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

The efficiency $\left(\eta \right)$ of Carnot engine and the coefficient of performance$\left(\beta \right)$ of a refrigerator are related as

$\beta =\frac{1-\eta }{\eta }$

Here, $\eta =\frac{1}{10}$

$\therefore \beta =\frac{1-\frac{1}{10}}{\left(\frac{1}{10}\right)}=9.$

Also, Coefficient of performance$\left(\beta \right)$ is given by $\beta =\frac{Q_2}{W},$ where$Q_2$ is the energy absorbed from the reservoir.

Or, $9=\frac{Q_2}{10}$

$\therefore Q_2=90J.$