Questions: Heat Pump Systems for Heating and Cooling

Question 1

A heat pump system has a COP_cooling of 2.8 when operating in air-conditioning mode. What is its COP_heating when operating in heating mode under the same conditions?

• A. 2.8 — COP is a property of the hardware and doesn't change with mode
• B. 1.8 — you subtract 1 because some energy is lost reversing the cycle
• C. 3.8 — COP_heating = COP_cooling + 1
• D. 1.0 — heat pumps in heating mode are equivalent to resistance heaters

Answer: (C)

The energy balance gives Q_H = Q_L + W_net. Dividing by W_net: COP_heating = Q_H/W_net = (Q_L/W_net) + 1 = COP_cooling + 1 = 3.8. The extra 1 unit comes from the compressor work itself being converted to heat at the condenser. This relationship always holds: heating COP is exactly 1 greater than cooling COP under the same operating conditions. Option D is the classic misconception — a heat pump always outperforms resistance heating (COP = 1) as long as any heat is harvested from the outdoor air.

Question 2

A homeowner's heat pump delivers 4 units of heat per unit of electricity on a mild 10°C winter day. On an extremely cold −20°C day, what would you expect?

• A. The same COP of 4 — the heat pump's efficiency is a fixed equipment specification
• B. A higher COP — colder outdoor air creates a larger temperature differential that drives more heat transfer
• C. A lower COP — the larger temperature differential between indoor and outdoor requires more compressor work
• D. COP exactly equals 1 — heat pumps cannot extract heat below freezing

Answer: (C)

COP depends on the temperature ratio between the hot and cold reservoirs (bounded above by Carnot COP). At −20°C, the evaporator must run even colder to absorb heat from the outdoor air, while the condenser still must heat the house — a much larger pressure ratio. More compressor work per unit of heat moved means lower COP. At extreme cold, COP can fall near 1, at which point supplemental resistance heating becomes necessary. Option B reverses the logic: a larger temperature difference always reduces thermodynamic efficiency, not improves it.

Question 3

A heat pump with COP_heating of 3.5 violates conservation of energy because it delivers more heat energy than it consumes in electrical energy.

• T. True
• F. False

Answer: False

No energy is created. The heat pump delivers Q_H = Q_L + W_net: 3.5 units of heat to the house, but 2.5 of those units come from outdoor heat extracted by the evaporator, and only 1 unit comes from the electrical input. Energy is fully conserved — the 'extra' heat comes from the environment, not from nothing. COP > 1 does not violate thermodynamics; it simply means the device moves more energy than it consumes as work.

Question 4

A heat pump's COP_heating always exceeds 1 (as long as the system is functioning), whereas an electric resistance heater always has COP_heating = 1.

• T. True
• F. False

Answer: True

This is the fundamental thermodynamic advantage of heat pumps. A resistance heater converts electrical work directly to heat with 100% conversion efficiency — exactly 1 unit of heat per unit of work, so COP = 1. A heat pump also converts the compressor work to heat, but additionally extracts heat from the outdoor environment, delivering Q_H = Q_L + W_net > W_net. As long as the system moves any heat at all from outdoors (Q_L > 0), COP_heating > 1. Even a poorly performing heat pump in very cold conditions is at least marginally better than resistance heat.

Question 5

Explain why a heat pump's COP_heating decreases as the outdoor temperature drops. What thermodynamic mechanism causes this?

Model answer: As outdoor temperature drops, the evaporator must operate at a lower temperature (below outdoor air) to absorb heat. Meanwhile, the condenser pressure stays high to deliver heat indoors. This wider temperature (and pressure) differential requires more compressor work per unit of heat transferred. Since COP_heating = Q_H / W_net, and W_net rises faster than Q_H as the temperature difference grows, COP falls. The upper bound is the Carnot COP = T_H / (T_H − T_L); as T_L drops, this bound decreases, pulling the real COP down with it.

The core insight is that COP is not a fixed equipment property — it depends on operating conditions. The compressor must overcome the pressure difference between the evaporator and condenser. Larger temperature differences mean larger pressure ratios, more compression work, and lower efficiency. This is why heat pumps are most efficient in mild climates and why cold-climate models require variable-speed compressors to maintain reasonable COP at low temperatures.