The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is $$ - {20^ \circ }C,$$  the temperature of the surroundings to which it rejects heat is

Solution :
Key Concept
Coefficient of performance $$\left( \beta \right)$$ of a refrigerator is defined as the ratio of quantity of heat removed per cycle $$\left( {{Q_2}} \right)$$ to the work done on the working substance per cycle to remove this heat.
Given, coefficient of performance of a refrigerator, $$\beta = 5$$
Temperature of surface, i.e. inside freezer,
$${T_2} = - {20^ \circ }C = - 20 + 273 = 253\,K$$
Temperature of surrounding, i.e. heat rejected outside $${T_1} = ?$$
$$\eqalign{ & {\text{So,}}\,\,\beta = \frac{{{T_2}}}{{{T_1} - {T_2}}} \Rightarrow 5 = \frac{{253}}{{{T_1} - 253}} \cr & \Rightarrow 5{T_1} - 1265 = 253 \cr & \Rightarrow 5{T_1} = 1518 \cr & {T_1} = \frac{{1518}}{5} = 303.6\,K \cr & {T_1} = 303.6 - 273 = {31^ \circ }C \cr} $$