A refrigerator has a coefficient of performance equal to 4.2. How much work must be done on the refrigerator in order to remove 250 J of heat from the interior?

59.52 J.

Explanation:

The Formula of Coefficient of performance is given as,

η = Q/W ... Equation 1

Where η = Coefficient of performance, Q = The heat from the interior, W = Work done by the refrigerator.

make W the subject of the equation

W = Q/η ... Equation 2

Given: Q = 250 J, η = 4.2

Substitute into equation 2

W = 250/4.2

W = 59.52 J.

Hence the work done by the refrigerator = 59.52 J.