Questions: Heat Pump Cycles and Heating Applications
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
A heat pump delivers 9 kW of heat to a building while its compressor consumes 3 kW of electricity. What is COP_heating, and which statement best explains why this doesn't violate energy conservation?
ACOP = 3.0; the heat pump creates three times as much heat as electricity consumed, which does violate energy conservation and indicates a faulty measurement
BCOP = 3.0; the first law is satisfied because the heat pump also extracts 6 kW of heat from the outdoor environment, which combines with the 3 kW of electrical work to produce 9 kW of heat delivered
CCOP = 0.33; all real heating devices must have efficiency below 1 by the second law
DCOP = 3.0; this performance is only achievable by a Carnot heat pump operating between ideal reservoirs
Q_hot = Q_cold + W_in by the first law. The heat pump extracts Q_cold = 6 kW from the outdoor air and adds W_in = 3 kW of compressor work, delivering Q_hot = 9 kW indoors. Energy is conserved — the 'extra' heat comes from the outdoor environment, not from nowhere. COP_heating = Q_hot / W_in = 9/3 = 3.0. No thermodynamic law is violated; the heat pump is a heat mover, not a heat creator.
Question 2 Multiple Choice
On a cold day (−15°C outdoors, 20°C indoors), an air-source heat pump achieves COP_heating = 1.5, while an electric resistance baseboard heater always has COP = 1.0. Which is the better choice for energy efficiency?
AThe resistance heater, because heat pumps become unreliable and inefficient below freezing
BThe heat pump, because COP 1.5 > 1.0 — it delivers 50% more heat per unit of electricity consumed than resistance heating, even at this degraded cold-weather performance
CThey are equivalent in practice; the 0.5 COP difference is within measurement uncertainty for real systems
DThe resistance heater, because the defrost cycles required at −15°C reduce the effective COP below 1.0
Even at COP = 1.5 on a very cold day, the heat pump delivers 1.5 kWh of heat per kWh of electricity — 50% more than resistance heating. As long as COP_heating > 1.0, the heat pump is more efficient. COP > 1 is guaranteed by COP_heating = COP_cooling + 1 ≥ 1 for any physically operating heat pump with non-negative COP_cooling. The degraded cold-weather performance is real but does not eliminate the efficiency advantage.
Question 3 True / False
The coefficient of performance of a heat pump in heating mode is always greater than 1, even for a very poorly performing real heat pump.
TTrue
FFalse
Answer: True
COP_heating = COP_cooling + 1. Since COP_cooling = Q_cold / W_in ≥ 0 (removing some heat from the cold reservoir is always non-negative for a physically operating system), COP_heating ≥ 1. This is why even a very inefficient heat pump delivers at least as much heat as resistance heating. The COP approaches 1 only in the limit where COP_cooling → 0 (no heat extracted from the cold reservoir, all heat from work alone — essentially resistance heating).
Question 4 True / False
A heat pump with COP = 4 creates heat from electrical energy, which is why it can deliver 4 kWh of thermal energy for nearly every 1 kWh of electricity consumed.
TTrue
FFalse
Answer: False
This is the central misconception. A heat pump does not create heat — it moves heat. The compressor uses 1 kWh of electricity to pump 3 kWh of heat from the cold outdoor environment to the warm indoor space, delivering 4 kWh total. The 'extra' 3 kWh was already present in the outdoor air; the compressor work provided the thermodynamic lift to move it against the temperature gradient. Energy is neither created nor destroyed — only relocated.
Question 5 Short Answer
Explain why a heat pump with COP = 3 does not violate energy conservation, even though it delivers 3 kWh of heat for every 1 kWh of electricity consumed.
Think about your answer, then reveal below.
Model answer: The first law of thermodynamics requires Q_hot = Q_cold + W_in. The heat pump delivers 3 kWh to the building: 1 kWh comes from the electrical work input (W_in) and 2 kWh comes from heat extracted from the outdoor environment (Q_cold). The compressor uses electricity to drive the refrigerant cycle, which absorbs heat from the cold outdoor air at the evaporator and rejects it at higher temperature to the building at the condenser. Total energy in (1 kWh electricity + 2 kWh from outdoor air) equals total energy out (3 kWh heat to building). No energy is created — it is moved from outside to inside.
The distinction between 'moving heat' and 'creating heat' is the conceptual heart of heat pump thermodynamics. The COP can exceed 1 precisely because the heat pump is not converting work to heat (which would give COP = 1) but using work to pump heat from a reservoir that already contains it.