Question: (II) What is the temperature inside an ideal refrigerator-freezer that operates with a COP = 7.0 in a 22°C room?

The ratio of the heat removed from the low-temperature area inside a refrigerator to the work done by the refrigerator to remove the heat is termed the coefficient of performance of the refrigerator (COP).

\({\rm{COP}} = \frac{{{Q_{\rm{L}}}}}{W}\)

For an ideal refrigerator, the expression for COP is

\({\rm{COP}} = \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}\).

Here, \({T_{\rm{L}}}\) is the temperature of the low-temperature region, i.e., inside of refrigerator, and \({T_{\rm{H}}}\)is the temperature of the high-temperature region, i.e., room in which the refrigerator is placed.

The coefficient of performance of the ideal refrigerator is COP &= 7.0.

The temperature of the high-temperature region, i.e., room is \({T_{\rm{H}}} = 22^\circ {\rm{C}} = 295\;{\rm{K}}\).

By equation (i), the temperature of the inside of refrigerator-freezer is calculated below:

\(\begin{aligned}{c}{\rm{COP}} &= \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}\\{\rm{COP}}\left( {{T_{\rm{H}}} - {T_{\rm{L}}}} \right) &= {T_{\rm{L}}}\\{\rm{COP}}\left( {{T_{\rm{H}}}} \right) &= \left( {1 + {\rm{COP}}} \right){T_{\rm{L}}}\\{T_{\rm{L}}} &= \frac{{{\rm{COP}}\left( {{T_{\rm{H}}}} \right)}}{{\left( {1 + {\rm{COP}}} \right)}}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{T_{\rm{L}}} &= \frac{{7.0\left( {295\;{\rm{K}}} \right)}}{{\left( {1 + 7.0} \right)}}\\ \approx 258\;{\rm{K}}\\ &= \left( {258 - 273} \right)^\circ {\rm{C}}\\ &= - 15^\circ {\rm{C}}\end{aligned}\)

Thus, the inside temperature of the ideal refrigerator-freezer is \( - 15^\circ {\rm{C}}\).