Question

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

Answer 1

First of all, we will use the concept of coefficient of performance of the refrigerator, which is the ratio of the heat absorbed to the work input. We will again convert the given temperature into kelvin. We will substitute the required values and manipulate accordingly to obtain the result.

Complete step by step solution:
In the given problem, we are supplied the following data:
There is an ideal refrigerator which has a freezer at a temperature of $- 13\,^\circ {\text{C}}$.
The coefficient of performance of the engine is given as $5$.
We are asked to find the temperature of the air to which heat is rejected.

To begin with, we will use the concept of coefficient of performance of the engine. We will make use of the concept of heat absorbed and heat rejected. Let us proceed to solve the numerical. We know that the coefficient is the ratio of the cooling effect to the work input.
So, we can use the formula:
$\beta = \dfrac{{{T_{\text{C}}}}}{{{T_{\text{H}}} - {T_{\text{C}}}}}$ …… (1)
Where,
$\beta$ indicates the coefficient of performance.
${T_{\text{C}}}$ indicates the temperature of the cold reservoir.
${T_{\text{H}}}$ indicates the temperature of the hot reservoir, here it is the air to which heat is rejected.

{T_{\text{C}}} = - 13^\circ {\text{C}} \\\ \Rightarrow {T_{\text{C}}} = \left( { - 13 + 273} \right)\,{\text{K}} \\\ \Rightarrow {T_{\text{C}}} = 260\,{\text{K}} \\\
Now, we substitute the required values in the equation (1) and we get:
\beta = \dfrac{{{T_{\text{C}}}}}{{{T_{\text{H}}} - {T_{\text{C}}}}} \\\ \Rightarrow 5 = \dfrac{{260}}{{{T_{\text{H}}} - 260}} \\\ \Rightarrow 5 \times \left( {{T_{\text{H}}} - 260} \right) = 260 \\\ \Rightarrow 5 \times {T_{\text{H}}} = 260 + 1300 \\\ \Rightarrow 5 \times {T_{\text{H}}} = 1560 \\\ \Rightarrow {T_{\text{H}}} = \dfrac{{1560}}{5} \\\ \Rightarrow {T_{\text{H}}} = 312\,{\text{K}} \\\ \Rightarrow {T_{\text{H}}} = \left( {312 - 273} \right)\,^\circ {\text{C}} \\\
$\therefore {T_{\text{H}}} = 39\,^\circ {\text{C}}$

Hence, the temperature of the air to which heat was rejected was $39\,^\circ {\text{C}}$. The correct option is C.

Note: While solving the problem, always remember that we need to convert the given temperature into kelvin scale. On failure to do so, the answer will come in negative and truly irrelevant and wrong. If you want you can use the temperature in degree centigrade in the above formula and see the difference.