The temperature inside a refrigerator is $t_{2}^{o} C$ and the room temperature is $t_{1}^{o} C$ . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

\frac{t_{1} + t_{2}}{t_{1} + 273}

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\frac{t_{1}}{t_{1} - t_{2}}

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\frac{t_{1} + 273}{t_{1} - t_{2}}

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\frac{t_{2} + 273}{t_{1} - t_{2}}

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Solution

The correct option is C $\frac{t_{1} + 273}{t_{1} - t_{2}}$
Heat delivered $= Q_{1}$
Coefficient of performance $= β = \frac{Q_{2}}{W} = \frac{Q_{1} - W}{W} = \frac{Q_{1}}{W} - 1 = \frac{T_{2}}{T_{1} - T_{2}}$
$\Rightarrow \frac{Q_{1}}{W} = 1 + \frac{t_{2} + 273}{t_{1} - t_{2}} = \frac{t_{1} + 273}{t_{1} - t_{2}}$

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