Arefrigerator has a coefficient of performance equal to 4.2. how much work must be done on the operating gas in the refrigerator in order to remove 250 j of heat from the interior compartment?

To solve this problem it is necessary to apply the concepts related to the coefficient of performance. The coefficient of performance allows us to relate both the Heat and the Work done on a heat pump or cooling system to know the percentage of losses or input energy.

Mathematically your relationship can be expressed as

Where,

Q = Heat

W = Work

Our values are given as

COP =4.2

Cooling

Replacing at our equation we have,

Therefore the work is 60J.

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.