Question

An ideal refrigerator does 240 J of work to remove 610 J as heat from its cold compartment. (a) What is the refrigerator's coefficient of performance? (b) How much heat per cycle is exhausted to the kitchen?

Answer 1

Step-by-step explanation:

It is given that,

Work done, W = 240 J

Heat removed, Q = 610 J

(a) The refrigerator's coefficient of performance is given by :

`COF=(Q)/(W)`

`COF=(610)/(240)`

COF = 2.54

(b) Let Q' is the heat per cycle is exhausted to the kitchen. It is given by :

Q' = Q + W

Q' = 610 + 240

Q' = 850 J

Hence, this is the required solution.