Q106 / 180 · NEET 2016 (Phase 2) · Physics

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

Solution

$K = \frac{Q_{2}}{W} = \frac{1}{\frac{t_{1}}{t_{2}} - 1}$

$Q_{2} = \frac{t_{2} W}{t_{1} - t_{2}}$ .

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

$= \frac{t_{1} W}{t_{1} - t_{2}}$ .

Since, $t_{1} & t_{2}$ are in $^{o} C & W = 1 J$ ,

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

Hence, the correct answer is option: 'B'.