Question

The temperature inside a refrigerator is ${t_2}^\circ C$ and the room temperature is ${t_1}^\circ C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:
A) $\dfrac{{{t_1} + {t_2}}}{{{t_1} + 273}}$
B) $\dfrac{{{t_1}}}{{{t_1} - {t_2}}}$
C) $\dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$
D) $\dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}}$

Answer 1

Hint: Treat the refrigerator as a Carnot refrigerator. Use the relation between the coefficient of performance of the refrigeration with the work done by the refrigerator and the heat delivered to the room.

Formula used:
$COP = \dfrac{{{Q_C}}}{W} = \dfrac{{{t_2}}}{{{t_1} - {t_2}}}$where ${Q_C}$ is the heat removed by the refrigerator from the system, $W$ is the work done by the refrigerator in the process, ${t_1}$ and ${t_2}$ are the temperatures inside and outside the refrigerator.

Complete step by step solution:
The coefficient of performance of a refrigerator (COP) can be calculated as:
$\Rightarrow COP = \dfrac{{{t_2}}}{{{t_1} - {t_2}}}$
It can also be defined as the ratio of the energy used by the refrigerator to the work done by the refrigerator i.e.
$\Rightarrow COP = \dfrac{{{Q_C}}}{W}$
To find the heat ejected to the room, we can use the work done by the system $W$ which can be calculated as the difference of heat ejected by the system to the energy used by the system. So, we have
$\Rightarrow W = {Q_H} - {Q_C}$
Rewriting the above equation, we get
$\Rightarrow {Q_H} = W + {Q_C}$
Dividing both sides by $W$, we get
$\Rightarrow \dfrac{{{Q_H}}}{W} = \dfrac{{{Q_C}}}{W} + 1$
The term $\dfrac{{{Q_C}}}{W}$ of the above equation is the COP and so, we can write
$\Rightarrow \dfrac{{{Q_H}}}{W} = \dfrac{{{t_2}}}{{{t_1} - {t_2}}} + 1$
However the value of temperature used to measure the COP must be in the Kelvin scale so substituting ${t_1} = {t_1} + 273$ and ${t_2} = {t_2} + 273$ to convert the temperatures in the Centigrade scale, we get
$\Rightarrow \dfrac{{{Q_H}}}{W} = \dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}} + 1$
$\dfrac{{{Q_H}}}{W} = \dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$ which corresponds to option (C).

Note:
The trick in this question is to use the relation of the coefficient of performance with the heat the refrigerator removes from the system with the work done by it. Since we’ve been asked to find the amount of heat delivered to the room for each joule of electrical energy we have to calculate the ratio of $\dfrac{{{Q_H}}}{W}$ and not ${Q_H}$. Also, we must not forget to take the temperatures of the system in Kelvin and not the Centigrade scale.