The coefficient of performance calculator above measures how much heat a machine moves for each unit of work it consumes. That ratio is the defining figure for heat pumps, refrigerators and air conditioners, and it is routinely greater than one — a well-matched heat pump might deliver three or four kilowatts of heat for every kilowatt of electricity. Nothing is being created. The machine is moving heat that already exists from one place to another, and paying only for the transport.
Arb Digital builds free calculators that put a number in context rather than leaving it bare. A COP of 3.6 means little on its own; against the theoretical ceiling for the temperatures it is working across, it becomes a judgement about the machine. This page reports both, plus the energy efficiency ratio used in equipment labelling and an optional running-cost figure. For the engine that produces work from heat rather than the reverse, see the Carnot efficiency calculator.
What This Coefficient of Performance Calculator Does
It divides the useful heat moved by the work consumed, in whichever direction the machine is being used. In heating mode the useful output is the heat delivered to the warm side; in cooling mode it is the heat removed from the cold side. Because the compressor's work also ends up in the warm side, the heating COP of a given machine is always exactly one greater than its cooling COP for the same conditions — a fact that catches people out when comparing two ratings that were measured in different modes.
Alongside the raw ratio it computes the Carnot coefficient of performance for the two temperatures you enter, which is the highest value any machine could achieve between them, and the second-law efficiency: the fraction of that ceiling your machine actually reaches. Real equipment typically lands somewhere between a quarter and a half. That figure, not the COP itself, is what tells you whether a machine is well engineered or merely working in easy conditions.
The energy efficiency ratio is the same information in the imperial convention: BTU per hour of cooling per watt of electrical input. It is numerically about 3.412 times the COP, and it appears on equipment labels in markets that use those units. The tool reports it so a specification sheet can be compared directly.
How to Use It
- Pick heating or cooling. This decides which side of the machine counts as useful output, and changes the Carnot expression used for the ceiling.
- Enter the heat moved and the work consumed. Both on the same basis — either as rates in kilowatts or as totals over the same period.
- Enter the two working temperatures. These are the actual source and sink temperatures the machine sees, not the room thermostat setting.
- Add hours and an electricity price if you want cost. The annual figure uses only the electrical input, since that is what you pay for.
- Read the second-law efficiency. It is the honest comparison, because it removes the advantage a machine gets from working across a narrow temperature gap.
The Formula: How COP Is Calculated
For a refrigerator the coefficient of performance is the heat removed from the cold reservoir divided by the work done, and for a heat pump it is the heat delivered to the hot reservoir divided by the work done. OpenStax University Physics Volume 2, section 4.3 on refrigerators and heat pumps, gives these as KR = Qc/W = Qc/(Qh − Qc) and KP = Qh/W = Qh/(Qh − Qc). Energy conservation supplies the link W = Qh − Qc, which is why the two coefficients differ by one.
Work the defaults. A heat pump delivering 9 kW of heat while drawing 2.5 kW of electricity has a heating COP of 9 ÷ 2.5 = 3.60. The remaining 6.5 kW came from the outdoor air, free of charge. In EER terms that is 3.60 × 3.412 = 12.28 BTU per hour per watt.
The ceiling comes from the reversible case. Running the Carnot cycle backwards gives a heating coefficient of Th/(Th − Tc) and a cooling coefficient of Tc/(Th − Tc), both with absolute temperatures. With a source at 7 °C (280.15 K) and a flow temperature of 35 °C (308.15 K), the gap is 28 K and the heating ceiling is 308.15 ÷ 28 = 11.01. A measured 3.60 is therefore 32.7 per cent of what thermodynamics permits.
Why the Temperature Gap Governs Everything
Look at the Carnot expression again: the temperature difference sits in the denominator. Narrow the gap and the ceiling rises steeply; widen it and the ceiling collapses. This single fact explains almost every observed behaviour of real heat pumps and refrigerators, and it is why manufacturer COP figures always come attached to a stated pair of temperatures.
An air-source heat pump delivering underfloor heating at 35 °C from 7 °C outdoor air works across 28 kelvin. The same machine feeding radiators at 55 °C on a night when the outside air is at −5 °C works across 60 kelvin, and its ceiling falls from about 11 to about 5.5. Its actual COP falls in proportion. That is not a fault; it is the physics, and it is the reason low-temperature emitters matter so much more for heat pumps than for boilers.
The same logic runs a domestic freezer. Keeping a cabinet at −18 °C in a 21 °C kitchen is a 39-kelvin gap against a cold reservoir of only 255 K, so the cooling ceiling is about 6.5 and a real appliance might manage 1.5 to 2. Push the kitchen to 30 °C in summer and both the gap and the running cost rise together.
COP, EER, SEER and Why Ratings Disagree
COP is an instantaneous ratio at one stated condition. EER is the same ratio in BTU per hour per watt, again at one condition. Seasonal figures such as SEER and SCOP are different animals: they are weighted averages across a modelled range of conditions over a whole season, including part-load operation and the cycling losses that come with it.
This is why a seasonal figure and a rated COP for the same unit will not match, and why comparing a SEER against an EER is comparing two different measurements. A machine that modulates smoothly at part load can score well seasonally even if its peak-condition rating is unremarkable, because most of its operating hours are nowhere near peak. When you are reading a specification, check which figure it is and which conditions it was measured at before drawing any conclusion.
Defrost cycles, auxiliary resistance heating and standby power all reduce delivered performance without appearing in a headline COP. Any comparison of running cost should be based on seasonal figures or on measured energy over a real period, which is what the electricity bill calculator handles once you have the consumption.
Why COP Above One Is Not a Free Lunch
An electric resistance heater converts electricity to heat with an efficiency of essentially one: a kilowatt in, a kilowatt out. A heat pump appears to beat that by a factor of three or four, which sounds like a violation of energy conservation until you notice that the extra heat was never created. It was pumped from outdoor air, from the ground, or from the inside of a fridge cabinet, and the electricity paid only for moving it.
The bar breakdown makes the split visible: with a COP of 3.6, seventy-two per cent of the delivered heat came from the source and twenty-eight per cent from the compressor's work. What the second law forbids is not a ratio above one; it is moving heat from cold to hot with no work at all, and the Carnot ceiling states exactly how little work you could get away with. Our Carnot efficiency calculator works the forward version of the same limit for engines.
Turning COP Into Sizing and Running Figures
The COP tells you how much electricity a given heat demand will require, so it becomes useful the moment you know the demand. Estimate a building's heat loss with the heat loss calculator, or a room's cooling load with the air conditioner BTU calculator, then divide by the COP to get the electrical input the machine will draw.
Improving the building envelope changes both sides of that sum. Better insulation lowers the demand and often lets the system run at a lower flow temperature, which raises the COP as well — a compounding gain that is easy to miss when only the demand is considered. If you need to move between kilowatts, BTU per hour and horsepower, the power converter and the energy converter handle the rescaling, and the temperature converter covers the inputs.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Comparing a heating COP with a cooling COP — for the same machine they differ by exactly one, because the compressor's work joins the heat on the warm side.
- Quoting a COP without its temperatures — the figure is meaningless without the conditions, since the ceiling itself moves with the temperature gap.
- Treating SEER or SCOP as interchangeable with COP — seasonal figures are weighted averages across many conditions, not a measurement at one point.
- Using the thermostat setting as the hot-side temperature — what matters is the refrigerant-side condition, typically the flow temperature, which is higher than room temperature.
- Assuming COP above one breaks physics — the machine relocates existing heat rather than generating it, and the work only pays for the transport.
Related Free Tools From Arb Digital
For the theoretical ceiling and the engine case use the Carnot efficiency calculator. Size the demand with the heat loss calculator, the air conditioner BTU calculator or the insulation calculator, and cost the result with the electricity bill calculator. Rescale units with the power converter, the energy converter and the temperature converter. Everything else is listed on the free online tools hub.
Frequently Asked Questions
That the machine moves 3.6 units of heat for every unit of work it consumes. The extra heat is not created; it is drawn from the source side, and the work pays only for moving it across the temperature gap.
Because it is a ratio of heat moved to work consumed, not of energy out to energy in. A heat pump relocates heat that already exists, so the useful output legitimately exceeds the electrical input without any energy being created.
By exactly one. The heat delivered to the warm side equals the heat taken from the cold side plus the work put in, so dividing both by the work gives figures that differ by a single unit.
They express the same ratio in different units. COP is dimensionless, while EER is BTU per hour of cooling per watt of input, so EER is roughly 3.412 times the COP for the same operating point.
Because the temperature gap it works across widens. The theoretical ceiling is the hot temperature divided by the difference between hot and cold, so colder outdoor air lowers the achievable coefficient before any equipment issue is involved.
No. That limit assumes a fully reversible cycle, which would transfer heat infinitely slowly and deliver no useful rate. Real equipment typically achieves somewhere between a quarter and a half of the ceiling.
Seasonal figures such as SCOP and SEER better reflect a full year of varying conditions and part-load running. A single-point COP is useful for understanding the physics and for comparing two machines measured at identical conditions.
This tool is provided for educational and estimating use. It performs a thermodynamic ratio calculation and does not model defrost cycles, part-load behaviour, auxiliary heating or equipment ratings, and its cost figures are illustrative arithmetic rather than financial or energy advice.
