Question

Short answer

The coefficient of performance is \(14.1\).

Step-by-step solution

Given Data

The temperature of the contents in the refrigerator is\({T_1} = 2.5^\circ {\rm{C}}\).

The temperature of the house is \({T_2} = 22^\circ {\rm{C}}\).

Understanding ideal coefficient of performance

To calculate the ideal coefficient of performance of the refrigerator, divide the lower temperature (temperature of contents) with the difference in temperature from high to low.

Calculation of the COP of the ideal refrigerator

The relation for COP is

\({\rm{COP}} = \left( {\frac{{{T_1}}}{{{T_2} - {T_1}}}} \right)\).

Put the values in the above relation.

\(\begin{aligned}{l}{\rm{COP}} &= \left( {\frac{{\left( {2.5^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( {\left( {22^\circ {\rm{C}} + 273} \right)\;{\rm{K}} - \left( {2.5^\circ {\rm{C}} + 273} \right)\;{\rm{K}}} \right)}}} \right)\\{\rm{COP}} &= 14.1\end{aligned}\)

Thus, \(14.1\) is the required COP.