Coefficient of Performance (COP) Calculator
This calculator finds the coefficient of performance (COP) for a refrigerator or heat pump. Choose between two methods: enter heat removed and work input for the actual COP, or enter hot and cold reservoir temperatures for the ideal Carnot COP. Switch the device type to see how the same system performs as a refrigerator versus a heat pump, since their COPs differ by exactly 1.
Formula
Worked example
A refrigerator removes 12 kJ of heat and rejects 16 kJ: W = 16 - 12 = 4 kJ; COP_r = 12/4 = 3.00. The same cycle as a heat pump: COP_hp = 16/4 = 4.00 = 3.00 + 1. Carnot check at Tc = 5 C (278 K) and Th = 35 C (308 K): COP_r,max = 278/30 = 9.27, so the actual cycle runs at about 32% of Carnot efficiency.
What is the coefficient of performance?
The coefficient of performance (COP) measures how efficiently a heat transfer device converts work (usually electrical energy) into useful thermal output. Unlike the thermal efficiency of a heat engine, which is always less than 100%, COP can be greater than 1 - because a heat pump or refrigerator moves heat rather than converting energy. A heat pump with COP 4 delivers 4 kJ of heat energy to the building for every 1 kJ of electricity it consumes. The extra energy comes from the environment (outdoor air, ground, or water), not from the electricity itself. This is why heat pumps are considered highly energy-efficient heating devices compared to direct electric resistance heaters, which have a COP of exactly 1.
How to calculate COP from heat values
The simplest COP calculation uses the first law of thermodynamics. For a refrigerator or air conditioner, COP equals the heat removed from the cold space (Qc) divided by the net work input (W). For a heat pump delivering warmth, COP equals the heat rejected to the warm side (Qh) divided by the same work input. Because energy is conserved, Qh = Qc + W, which means the heating COP is always exactly 1 greater than the cooling COP for the same cycle. If you know Qc and Qh but not W separately, just subtract: W = Qh - Qc. For example, if a cycle removes 12 kJ of heat and rejects 16 kJ, then W = 4 kJ, the cooling COP is 12/4 = 3.0, and the heating COP is 16/4 = 4.0.
Carnot COP and why it matters
The Carnot COP is the theoretical upper limit set by the second law of thermodynamics. No real device can exceed it for given reservoir temperatures. For a refrigerator, the Carnot COP is Tc / (Th - Tc), and for a heat pump it is Th / (Th - Tc), where temperatures must be in Kelvin (add 273.15 to Celsius values). The key insight is that COP improves as the temperature difference between hot and cold reservoirs shrinks. A heat pump operating between an outdoor temperature of 5 C (278 K) and an indoor set-point of 20 C (293 K) has a Carnot COP of 293/15 = 19.5, but real systems reach only about 3-5 due to irreversibilities in the compressor, heat exchangers, and refrigerant flow. Comparing the actual COP to the Carnot limit gives the second-law efficiency, which tells engineers how much room remains for improvement.
Practical factors that affect COP
Several factors lower real COP below the Carnot ideal. Compressor isentropic efficiency (typically 70-85%) is the largest single loss. Heat exchanger size determines how close the refrigerant gets to the reservoir temperature - undersized exchangers force a larger temperature lift, which reduces COP. Refrigerant selection, expansion device losses, and piping pressure drops also contribute. Ambient temperature is especially important for air-source heat pumps: as outdoor temperature drops from 10 C to -5 C, COP can fall from around 4 to below 2, because the temperature lift increases and frosting of the outdoor coil adds defrost energy losses. Ground-source heat pumps avoid this variability because soil temperature stays relatively stable year-round at about 10-15 C in temperate climates.
Typical COP values for common systems
| System type | Typical COP range | Notes |
|---|---|---|
| Household refrigerator | 1.5 - 2.5 | Measured at 0 C interior, 25 C ambient |
| Commercial freezer (food prep) | 2.3 - 3.0 | Evaporator near 0 C |
| Frozen food / ice cream storage | 1.0 - 1.5 | Evaporator at -25 C or lower |
| Room air conditioner | 2.5 - 4.0 | SEER ratings map to steady-state COP |
| Central air conditioning | 3.0 - 5.0 | At AHRI standard conditions |
| Air-source heat pump (heating) | 2.0 - 4.5 | Drops significantly below 0 C outdoors |
| Ground-source heat pump (heating) | 3.0 - 6.0 | More stable; ground stays ~10-15 C |
| Industrial refrigeration | 1.5 - 4.0 | Varies widely with process temperatures |
Real-world COP ranges depend on operating conditions. These figures are representative steady-state values at standard rating conditions.
Frequently asked questions
What is a good COP for a heat pump?
For an air-source heat pump in heating mode, a COP between 3.0 and 4.5 is considered good at standard rating conditions (typically 7 C outdoor, 21 C indoor). Ground-source heat pumps regularly achieve 3.5-6.0. Seasonal or annual COPs (sometimes called HSPF in North America or SCOP in Europe) account for the full range of outdoor temperatures and are typically lower than the steady-state COP at a single test point.
Can COP be greater than 1?
Yes, and for heat pumps and refrigerators it usually is, often by a large margin. A COP greater than 1 does not violate energy conservation. The device uses work (electricity) to move heat from a cold reservoir to a hot reservoir, so the useful output includes both the electrical energy and the thermal energy extracted from the environment. A COP of 1 would correspond to direct electrical resistance heating, which converts all electricity to heat with no boost from the environment.
What is the difference between COP and EER or SEER?
EER (Energy Efficiency Ratio) and SEER (Seasonal Energy Efficiency Ratio) are North American rating metrics for air conditioners. EER is the cooling output in BTU/h divided by the electrical input in watts at a single test condition. SEER averages this over a notional cooling season. COP is the dimensionless ratio of the same quantities but expressed in consistent energy units (both in kJ or kWh). To convert: COP = EER / 3.412. A modern room air conditioner with EER 12 has a COP of about 3.5.
Why is the heat pump COP always 1 greater than the refrigerator COP?
Because both are describing the same thermodynamic cycle, just accounting for a different output. For the refrigerator, the useful output is the cold-side heat Qc. For the heat pump, it is the hot-side heat Qh = Qc + W. Dividing both by W gives COP_hp = Qc/W + W/W = COP_r + 1. This is a strict identity that holds for any cycle, reversible or not, as long as the device is the same.
How do reservoir temperatures affect the Carnot COP?
The Carnot COP rises as the temperature difference between the hot and cold reservoirs shrinks. For refrigeration (COP_r = Tc / (Th - Tc)), cutting the temperature lift in half roughly doubles the COP. This is why efficient refrigerators use the minimum practical condenser temperature and the maximum practical evaporator temperature. In practice, reducing the condensing temperature by 5 C or raising the evaporating temperature by 5 C can improve real system COP by 15-30%.
What is second-law efficiency for a refrigerator or heat pump?
Second-law efficiency (also called exergetic efficiency) is the actual COP divided by the Carnot COP for the same reservoir temperatures. It expresses how close a real device comes to the thermodynamic ideal. A system with actual COP 3.0 operating between reservoirs that give a Carnot COP of 9.0 has a second-law efficiency of 33%. Most commercial refrigeration equipment achieves 20-50% second-law efficiency, with advanced systems pushing toward 60-70%.