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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:
A
$\frac{t_2 + 273}{t_1 - t_2}$
B
$\frac{t_1 + t_2}{t_1 + 273}$
C
$\frac{t_1}{t_1 - t_2}$
D
$\frac{t_1 + 273}{t_1 - t_2}$
Explanation
$Q_1/W = T_1/(T_1 - T_2)$.
Detailed Solution
Heat delivered to the room $= Q_1$
COP $\beta = \frac{t_2 + 273}{t_1 - t_2} = \frac{Q_2}{W} = \frac{Q_1 - W}{W} = \frac{Q_1}{W} - 1$
$\Rightarrow \frac{Q_1}{W} = 1 + \frac{t_2 + 273}{t_1 - t_2} = \frac{t_1 + 273}{t_1 - t_2}$
COP $\beta = \frac{t_2 + 273}{t_1 - t_2} = \frac{Q_2}{W} = \frac{Q_1 - W}{W} = \frac{Q_1}{W} - 1$
$\Rightarrow \frac{Q_1}{W} = 1 + \frac{t_2 + 273}{t_1 - t_2} = \frac{t_1 + 273}{t_1 - t_2}$