Question

a refridgerator has a coefficient of performance equal to 4.2 how much work must be done on the refridgerator in order to remove 250 j of heat from the interior

Answer 1

Answer:

60 J

Explanation:

Given That

Coefficient of performance, CoP = 4.2

Quantity of heat of the refrigerator, Q = 250 J

Work done by the refrigerator, W = ?

First, it should be noted that the Coefficient of Performance can be said to be, the ratio of the Heat required to the Work done by the system. Mathematically written as,

CoP = Q / W

Since we are already given the values from our question, we can plug it in as it's a pretty straightforward question

4.2 = 250 / W, making W subject of formula, we have

W = 250 / 4.2

W = 60J

Thus, the Work done by the refrigerator is 60 J