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 = (T_2)/(T_1 - T_2) = (Q_2)/(W)`

here we know that

`T_2 = 288`

`T_1 = 258`

`Q_2 = 300 kJ/min`

now from above formula

`(300)/(W) = (288)/(288 - 258)`

`(300)/(W) = (288)/(30)`

`W = 31.25 kJ/min`

`W = 520.8 W`

so its work done is 520.8 W