Question

A refrigerator is to remove heat from the cooled space at a rate of 300 kj/min to maintain its temperature at 288c. if the air surrounding the refrigerator is at 258c, determine the minimum power input required for this refrigerator.

Answer 1

efficiency of refrigerator is given as

$\eta = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$

here we know that

$T_2 = 288$

$T_1 = 258$

$Q_2 = 300 kJ/min$

now from above formula

$\frac{300}{W} = \frac{288}{288 - 258}$

$\frac{300}{W} = \frac{288}{30}$

$W = 31.25 kJ/min$

$W = 520.8 W$

so its work done is 520.8 W