A refrigerator works between $4^{\circ} \mathrm{C}$ and $30^{\circ} \mathrm{C}$. It is required to remove $\mathbf{6 0 0}$ calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = $4.2$ Joules) A $\quad 2.365 \mathrm{~W}$ B $\quad 23.65 \mathrm{~W}$ C $\quad$ 236.5W D $\quad 2365 \mathrm{~W}$

Answer: C

Solution:
Given,

$\mathrm{T}_{1}=303$
$\mathrm{~T}_{2}=277$

Coefficient of performance

$\beta=\frac{\mathrm{Q}_{2}}{\mathrm{~W}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}-\mathrm{Q}_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}} \quad\left(\mathrm{Q}_{1}=\mathrm{W}+\mathrm{Q}_{2}\right)$
Substituting the value,

$\frac{Q_{2}}{\mathrm{~W}}=\frac{277}{26}$

$\mathrm{W}=\mathrm{Q}_{2} \frac{26}{277}=\frac{600 \times 4.2 \times 26}{277}$

$=236.5 \mathrm{~J}$