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 $$10\,J,$$  the amount of energy absorbed from the reservoir at lower temperature is

Solution :
$$\eqalign{ & {\text{As,}}\,\,{Q_1} + W = {Q_2} \cr & {\text{Given,}}\,\,\eta = \frac{1}{{10}} \cr & {\text{Now,using}}\,\,\eta = 1 - \frac{{{T_1}}}{{{T_2}}} \cr & {\text{So,}}\,\,\frac{1}{{10}} = 1 - \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \frac{9}{{10}} \cr & {\text{Now}}\,\,\frac{{{Q_1}}}{{{Q_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \frac{{{Q_1}}}{{{Q_1} + W}} = \frac{9}{{10}} \cr & \Rightarrow 10{Q_1} = 9{Q_1} + 9W \cr & \Rightarrow {Q_1} = 9W = 9 \times 10 = 90\;J \cr} $$