Question

Question: (II) (a) What is the coefficient of performance of an ideal heat pump that extracts heat from 6°C air outside and deposits heat inside a house at 24°C? (b) If this heat pump operates on 1200 W of electrical power, what is the maximum heat it can deliver into the house each hour? See Problem 35.

Short answer

(a) The coefficient of performance of an ideal heat pump is 16.5.

(b) The maximum heat that can be delivered by the heat pump into the house each hour is \(7.1 \times {\rm{1}}{{\rm{0}}^7}{\rm{ J}}\).

Step-by-step solution

Understanding the coefficient of performance of heat pump

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

The expression for the COP is

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

For an ideal heat pump, the COP is

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

Here, \({T_{\rm{L}}}\) is the temperature of the low-temperature region, i.e., outside the house, and \({T_{\rm{H}}}\) is the temperature of the high-temperature region, i.e., inside the house.

Given information

The temperature of low-temperature region is \({T_{\rm{L}}} = 6^\circ {\rm{C}} = \left( {6 + 273} \right)K = 279\;{\rm{K}}\).

The temperature of high temperature region is\({T_{\rm{H}}} = 24^\circ {\rm{C}} = \left( {24 + 273} \right)\;{\rm{K}} = 297\;{\rm{K}}\).

The electrical power of heat pump is P = 1200 W.

Time taken is \(t = 1\;{\rm{h}} = 36{\rm{00}}\;{\rm{s}}\).

(a) Determination of COP of an ideal heat pump

The coefficient of performance of an ideal heat pump is

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

Substitute the values in the above expression.

\(\begin{aligned}{c}{\rm{COP}} &= \frac{{297\;{\rm{K}}}}{{\left( {297 - 279} \right)\;{\rm{K}}}}\\ &= 16.5\end{aligned}\)

Thus, the coefficient of performance of an ideal heat pump is 16.5.

(b) Determination of heat delivered by the heat pump in an hour

The electrical power of the heat pump is

\(\begin{aligned}{c}P &= \frac{W}{t}\\W &= P \times t.\end{aligned}\)

The expression for the coefficient of performance of an ideal heat pump is

\(\begin{aligned}{c}{\rm{COP}} &= \frac{{{Q_{\rm{H}}}}}{W}\\ &= \frac{{{Q_{\rm{H}}}}}{{P \times t}}.\end{aligned}\)

The heat delivered by the heat pump can be calculated as shown below:

\(\begin{aligned}{c}{Q_{\rm{H}}} &= P \times t \times {\rm{COP}}\\ &= \left( {1200\;{\rm{J}}} \right) \times \left( {3600\;{\rm{s}}} \right) \times \left( {16.5} \right)\\ &= 7.1 \times {\rm{1}}{{\rm{0}}^7}{\rm{ J}}\end{aligned}\)

Thus, the maximum heat that can be delivered by the heat pump into the house each hour is \(7.1 \times {\rm{1}}{{\rm{0}}^7}{\rm{ J}}\).