Question

A Carnot engine, having an efficiency of $\eta =\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10\text{J}$, the amount of energy absorbed from the reservoir at lower temperature is (in $\text{J}$)

Answer 1

Given:
Efficiency of engine, $\eta =\frac{1}{10}$
Work done on system $W=10\text{J}$
Coefficient of performance of refrigerator
$\beta =\frac{Q}{W}=\frac{1-\eta }{\eta }$
$\beta =\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}=9$
Energy absorbed from reservoir –
$Q=\beta W=9\times 10=90\text{J}$