Question

A refrigerator has COP o 3.How much work must be supplied to a refrigerator in order to remove 200J of heat from its interior?

Answer 1

Answer:

Work that is supplied by the refrigerator in order to remove 200 J energy from refrigerator is 66.7 J

Explanation:

As we know that COP of refrigerator is defined as the ratio of heat removed from the refrigerator and input work

So here we can say

$COP = \frac{Heat}{Work}$

so we have

$3 = \frac{Q}{W}$

Q = 200 J

$3 = \frac{200}{W}$

$W = \frac{200}{3} J$

#Learn

Topic : 2nd Law of thermodynamics

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