Question

The coefficient of performance of a refrigerator is $5$. If the temperature inside the freezer is $- {20^ \circ }C$, what is the temperature of the surroundings to which it rejects heat?
A. ${21^ \circ }C$
B. ${31^ \circ }C$
C. ${41^ \circ }C$
D. ${11^ \circ }C$

Answer 1

A refrigerator transports heat from low temperature place to high temperature place using work so that temperature of the colder place further reduces. The efficiency here is the heat removed from cold places per unit work. This is called the coefficient of performance of a refrigerator.
The coefficient of performance can be defined as the ratio of heat extracted from the refrigerator to the work that is done on the refrigerator.

Formula Used:
The coefficient of performance of a refrigerator is given by the equation,
$\beta = \dfrac{{{T_C}}}{{{T_H} - {T_C}}}$
Where ${T_C}$ is the colder temperature and ${T_H}$ is the hotter temperature.
${T_H} - {T_C}$ is the work done since the change in heat energy is transformed into useful work.

Complete step by step answer:
Given,
The temperature of the freezer$= - {20^ \circ }C$
To convert it into Kelvin we need to add $273$ .
Therefore, the temperature of the freezer
A refrigerator transports heat from low temperature place to high temperature place using work so that temperature of the colder place further reduces. The efficiency here is the heat removed from cold places per unit work. It is called the coefficient of performance of a refrigerator.
The coefficient of performance can be defined as the ratio of heat extracted from the refrigerator to the work that is done on the refrigerator
Coefficient of performance of a refrigerator $\beta$ is given by the equation,
$\beta = \dfrac{{{T_C}}}{{{T_H} - {T_C}}}$ (1)
${T_C}$ is the colder temperature and ${T_H}$ is the hotter temperature.
${T_H} - {T_C}$ is the work done since the change in heat energy is transformed into useful work.
Given, the coefficient of performance of a refrigerator, $\beta = 5$
Value of ${T_C}$ is temperature of freezer, that is $253\,K$
Substituting the given values in equation (1), we get
$5 = \dfrac{{253\,K}}{{{T_H} - 253\,K}}$
$\Rightarrow 5 \times \left( {{T_H} - 253\,K} \right) = 253\,K$

$$\Rightarrow 5{T_H} - 1265K = 253\,K \\\ \Rightarrow 5{T_H} = 1265K + 253\,K \\\ \Rightarrow 5{T_H} = 1518\,K \\\ \Rightarrow {T_H} = 303.6\,K \\\$$

Therefore,
${T_H} = 303.6 - 273 = {30.6^ \circ }C \cong {31^ \circ }C$
So, the answer is option B.

Note: While calculating the coefficient of performance remember to convert the temperature given in $^ \circ C$ to Kelvin by adding $273$.The final answer we get is in kelvin. Answer options are given in $^ \circ C$. So, convert the final answer into corresponding $^ \circ C$ value.