An electric resistance heater turns one kilowatt of electricity into one kilowatt of heat. A heat pump, fed the same kilowatt, can deliver three or four. It looks like a violation of energy conservation, and the first time most people meet the number they assume something has been counted twice. Nothing has. The heat pump is not creating energy — it is moving it, pumping warmth from the cold outdoors into a warmer room, and the electricity only pays for the pumping.
This article explains the single ratio that measures how well a heat pump does that job, where its theoretical ceiling comes from, and why the same unit performs worse on the coldest day, exactly when you need it most.
Why this calculation matters
The coefficient of performance, or COP, is the number that decides whether a heat pump is worth installing. It is the ratio of useful heat delivered to electrical work consumed. A COP of 4 means every unit of electricity yields four units of heat — and four times less electricity on the meter than a resistance heater doing the same job.
COP drives the economics of home heating, the running cost of refrigeration and air conditioning, and the carbon footprint of buildings shifting away from gas. But COP is not a fixed badge stamped on a machine. It depends on the temperatures the heat pump works between, and those change with the weather and the thermostat setting. Understanding what raises and lowers COP is the difference between an estimate that holds up and a heating bill that surprises you.
The core formula
A heat pump moves heat from a cold reservoir at temperature T_c to a hot one at T_h. Thermodynamics sets a hard ceiling on how efficiently any device can do this — the Carnot limit. For heating, the ideal coefficient of performance is:
COP_ideal = T_h / (T_h - T_c)
Both temperatures must be absolute, in kelvin. The structure of the formula carries the key insight. The denominator is the temperature lift, the gap the heat pump has to bridge. When that gap is small, the COP is large; when the gap widens, COP collapses. Pumping heat across a small temperature difference is cheap; pumping it across a large one is expensive.
No real machine reaches the Carnot value. Compressors, heat exchangers, and refrigerant flow all introduce losses. Engineers fold these into a Carnot efficiency factor, often somewhere around 25% to 50% for typical equipment, so the real COP is:
COP_real = eta_Carnot * COP_ideal
Once you know the real COP, the electrical power needed to deliver a given heat output Q_h follows directly:
W = Q_h / COP_real
That is the whole chain: temperatures set the ideal ceiling, the efficiency factor scales it down to reality, and the real COP converts a heat demand into an electricity demand.
A worked example
Consider a heat pump keeping a room at T_h = 20 C while drawing heat from outdoor air at T_c = 0 C. First convert both temperatures to kelvin: T_h = 293 K and T_c = 273 K.
Step 1 — the ideal (Carnot) COP.
COP_ideal = T_h / (T_h - T_c)
COP_ideal = 293 / (293 - 273)
COP_ideal = 293 / 20 = 14.6
In a perfect, lossless world this heat pump would deliver almost 15 units of heat per unit of electricity. That number is a thermodynamic ceiling, not a sales figure.
Step 2 — the real COP. A practical machine reaches only a fraction of Carnot. At about 25% of the Carnot value:
COP_real = 0.25 * 14.6 = 3.7
A real COP of 3.7 is a realistic figure for a heat pump working across this modest 20 K lift.
Step 3 — electrical power for a given heat demand. Suppose the room needs 5 kW of heat:
W = Q_h / COP_real
W = 5 / 3.7 = 1.35 kW
The compressor draws just 1.35 kW of electrical power to deliver 5 kW of heat. A resistance heater would draw the full 5 kW for the same result. That gap — 1.35 kW versus 5 kW — is the entire commercial case for heat pumps, and it is why the same building can be far cheaper to heat after switching.
Common mistakes
Forgetting to convert to kelvin. The Carnot formula needs absolute temperature. Plugging in 20 and 0 instead of 293 and 273 gives a wildly wrong answer — and a negative or infinite COP if T_c happens to be entered as zero. Always convert first.
Quoting the Carnot COP as the real one. The ideal figure of 14.6 is a ceiling no hardware reaches. Treat it as the reference against which real performance is measured, never as the expected number.
Assuming COP is constant year-round. As outdoor air drops toward freezing and below, the temperature lift grows and COP falls. A unit rated at COP 4 in mild weather may run nearer 2 on a hard winter night — the worst possible time for the drop.
Ignoring defrost and auxiliary heat. In cold, damp conditions an air-source heat pump periodically reverses to clear frost from the outdoor coil, and a backup resistance element may cut in. Both pull the seasonal average COP below the steady-state value.
Confusing heating and cooling COP. For cooling, the useful output is the heat removed from the cold side, so the formula uses T_c in the numerator. Heating and cooling COP for the same machine and temperatures are not equal; they differ by exactly one.
Try the interactive NovaSolver calculator
Carnot COP is quick to compute once, but the interesting question is how performance moves as the weather and the thermostat change. The Heat Pump & Refrigeration COP Calculator on NovaSolver lets you switch between heating, cooling, and refrigeration modes, set the hot-side and cold-side temperatures and the Carnot efficiency factor, and read off the actual COP, the Carnot COP, the heat transfer rates, the temperature lift, and an estimated annual energy cost — all updating live as you drag the sliders.
Related calculators
- Heat pump cycle calculator — steps through the refrigerant cycle on a property diagram, showing where the compressor work and heat exchange actually happen.
- Heat pump calculator — a broader sizing view for matching a heat pump to a building's heat demand.
- Heat exchanger NTU calculator — for the evaporator and condenser performance that sets the real temperature lift.
Explore the full set in the thermal engineering tools hub.
Closing note
A heat pump's COP is the most honest single number you can ask about it. Keep the chain of reasoning clear: absolute temperatures set the Carnot ceiling, the lift in the denominator is what punishes you in cold weather, a real efficiency factor scales the ceiling down to something achievable, and the real COP converts heat demand straight into an electricity bill. Compute the lift first, be honest about the efficiency factor, and you will know what a heat pump will really cost to run — in mild weather and on the coldest night alike.
Top comments (0)