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 :
The efficiency $$\left( \eta \right)$$ of a Carnot engine and the co - efficient of performance $$\left( \beta \right)$$ of a refrigerator are related as
$$\eqalign{ & \beta = \frac{{1 - \eta }}{\eta } \cr & {\text{Here, }}\eta = \frac{1}{{10}} \cr & \therefore \,\,\beta = \frac{{1 - \frac{1}{{10}}}}{{\left( {\frac{1}{{10}}} \right)}} \cr & = 9. \cr} $$
Also, Co - efficient of performance $$\left( \beta \right)$$  is given by $$\beta = \frac{{{Q_2}}}{W},$$
where $${{Q_2}}$$ is the energy absorbed from the reservoir.
$$\eqalign{ & {\text{or, }}9 = \frac{{{Q_2}}}{{10}} \cr & \therefore \,\,{Q_2} = 90\,J. \cr} $$