6-44Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.

The correct answer will be "8640 kWh".

Explanation:

The given values are:

Rate (Ql)

= 5040 kJ/h

As we know,

⇒  

On putting the values, we get

⇒          

⇒          

Now,

⇒  

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