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# A refrigerator works between ${4^0}C$ and ${30^0}C$. It is required to remove 600 calories of heat every second in order to keep the temperature of refrigerated space constant. The power required is: (Take $1cal = 4.2Joules$)A) $2.365 W$B) $23.65 W$C) $236.5 W$D) $2365 W$

Last updated date: 13th Jun 2024
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Hint: We know that the coefficient of performance of the refrigerator is the ratio of heat removed to the work done by the refrigerator. Here work done is equal to energy supplied to refrigeration. coefficient of performance of refrigerator lower working temperature to the size of range of working.

Compete step by step solution:
Given, a refrigerator works between ${4^0}C$ and ${30^0}C$.
Then, ${T_1} = {30^0}C = 303K$ and ${T_2} = {4^0}C = 277K$.
We need to remove 600 calories of heat then, ${Q_2} = 600Cal = 600 \times 4.2 Joules$.
Let work done by the refrigerator is $W$.
We know that coefficient of performance is given by
$\beta = \dfrac{{{Q_2}}}{W} = \dfrac{{{T_2}}}{{{T_1} - {T_2}}}$
Then $W = {Q_2} \times \dfrac{{{T_1} - {T_2}}}{{{T_2}}}$
Putting values in above equation we get
$W = 600 \times 4.2 \times \dfrac{{303 - 277}}{{277}} = 237.3 Joules$.
We know that 600 calories removed per second, then $W$ is work done per second means it is equal to power.
$Power = W = 236.5 Watt$.

Hence, the correct answer is C.

Note:The net heat energy released by the refrigerator in the surrounding is the sum of heat removed and work done by the refrigerator. Efficiency of the refrigerator increases when the surrounding temperature is lower. Refrigerators stop work above some temperature because surrounding temperature is very high and efficiency becomes very low.