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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?

User Joche Wis
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1 Answer

2 votes

Answer:

59.52 J.

Step-by-step 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.

User DreamingOfSleep
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