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

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

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Solution

The correct option is B $\frac{t_{1} + 273}{t_{1} - t_{2}}$

Coefficient of Performance of Refrigerator $C O P = \frac{Q_{c}}{W} = \frac{t_{2} + 273}{t_{1} - t_{2}}$

Temperatures should be written in kelvin.

We have to find heat given to room per unit electrical energy.
Room is at higher temperature and act as hot reservoir. And electrical energy is the input energy shown as W.

So, we have to find $\frac{Q_{H}}{W}$

Again, $Q_{H} = Q_{c} + W$

$⟹ \frac{Q_{H}}{W} = \frac{Q_{c}}{W} + 1$

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

$⟹ \frac{Q_{H}}{W} = \frac{t_{1} + 273}{t_{1} - t_{2}}$

Answer-(B)

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