Question

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.

Answer 1

The correct answer will be "8640 kWh".

Step-by-step explanation:

The given values are:

Rate (Ql)

= 5040 kJ/h

As we know,

⇒
`COP = (Cooling)/(heat \ required)`

On putting the values, we get

⇒
`=(5040 \ kJ/h)/(1 \ kW)`

⇒
`=1.4`

Now,

⇒
`Q = Ql + W`

`= 1 + 1.4`

`=2.4 \ kW`

`=2.4* 3600`

`=8640 \ kWh`