Question: (II) A heat pump is used to keep a house warm at 22°C. How much work is required of the pump to deliver 3100 J of heat into the house if the outdoor temperature is (a) 0°C, (b) \({\bf{ - 15^\circ C}}\)? Assume a COP of 3.0. (c) Redo for both temperatures, assuming an ideal (Carnot) coefficient of performance \({\bf{COP = }}{{\bf{T}}_{\bf{L}}}{\bf{/}}\left( {{{\bf{T}}_{\bf{H}}}{\bf{ - }}{{\bf{T}}_{\bf{L}}}} \right)\).

The ratio of the heat delivered inside the room to the work done by the heat pump to deliver the heat is termed the coefficient of performance of the heat pump (COP).

The expression for COP is

\({\rm{COP}} = \frac{{{Q_{\rm{H}}}}}{W}\). … (i)

Heat delivered into the house is\({Q_{\rm{H}}} = 3100\;{\rm{J}}\).

The coefficient of performance of the heat pump is COP = 3.0.

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

From equation (i), the work required to be done by the heat pump is

\(\begin{aligned}{c}W &= \frac{{{Q_{\rm{H}}}}}{{{\rm{COP}}}}\\ &= \frac{{3100\;{\rm{J}}}}{{3.0}}\\ \approx 1.0 \times {\rm{1}}{{\rm{0}}^3}\;{\rm{J}}{\rm{.}}\end{aligned}\)

Since the expression of work does not use the outside temperature, \({T_{\rm{L}}}\), it will be constant for all temperatures.

Thus, the work required to be done by the heat pump to deliver 3100 J of heat is \(1.0 \times {\rm{1}}{{\rm{0}}^3}\;{\rm{J}}\).

The coefficient of performance for an ideal heat pump is

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

Equate equations (i) and (ii).

\(\begin{aligned}{l}\frac{{{Q_{\rm{H}}}}}{W} &= \frac{{{T_{\rm{H}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}\\W &= \left( {\frac{{{T_{\rm{H}}} - {T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right){Q_{\rm{H}}}\end{aligned}\)

For \({T_{\rm{L}}} = 0^\circ {\rm{C }} = 273\;{\rm{K}}\), the work required to be done by the ideal heat pump is calculated below:

\(\begin{aligned}{c}W &= \left( {\frac{{\left( {295 - 273} \right)\;{\rm{K}}}}{{295\;{\rm{K}}}}} \right)3100\;{\rm{J}}\\ &= 231.{\rm{19}}\;{\rm{J}}\\ &= 2.3 \times 1{{\rm{0}}^2}\;{\rm{J}}\end{aligned}\)

For \({T_{\rm{L}}} = - 15^\circ {\rm{C}} = 258\;{\rm{K}}\), the work required to be done by the ideal heat pump is calculated below:

\(\begin{aligned}{c}W &= \left( {\frac{{\left( {295 - 258} \right)\;{\rm{K}}}}{{295\;{\rm{K}}}}} \right)3100\;{\rm{J}}\\ &= 388.81\;{\rm{J}}\\ &= 3.9 \times 1{{\rm{0}}^2}\;{\rm{J}}\end{aligned}\)

Thus, the work required to be done by an ideal heat pump when the outdoor temperature is \(0^\circ {\rm{C}}\) is \(2.3 \times {10^2}\;{\rm{J}}\) and when it is \( - 15^\circ \) is \(3.9 \times {10^2}\;{\rm{J}}\).