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. $\dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}}$
B. $\dfrac{{{t_1} + {t_2}}}{{{t_1} + 273}}$
C. $\dfrac{{{t_1}}}{{{t_1} - {t_2}}}$
D. $\dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$

Answer 1

Hint:-
In order to solve the problem we are using the principle of conservation of energy that is ( Energy can neither be created nor be destroyed but it can be transformed from one form to another ) . Based on this principle the electrical energy is used to deliver the heat to the room because refrigerator is the device in which the inside temperature is less than the temperature of the room in which it is kept, so the equation becomes ${Q_{Supplied}} = {Q_{REJ}} + W$ & we want the value of amount of heat delivered to the room for each joule of electrical energy consumed ideally ,that means we want the value of $\dfrac{{{Q_{Supplied}}}}{W}$. On simplifying the relation we get $\dfrac{{{Q_{Supplied}}}}{W} = \dfrac{{{Q_{REJ}}}}{W} + 1$.
Formula used:-
${Q_{Supplied}} = {Q_{REJ}} + W$; where
${Q_{Supplied}}$= heat energy supplied by the electrical means.
${Q_{REJ}}$= heat rejected by the refrigerator.
$W$= Work done

Complete step-by-step solution :
Since the coefficient of performance of the refrigerator is ratio of heat rejected and work done
Coefficient of performance ($COP$)
$\Rightarrow COP = \dfrac{{{Q_{REJ}}}}{W}$
$\Rightarrow \dfrac{{{Q_{REJ}}}}{W} = \dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}}$
Since the room is at a higher temperature than the refrigerator , it will act as a hot reservoir .
The electric energy is the input energy or the energy supplied to do the work ($W$)
${Q_{Supplied}} = {Q_{REJ}} + W$
amount of heat delivered to the room for each joule of electrical energy is
$\Rightarrow \dfrac{{{Q_{Supplied}}}}{W} = \dfrac{{{Q_{REJ}}}}{W} + 1$
Putting the value of $\dfrac{{{Q_{REJ}}}}{W}$in the equation
$\Rightarrow \dfrac{{{Q_{Supplied}}}}{W} = \dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}} + 1$ since $\Rightarrow \dfrac{{{Q_{REJ}}}}{W} = \dfrac{{{t_2} + 273}}{{{t_1} - {t_2}}}$
Further solving the equation.
$\Rightarrow \dfrac{{{Q_{Supplied}}}}{W} = \dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$
Hence option (D) is the correct answer.
The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be$\dfrac{{{t_1} + 273}}{{{t_1} - {t_2}}}$.

Note:-
In physical significance of the coefficient of performance ($COP$) we mean that we take the input depending upon the output which we desired , while in physical significance of efficiency the output depends on the input which we have supplied .
When we kept open the door of the refrigerator in the kitchen the overall temperature of the kitchen rises.