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

For Carnot engine 

$\text{ Efficiency, }\eta =1-\frac{T_1}{T_2};\frac{1}{10}=1-\frac{T_1}{T_2}$

$\frac{T_1}{T_2}=1-\frac{1}{10}=\frac{9}{10} . . . . \left(i\right)$

For refrigerator

$\therefore \frac{Q_2}{Q_1}=\frac{T_2}{T_1}$

$\text{ or }\frac{Q_1+W}{Q_1}=\frac{T_2}{T_1}$

$\frac{Q_1+10}{Q_1}=\frac{10}{9}$

$1+\frac{10}{Q_1}=\frac{10}{9};\frac{10}{Q_1}=\frac{10}{9}-1=\frac{1}{9}$

$Q_1=90 \mathrm{J}$

$\text{ So, }90 \mathrm{J}\text{ heat is absorbed at lower temperature. }$