Question

An ideal refrigerator is working between temperature ${27}^{0}C$ and ${127}^{0}C$ .If it expells 120 calorie of heat in one second then calculate its wattage.

Answer 1

$T_1=127°C=400k$ $T_2=27°C=300k$

$\Rightarrow \frac{Q}{Time}=120cal.=502.1J$

$\Rightarrow$ Coefficient of poerformance (k) for ideal refrigerator $=\frac{Heatexpell}{Workdone}=\frac{Q}{W}$

In terms of temperature

$\Rightarrow k=\frac{T_2}{T_1-T_2}$

So, $\frac{Q}{W}=\frac{T_2}{T_1-T_2}$

For per unit time

$\Rightarrow \frac{\frac{Q}{time}}{\frac{Work}{Time}}=\frac{\frac{Q}{time}}{Wattage}=\frac{T_2}{T_1-T_2}$

$\Rightarrow$ Wattage $=\frac{502.1}{\frac{300}{400-300}}=\frac{502.1\times 100}{300}$

$\Rightarrow$ Wattage $=167.4J/s.$

Hence, the answer is $167.4J/s.$