An ideal refrigerator has a freezer at a temperature of \[ - {13^ \circ }C\]. The coefficient of the performance of the engine used is \[5\]. The temperature of air (to which heat is rejected) will be
A. \[{325^ \circ }C\]
B. \[{325^ \circ }K\]
C. \[{39^ \circ }C\]
D. \[{320^ \circ }C\]

Hint:Using the formulae for the coefficient of performance of the engine of an ideal refrigerator in a Carnot cycle, the temperature of air(to which heat is rejected) is calculated after converting the given temperature of the freezer from degree Celsius (\[^ \circ C\]) to degree Kelvin (\[^ \circ K\]).

Formulae Used:
For a Carnot cycle, the coefficient of performance of the engine is,
\[\beta = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}\]
Where, \[\beta = \] the coefficient of performance of the engine, \[{T_1} = \] the temperature of air (to which heat is rejected) in \[^ \circ K\] and \[{T_2} = \] the temperature of freezer in \[^ \circ K\].

Any temperature in degree Celsius (\[^ \circ C\]) is converted to degree Kelvin (\[^ \circ K\]) by adding \[273\]to the numerical value in degree Celsius i.e. \[{t^ \circ }C = {(t + 273)^ \circ }K\].

Complete step by step solution:
The refrigerator used is the ideal type. So, the engine of the refrigerator is Carnot’s engine.
We have been given that the temperature of the freezer is \[ - {13^ \circ }C\] and the coefficient of the performance of the engine is \[5\].

We will use the formulae \[\beta = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}\] to calculate the efficiency of the engine.
First, we will convert \[ - {13^ \circ }C\] to degree Kelvin by applying the formulae \[{t^ \circ }C = {(t + 273)^ \circ }K\].
Thus, we have \[ - {13^ \circ }C = {( - 13 + 273)^ \circ }K = {260^ \circ }K\]

As per given data, we have;
The coefficient of performance of the engine\[ = 5\]
The temperature of freezer in degree Kelvin\[ = 260\]
Let, the temperature of air (to which heat is rejected) in degree Kelvin\[ = {T_1}\]
Assigning the given values to the variables, we have
\[\beta = 5\]
\[{T_2} = 260\]
Substituting the values of the variables in the formulae
\[\beta = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}\]
\[ \Rightarrow 5 = \dfrac{{260}}{{{T_1} - 260}}\]
\[ \Rightarrow 5{T_1} = 1300 + 260\]
\[ \Rightarrow {T_1} = 312\,K\]
So, the temperature of air (to which heat is rejected) is \[{312^ \circ }K\].

Now, we will convert \[{312^ \circ }K\] to degree Celsius by subtracting \[273\] from the numerical value in degree Kelvin.
Thus, \[{T_1} = {312^ \circ }K = {(312 - 273)^ \circ }C = {39^ \circ }C\]
Hence, , the temperature of air (to which heat is rejected) is \[{39^ \circ }C\].

Therefore, option C. ${39^ \circ }C$ is the correct answer.

Note:The formulae \[\beta = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}\] can be used to calculate the coefficient of performance of the engine of an ideal refrigerator in a Carnot cycle, if the units of temperatures are in degree Kelvin. This formula is not valid, if the units of temperatures are considered in degrees Celsius.