Question

A residential air conditioner is removing 850 kJ/min of heat from the house to koop the house at 20 °C. If the COP of the nutrigerator is 2.08, how much heat is rejected to the ambient air outside the house? O A 20.07 kW OB. 14.16 KW O 0.29.46 KW OD.6.81 kW

Answer 1

The coefficient of performance of the refrigerator is defined as the heat removed from the cold reservoir(which is the inside of the house for our problem) divided by the amount of work required to do so.

The formula is given by:

$COP = \frac{Q_{cold}}{Work}$

Also from the first law of thermodynamics we know that where $Q_{hot}$ is the heat rejected to the ambient air.

So now the formula becomes:

$COP = \frac{Q_{cold}}{Q_{hot}-Q_{cold}}$

Also we are given the value of COP as 2.08 and the value of $Q_{cold}$ is given as 850KJ/min.

Converting the value of $Q_{cold}$ in KW we have $Q_{cold} = \frac{850}{60} = 14.16667\ KW$

Substituting the values in the formula of COP, we have

$2.08 = \frac{14.16667}{Q_{hot}-14.16667}\Rightarrow Q_{hot} = \frac{14.16667}{2.08}+14.16667$

From here we get the value of $Q_{hot}$ to be equal to 20.97KW.

A) 20.97KW