Question

Determine the COP of a heat pump that supplies energy to a house at a rate of \(8000 \mathrm{kJ} / \mathrm{h}\) for each \(\mathrm{kW}\) of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

Short answer

Answer: The Coefficient of Performance (COP) of the heat pump is 2.22 and the rate of energy absorption from the outdoor air is 1.22 kW.

Step-by-step solution

Convert energy rate to standard units

The energy rate at which the heat pump supplies energy to the house is given as 8000 kJ/h. We need to convert this to standard SI units, which are J/s or W (watts). To do this, we'll use the conversion factors: 1 hour = 3600 seconds (s) 1 kJ = 1000 J So, the energy rate in watts is: \(Q_{out} = \frac{8000 \mathrm{kJ}}{\mathrm{h}} \cdot \frac{1000 \mathrm{J}}{1 \mathrm{kJ}} \cdot \frac{1 \mathrm{h}}{3600 \mathrm{s}} = 2.22\overline{2} \mathrm{ kW}\)

Calculate the COP of the heat pump

Now we can calculate the COP of the heat pump using the given energy rate and the electrical power input: \(\text{COP} = \frac{Q_{out}}{W_{in}}\) Given that the heat pump supplies energy to the house at a rate of 2.22 \(\mathrm{kW}\) for each \(\mathrm{kW}\) of electric power it draws, we have: \(\text{COP} = \frac{2.22 \mathrm{kW}}{1 \mathrm{kW}} = 2.22\) So, the COP of the heat pump is 2.22.

Calculate the rate of energy absorption from the outdoor air

Recall that the heat pump's COP is the ratio of the heat output (Q_out) to the power input (W_in), and the heat pump absorbs energy from the air at a rate of (Q_abs). We can rearrange the COP equation to solve for the rate of energy absorption from the outdoor air: \(Q_{abs} = Q_{out} - W_{in}\) Now we can calculate the rate of energy absorption from the air: \(Q_{abs} = 2.22\overline{2} \mathrm{kW} - 1 \mathrm{kW} = 1.22\overline{2} \mathrm{kW}\) The rate of energy absorption from the outdoor air is 1.22 \(\mathrm{kW}\). In summary, the COP of the heat pump is 2.22, and the rate of energy absorption from the outdoor air is 1.22 \(\mathrm{kW}\).