Question

A heat pump is absorbing heat from the cold outdoors at \(5^{\circ} \mathrm{C}\) and supplying heat to a house at \(25^{\circ} \mathrm{C}\) at a rate of \(18,000 \mathrm{kJ} / \mathrm{h}\). If the power consumed by the heat pump is \(1.9 \mathrm{kW},\) the coefficient of performance of the heat pump is \((a) 1.3\) (b) 2.6 \((c) 3.0\) \((d) 3.8\) \((e) 13.9\)

Short answer

a) 1.6 b) 2.6 c) 3.6 d) 4.6 Answer: b) 2.6

Step-by-step solution

Convert heat transfer rate to power

We need to convert the given heat transfer rate from kJ/h to kW to match the units of the power consumption. Use the conversion factor 1 h = 3600 s. \(18,000 \ \frac{\text{kJ}}{\text{h}} \times \frac{1 \ \text{kW}}{1,000 \ \text{kJ/s}} \times \frac{1 \ \text{h}}{3600 \ \text{s}} = 5 \ \text{kW}\) So, the heat pump is supplying heat to the house at a rate of 5 kW.

Calculate the coefficient of performance (COP)

Now, we can calculate the COP using the formula: COP \(= \frac{\text{Heat supplied to the house}}{\text{Power consumed}}\) Plug the values we found in step 1: COP \(= \frac{5 \ \text{kW}}{1.9 \ \text{kW}} = 2.63\)

Choose the correct answer

The calculated COP value is approximately 2.6. So, the correct answer is (b) 2.6.