Question

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

Answer 1

Temperature inside the refrigerator:

$T_H={27}^{\circ }$

$=27+273=300K$

Temperature outside the refrigerator:

$T_H={127}^{\circ }$

$=127+273=400K$

Amount of heat expelled outisde:

$Q_H=120$ calorie $=502J$

We have:

$CoP=\frac{T_H}{T_C}$ where $CoP is coefficient of performance.

$CoP=\frac{300}{100}=3$

Also,$CoP=\frac{Q_C}{W}=3$ where,

$Q_C$ is amount of heat taken from lower temperature region and W is the work done to shove heat into higher temperature region.

$Q_C=3W$

Now, $Q_C+W=Q_H$

$3W+W=502$

$4W=502$

$W=\frac{502}{4}=125.5J$

Power $=\frac{W}{t}=125.5watt$