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

Answer:

The correct answer will be "8640 kWh".

Explanation:

The given values are:

Rate (Ql)

= 5040 kJ/h

As we know,

⇒  $COP = \frac{Cooling}{heat \ required}$

On putting the values, we get

⇒           $=\frac{5040 \ kJ/h}{1 \ kW}$

⇒           $=1.4$

Now,

⇒  $Q = Ql + W$

        $= 1 + 1.4$

        $=2.4 \ kW$

        $=2.4\times 3600$

        $=8640 \ kWh$