Questions: Coefficient of Performance: Heat Pumps and Refrigerators
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
A heat pump has a COP_heat of 3. For every 1 kJ of electrical work supplied, how much heat energy is delivered to the warm space?
A1 kJ — equal to the work input, since energy is conserved
B2 kJ — only the heat drawn from the cold reservoir is delivered
C3 kJ — the work plus the heat drawn from the cold reservoir
DMore than 3 kJ — a Carnot heat pump always exceeds its COP rating
COP_heat = Q_H/W, so Q_H = COP × W = 3 × 1 kJ = 3 kJ. This is not a violation of energy conservation: the heat pump moves Q_C = 2 kJ from the cold reservoir and adds the 1 kJ of work as heat, delivering Q_H = 3 kJ total. Option A is the resistance-heater answer — it conflates moving heat with creating it.
Question 2 Multiple Choice
A ground-source heat pump draws heat from underground soil at 10°C and delivers it to a house at 22°C. An air-source heat pump delivers heat to the same house but draws from outdoor air at −5°C. Which has a higher Carnot COP, and why?
AThe air-source pump, because colder outdoor air provides more thermal energy to extract
BBoth are identical — Carnot COP depends only on the indoor delivery temperature
CThe ground-source pump, because the smaller temperature difference between source and sink increases the Carnot COP
DThe ground-source pump, because underground heat is renewable and Carnot bounds only apply to fossil fuels
Carnot COP_heat = T_H/(T_H − T_C). The ground-source pump has T_H = 295 K, T_C = 283 K, giving 295/12 ≈ 24.6. The air-source pump has T_H = 295 K, T_C = 268 K, giving 295/27 ≈ 10.9. A smaller temperature difference means less work is required per unit of heat delivered — the thermodynamic bound is simply more favorable.
Question 3 True / False
The coefficient of performance of a refrigerator (COP_ref = Q_C/W) can be greater than 1.
TTrue
FFalse
Answer: True
Yes — COP_ref measures useful cooling output (heat removed from the cold space) per unit of work input. Since Q_C can be much larger than W, COP values of 2–5 are typical for modern refrigerators. The term 'coefficient of performance' was chosen precisely because 'efficiency' (usually capped at 100%) would be misleadingly restrictive for devices that move heat rather than convert it.
Question 4 True / False
A refrigerator and a heat pump operating between the same two temperatures have equal coefficients of performance.
TTrue
FFalse
Answer: False
COP_heat = Q_H/W and COP_ref = Q_C/W. Since Q_H = Q_C + W, we have COP_heat = COP_ref + 1. A heat pump always has a COP one unit higher than the refrigerator operating between the same temperatures. The heat pump's useful output includes both the heat drawn from the cold reservoir and the work input itself, while the refrigerator's useful output is only the heat removed from the cold space.
Question 5 Short Answer
Why can the COP of a heat pump exceed 1, even though no energy conversion device can be more than 100% efficient?
Think about your answer, then reveal below.
Model answer: A heat pump does not convert work into heat — it uses work to move heat from one place to another. The work input plus the heat drawn from the cold reservoir both end up delivered to the warm space. 'Efficiency' (output/input of the same energy form) caps at 1; COP measures a different ratio — useful heat delivered per unit of work — which can exceed 1 because the system harvests heat from the environment.
This is the key conceptual shift from heat engines. An electric resistance heater converts work to heat at COP = 1 (exactly). A heat pump uses the same work to leverage free environmental heat, delivering far more total heat. COP > 1 is not magic — it reflects the second law: moving heat from a warmer reservoir to an even warmer one requires relatively little work when the temperature difference is small.