A cooling unit for a new freezer has an inner surface area of\({\bf{8}}{\bf{.0}}\;{{\bf{m}}{\bf{2}}}\), and is bounded by walls 12 cm thick with a thermal conductivity of\({\bf{0}}{\bf{.050}}\;{\bf{W/m}} \cdot {\bf{K}}\). The inside must be kept at -15°C in a room that is at 22°C. The motor for the cooling unit must run no more than 15% of the time. What is the minimum power requirement of the cooling motor?

The ratio of the heat removed from a low temperature area\(\left( {{{\bf{Q}}_{\bf{L}}}} \right)\)to work done (W) to remove the heat.

The expression for the COP is given as:

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

Here,\({Q_{\rm{L}}}\)is the heat removed and W is the work done.

The temperature in the freezer is, \({T_{\rm{L}}} = - 15\circ {\rm{C}}\).

The temperature in the room is, \({T_{\rm{H}}} = 22\circ {\rm{C}}\).

The inner surface area is, \(A = 8\;{{\rm{m}}2}\).

The thickness is, \(d = 12\,{\rm{cm}}\).

The thermal conductivity is, \(k = 0.050\;{\rm{W/m}} \cdot {\rm{K}}\).

The relation to find the rate of change of heat is given by,

\(\frac{{{Q_{\rm{L}}}}}{t} = kA\frac{{{T_{\rm{H}}} - {T_{\rm{L}}}}}{d}\)

Substitute the values in the above expression.

\(\begin{array}{c}\frac{{{Q_{\rm{L}}}}}{t} = \left( {0.050\;{\rm{W/m}} \cdot {\rm{K}}} \right)\left( {8\;{{\rm{m}}2}} \right)\frac{{\left( {22 + 273} \right)\;{\rm{K}} - \left( { - 15 + 273} \right)\;{\rm{K}}}}{{12\;{\rm{cm}} \times \frac{{{{10}{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}\\ = 0.40 \times \frac{{37}}{{0.12}}\\ = 123.33\,{\rm{J/s}}\end{array}\)

The expression for COP is given as:

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

Since 15% of the time is taken for the cooling unit of the motor then,

\({\rm{COP}} = \frac{{{Q_{\rm{L}}}{\rm{/}}t}}{{W/0.15t}}\) … (i)

The expression for the COP for an ideal refrigerator is given as:

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

Equate expression (i) and (ii).

\(\begin{array}{c}\frac{{{Q_{\rm{L}}}{\rm{/}}t}}{{W/0.15t}} = \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}\\\frac{W}{t} = \frac{{{Q_{\rm{L}}}{\rm{/}}t}}{{0.15}} \times \frac{{{T_{\rm{H}}} - {T_{\rm{L}}}}}{{{T_{\rm{L}}}}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}\frac{W}{t} = \left( {\frac{{123.33\,{\rm{J/s}}}}{{0.15}}} \right)\left( {\frac{{\left( {22 + 273} \right)\;{\rm{K}} - \left( { - 15 + 273} \right)\;{\rm{K}}}}{{\left( { - 15 + 273} \right)\;{\rm{K}}}}} \right)\\ \approx 118\;{\rm{W}}\end{array}\)

Thus, the minimum power requirement of the cooling motor is\(118\;{\rm{W}}\).