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

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Solution

Given:
$T_{1} = 27 ° C = (27 + 273) K = 300 K$
$T_{2} = 127 ° C = (127 + 273) K = 400 K$
Heat expelled, $Q_{1} = 120 c a l$
To find:
Wattage of refrigerator, $P = ?$
The coefficient of performance of an ideal refrigerator can be calculated using the formula,
$β = \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} ⟹ \frac{Q_{2}}{Q_{1} - Q_{2}} = \frac{T_{2}}{T_{1} - T_{2}} ⟹ \frac{Q_{2}}{120 - Q_{2}} = \frac{400}{400 - 300} ⟹ Q_{2} = 480 - 4 Q_{2} ⟹ 5 Q_{2} = 480 ⟹ Q_{2} = 96 c a l$

We also know,
Work done, $W = Q_{1} - Q_{2} = 120 - 96 = 24 c a l = (24 \times 4.184) J = 100.42 J$ (as 1cal =4.184J).

Now, the amount of power equal to one joule of energy per second,
Hence the wattage of the refrigerator is 100.42W

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