Question

A reversible engine takes in heat from a reservoir of heat at ${527}^{o}C$ and gives heat to the sink at ${127}^{o}C$ . How many calorie/s shall it take from the reservoir to do a work of $750W$?

• A. $257cal-s^{-1}$
• B. $357cal-s^{-1}$
• C. $1500cal-s^{-1}$
• D. none of these

Answer 1

The correct option is B $357cal-s^{-1}$

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}=\frac{W}{Q_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. $Q_1$ is the heat taken up from the source and $W$ is the work done in the process. Thus we get $\eta =1-\frac{400}{800}=\frac{750}{Q_1}=\frac{1}{2}$
Thus we calculate $Q_1$ as $1500J=357.1cal-s^{-1}$