A Carnot engine having an efficiency 1/5 as a heat engine, is used as a refrigerator. If the work done on…
A Carnot engine having an efficiency 1/5 as a heat engine, is used as a refrigerator. If the work done on the system is 50 J, then the
amount of energy absorbed from the reservoir at lower temperature is
90 J
99 J
200 J
1 J
Solution
Efficiency of Carnot engine, $\eta=\frac{1}{5}$
Work done on the system, W = 50 J
Coefficient of performance of a refrigerator,
$\begin{aligned} \beta & =\frac{1-\eta}{\eta} \\ & =\frac{1-\frac{1}{5}}{\frac{1}{5}}=\frac{4}{5} \times \frac{5}{1}=4\end{aligned}$
$\begin{aligned} & \text { Also, } \quad \beta=\frac{\text { Heat absorbed from reservoir }}{\text { Work done }} \\ & \Rightarrow \quad \beta=\frac{Q}{50} \\ & \Rightarrow \quad 4 \times 50=Q \\ & \Rightarrow \quad Q=200 \mathrm{~J} \\ & \end{aligned}$