Quick answer The Carnot COP sets the theoretical ceiling: COP_H = T_H/(T_H − T_L) for heating and COP_cool = T_L/(T_H − T_L) for cooling (absolute temperatures in K). Real machines achieve COP_real = η·COP_Carnot, so a smaller temperature lift means a higher COP.
Adjust hot-side and cold-side temperatures plus the Carnot efficiency factor to instantly compute COP, heat transfer rates, and annual energy costs for heat pumps, refrigerators, and air conditioners.
Mode
Presets
Temperatures
Hot side TH
°C
Cold side TL
°C
Equipment
Carnot efficiency factor η
Power input W
kW
Results
Actual COP—
Carnot COP (theoretical max)—
Electricity assumptions
Energy=W×hours; cost=energy×price. With W fixed, changing COP or η alone does not change this bill. The defaults 8760 h and 27 JPY/kWh are assumptions, not a current tariff.
While paused, moving a slider updates the results immediately.
Results
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Actual COP
—
Carnot COP
—
QH (kW)
—
QL (kW)
—
ΔT (K)
—
Annual cost (×10,000 JPY)
Cycle
COP comparison
Curves vary the cold-side temperature at fixed hot-side temperature and η. The dot marks the current inputs. Moving particles are schematic, not a refrigerant mass-flow or phase calculation.
η is the selected mode COP divided by its Carnot COP. Separate heating/cooling calculations at the same η need not differ by 1. Within a single cycle the heat balance is always Q_H−Q_L=W.
What is Coefficient of Performance (COP)?
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What exactly is COP? I know it measures efficiency, but how is it different from just saying "it's 90% efficient"?
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Great question! Basically, COP is a multiplier, not a percentage. For instance, an electric space heater has a COP of 1—it turns 1 kWh of electricity directly into 1 kWh of heat. A heat pump with a COP of 4 delivers 4 kWh of heat for that same 1 kWh of electricity by moving existing heat from outside. Try setting the hot-side temperature to 20°C and cold-side to 0°C in the simulator above. You'll see the theoretical Carnot COP for heating is quite high!
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Wait, really? So a COP can be greater than 1? That seems like getting more energy out than you put in. Is that possible?
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It does seem like magic, but it's not creating energy. It's moving it. The electricity powers the compressor to pump thermal energy from a cold source (like outside air) to a warm sink (your house). The key is that most of the delivered heat is "free" energy harvested from the environment. In practice, real machines aren't perfect. That's what the "Carnot efficiency factor (η)" slider represents—it bridges ideal theory and real-world losses. Slide it from 0.70 down to 0.30 to see how the real COP drops.
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Okay, that makes sense. But the simulator shows both a heating COP and a cooling COP for the same machine. How can one device have two different COPs?
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For the same cycle at the same operating point, Q_H=Q_L+W implies COP_H−COP_L=1. Summer and winter operation do not generally have the same conditions. This tool separately applies η to each selected mode, so switching modes at unchanged η need not give a difference of 1.
Physical Model & Key Equations
The maximum theoretically possible efficiency for a heat pump or refrigerator is given by the Carnot cycle, which depends only on the absolute temperatures (in Kelvin) of the hot and cold reservoirs.
Where $T_H$ is the hot-side absolute temperature (e.g., indoor coil), $T_L$ is the cold-side absolute temperature (e.g., outdoor air), and $\text{COP}_\text{H, Carnot}$ is the ideal Coefficient of Performance for heating.
For the cooling (refrigeration) mode of the same cycle, the useful output is the heat extracted from the cold space. The ideal Carnot COP is:
Real devices have losses (friction, heat leaks, compressor inefficiency). A Carnot efficiency factor ($\eta$, between 0 and 1) is used to estimate real-world performance from the ideal limit.
The heat transfer ($Q$) and electrical work ($W$) are then related by: $Q = \text{COP}_\text{real} \times W$.
Frequently Asked Questions
This tool automatically converts Celsius to absolute temperature (Kelvin) internally for calculations. However, since the COP becomes extremely large when the temperature difference is close to zero, it is recommended to input values within a realistic range (e.g., high temperature side 40°C, low temperature side 0°C).
η is a model input: estimated COP divided by the Carnot COP for the selected mode. Its input range is 0.30–0.70, not a rating scale for machine age or compressor type. Calibrate it only as a simplified comparison when measured performance at matching reservoir temperatures and operating conditions is available.
Annual cost is electrical input W×equivalent full-load hours×entered JPY/kWh. It does not infer a thermal load or required W from COP. At fixed W, changing η or temperatures alone does not change the bill. Defaults of 8760h and 27 JPY/kWh are assumptions, not a current tariff.
For heating, heat is supplied to the high-temperature side (indoor), so COP = T_H/(T_H−T_L). For cooling, heat is removed from the low-temperature side (indoor), so COP = T_L/(T_H−T_L). This tool automatically applies the appropriate formula by selecting the operation mode.
Real-World Applications
Residential Heating & Air Conditioning: Modern air-source heat pumps are the primary application. Engineers use COP calculations to select equipment and predict seasonal performance (SPF/APF). For instance, a unit might have a rated COP of 4 at 7°C outdoor temperature, but this can drop to near 2 at -10°C, which is simulated by changing the cold-side T and η factor.
Commercial Refrigeration: Supermarket freezer aisles and walk-in coolers are essentially large refrigerators. The cooling COP dictates operational costs. Engineers optimize the system by balancing the temperature lift ($T_H - T_L$) against compressor power, exactly as shown in the calculator when you adjust the two temperature sliders.
Industrial Heat Recovery: Heat pumps can upgrade waste heat from industrial processes to a usable temperature. For example, recovering low-grade heat from wastewater at 10°C and boosting it to 70°C for space heating. The viability of such a project hinges on achieving a high enough real COP to be cost-effective, which is sensitive to the Carnot efficiency factor η.
Electric Vehicle Cabin Conditioning: In cold climates, using a heat pump to warm an EV cabin is far more efficient than a resistive heater, directly extending driving range. Automotive engineers simulate performance across a range of ambient temperatures (cold-side T) to ensure adequate cabin heating and defrosting power while maximizing the system COP to preserve battery charge.
Common Misconceptions and Points to Note
Temperature units: Use absolute temperature in the numerator: 20°C=293.15K and 5°C=278.15K. The difference is 15K, but replacing the numerator 293.15 with 20 gives the wrong COP. The calculator converts using 273.15.
Choosing η: This input assumes a selected-mode ratio to Carnot COP. It is not an age or quality scale: 0.3 does not identify an old machine and 0.7 does not certify a new one. A measured COP of 5 divided by Carnot COP 10 gives 0.5 only at matching reservoir temperatures and operating conditions. No constant value is guaranteed when those conditions change.
Fixed input versus fixed thermal demand: This bill holds entered electrical power W fixed, so changing COP alone cannot change it. To compare equipment meeting the same thermal demand, separately calculate W=thermal demand/COP and match operating hours and price. Building loads and seasonal variations are not computed.
Choose heating, air conditioning or refrigeration. Cooling and refrigeration use the same useful-output definition: heat removed from the cold side.
Set hot side 20–80°C, cold side −30–20°C, η=0.30–0.70 and electrical input 0.5–20kW. Keep the hot side hotter than the cold side.
Compare both heat flows: Q_H−Q_L=W. A heating setting without positive cold-side heat uptake is marked outside the heat-pumping model.
Open Electricity assumptions to enter equivalent full-load hours and JPY/kWh. Shared URLs retain the mode and these assumptions.
Reproducible worked example
Choose Refrigeration with hot side 40°C, cold side −20°C, η=0.70 and electrical input 20kW. The cold side is 253.15K and the lift is 60K: Carnot COP=253.15/60≈4.22, estimated COP≈2.95, cold-side heat Q_L≈59.07kW and rejected heat Q_H≈79.07kW. Their difference is the 20kW electrical input.
At 2000 equivalent full-load hours and 30 JPY/kWh, cost=20×2000×30=1,200,000 JPY; the card displays 120.0 in units of 10,000 JPY. The original defaults 8760h and 27 JPY/kWh give 473.0. Increasing η at fixed W increases heat output, not this bill. A fixed thermal-demand comparison requires a separate electrical-input calculation W=thermal demand/COP.
Assumptions and limits
Temperatures describe simplified constant hot/cold reservoirs. Refrigerant properties, compressor maps, exchanger approach temperatures, pipes, defrost and seasonal variations are not separately calculated.
η is an assumed selected-mode ratio to Carnot COP, not a compressor isentropic efficiency or a measured equipment rating. No machine type or age is certified by a chosen value.
For a single cycle, Q_H/W−Q_L/W=1. Independently applying the same η to heating and cooling is not a comparison of one unchanged physical cycle.
Cost is electrical input×equivalent full-load hours×entered tariff. It excludes seasonal efficiency, building loads and fixed contract charges; it is not an equipment-sizing or billing guarantee.
