A heat engine operates in a Carnot cycle between reservoirs of temperatures 1000 K and 400 K.It is found to discharge 20 J of heat per cycle to the cold reservoir.What is the work output per cycle?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $ \eta = 1 - \frac{T_c}{T_h} $, where $ T_c $ is the temperature of the cold reservoir and $ T_h $ is the temperature of the hot reservoir. Plugging in the given temperatures (in Kelvin), we get $ \eta = 1 - \frac{400}{1000} = 0.6 $. The work output ($ W $) of the engine can be calculated using the efficiency formula $ \eta = \frac{W}{Q_h} $, where $ Q_h $ is the heat absorbed from the hot reservoir. Since $ Q_c = 20 $ J is discharged to the cold reservoir and $ \eta = 0.6 $, we can use the first law of thermodynamics which states $ Q_h = W + Q_c $. Rearranging the efficiency formula gives $ W = \eta \cdot Q_h $. To find $ Q_h $, we use $ Q_h = \frac{Q_c}{1 - \eta} = \frac{20}{1 - 0.6} = 50 $ J. Therefore, $ W = \eta \cdot Q_h = 0.6 \cdot 50 = 30 $ J.

A Heat Engine Operates in a Carnot Cycle Between Reservoirs