Question

The temperature inside the 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
$\begin{aligned} & a)\dfrac{{{t}_{1}}}{{{t}_{1}}-{{t}_{2}}} \\\ & b)\dfrac{{{t}_{1}}+273}{{{t}_{1}}-{{t}_{2}}} \\\ & c)\dfrac{{{t}_{2}}+273}{{{t}_{1}}-{{t}_{2}}} \\\ & d)\dfrac{{{t}_{1}}+{{t}_{2}}}{{{t}_{1}}+273} \\\ \end{aligned}$

Answer 1

In the above question we are asked to determine the amount of heat delivered to the room for each joule of electrical energy consumed ideally. This basically means we have to determine the ratio of heat delivered to the hot reservoir to the mechanical work done per cycle by the refrigerator. Hence we will use the definition of coefficient of performance of the refrigerator to determine the following.
Formula used:
$\beta =\dfrac{{{Q}_{2}}}{W}$
$\beta =\dfrac{{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}}$

Complete answer:
The coefficient of performance ($\beta$ ) of a refrigerator is defined as the ratio of the amount of heat removed (${{Q}_{2}}$) per cycle to the mechanical work(W) required to be done on it. Mathematically this can be represented as,
$\beta =\dfrac{{{Q}_{2}}}{W}...(1)$
Hence we can say that the heat (${{Q}_{1}}$) delivered to the reservoir or the surroundings is equal to,
${{Q}_{1}}={{Q}_{2}}+W$
Let us say the temperature of the heat reservoir is ${{t}_{1}}$ and the temperature of the refrigerator is ${{t}_{2}}$. Hence the coefficient of performance can also be written as,
$\beta =\dfrac{{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}}...(2)$
Equating equation 1 and 2, the heat delivered to the surrounding for each joule of consumption of electrical energy we get as,
$\begin{aligned} & \beta =\dfrac{{{Q}_{2}}}{W}=\dfrac{{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}} \\\ &\Rightarrow\dfrac{{{Q}_{2}}}{W}=\dfrac{{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}}\text{, }\therefore {{Q}_{1}}={{Q}_{2}}+W \\\ & \Rightarrow \dfrac{{{Q}_{1}}-W}{W}=\dfrac{{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}} \\\ \end{aligned}$
Expressing the temperature of the sink and the source in terms of Kelvin in the above equation we get,
$\begin{aligned} & \dfrac{{{Q}_{1}}-W}{W}=\dfrac{{{t}_{2}}+273}{273+{{t}_{1}}-(273+{{t}_{2}})} \\\ & \Rightarrow \dfrac{{{Q}_{1}}}{W}-1=\dfrac{{{t}_{2}}+273}{{{t}_{1}}-{{t}_{2}}} \\\ &\Rightarrow\dfrac{{{Q}_{1}}}{W}=\dfrac{{{t}_{2}}+273}{{{t}_{1}}-{{t}_{2}}}+1=\dfrac{{{t}_{2}}+273+{{t}_{1}}-{{t}_{2}}}{{{t}_{1}}-{{t}_{2}}}=\dfrac{273+{{t}_{1}}}{{{t}_{1}}-{{t}_{2}}} \\\ & \Rightarrow \dfrac{{{Q}_{1}}}{W}=\dfrac{{{t}_{1}}+273}{{{t}_{1}}-{{t}_{2}}} \\\ \end{aligned}$

So, the correct answer is “Option B”.

Note:
It is mentioned in the question that electrical energy is consumed ideally by the refrigerator. Hence we could imply that work done is equal to the total supply of electrical energy to it. The only trick in the above question was that one had to express the temperature of the source (surrounding)and the sink (refrigerator) in terms of Kelvin temperature scale.