Looking for classes? Ksquare Career Institute, Bengaluru →
A Carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
Explanation
COP of the reversed engine is $(1-\eta)/\eta = 9$.
Detailed Solution
Coefficient of performance $\beta = \frac{1 - \eta}{\eta} = \frac{1 - \frac{1}{10}}{\frac{1}{10}} = 9$
$\beta = \frac{Q_2}{W}$
$Q_2 = 9\times10 = 90$ J