Question

An refrigerator with an average COP 2.8 is used to cool a well insulated container whose contents are equivalent to 12 kg of water from 40 C to 10 C. when the running , the refrigerator consumes 400 W of electric power. Calculate the time required for the refrigerator to accomplish this cooling.

Answer 1

Answer:

It is going to take about 22.43 minutes

Explanation:

The Coefficient of Performance of a refrigerator is the ratio of useful cooling to work required to achieve it. The formula would be:

COP = Cooling Effect/Power Input

The COP is given as 2.8 and the power rating is 400 Watts. We can find the cooling effect as follows:

2.8 = Cooling Effect/400

Cooling Effect = 2.8 * 400 = 1120 Watts

Now,

The cooling effect can be done using the formula:

$Q=mc \Delta T$

Where

Q is the thermal energy (or work)

m is the mass

c is the specific heat capacity

$\Delta T$ is the temperature change

We know

m = 12 kg

c is the specific heat of water, 4187 J/kg*C

$\Delta T$ is 30, from 40C to 10C

Substituting, we get:

$Q=mc \Delta T\\Q=12*4187*30\\Q=1507320$

This thermal energy is the Work, which is Power * Time required

Thus, we can say:

Time Required = Q/Power

So,

Time Required = 1507320/1120 = 1346 Seconds

To minutes, we can say:

1346/60 = 22.43 minutes