73.2 kJ
C) 39.2 kJ
D) 800 kJ
E) 20.5 kJ

Question

Quizplus · Accepted Answer

The efficiency of a Carnot cycle engine is given by:$$e = 1 - \frac{T_L}{T_H}$$where $T_L$ is the temperature of the low temperature reservoir and $T_H$ is the temperature of the high temperature reservoir.In this case, $T_L = 20°C + 273.15 = 293.15 K$ and $T_H = 800°C + 273.15 = 1073.15 K$. Therefore, the efficiency is:$$e = 1 - \frac{293.15 K}{1073.15 K} = 0.729$$The work output per cycle is given by:$$W = eQ_H$$where $Q_H$ is the heat transferred to the engine from the high temperature reservoir.In this case, $W = 20.0 kJ$. Therefore, the heat transferred to the engine from the high temperature reservoir is:$$Q_H = \frac{W}{e} = \frac{20.0 kJ}{0.729} = 27.5 kJ$$

A Carnot Cycle Engine Operates Between a Low Temperature Reservoir