Work can move heat from cold to hot. Performance greater than one is possible because most delivered heat is moved, not created.
By the end: calculate cooling and heating coefficients of performance without confusing them with heat-engine efficiency.
Reverse the direction and account for work input
A refrigerator takes heat Qc from a cold region and rejects Qh to a hotter region while receiving work Win. These are positive magnitudes. Over a complete cycle, the rejected heat equals the extracted heat plus work input. In the earlier work-output sign convention, the device’s net W would be negative. A refrigerator left open in a room cannot cool that whole isolated room: its work input ultimately adds energy to the room.
Cooling coefficient of performance, or COP, is heat extracted from the cold region divided by work input. Heating COP instead counts heat delivered to the hot region per work input. For the same complete cycle, heating COP exceeds cooling COP by one. Neither ratio is the fraction of input heat converted into work, so COP can exceed one without violating conservation.
Follow energy across the cooling boundary
Cooling a region means reducing its internal energy or removing energy that continually enters it. Neither task requires destroying energy. Choose the region being cooled as one system, and the cyclic cooling device as another. Heat leaving the region enters the device, then leaves the device at its hotter side. The device also receives work. Its working substance can change temperature and phase during a cycle, but returning to its starting state removes any net internal-energy accumulation from the cycle account.
Consider a proposed cycle that removes 600 J from a cold reservoir and receives 200 J of work. The device must reject 800 J to the hot reservoir. Using positive heat magnitudes makes that addition easy, but the course's signed first law gives the same result. Net heat into the device is 600 minus 800, or −200 J. Work done by the device is −200 J because work enters. Subtracting that negative work from net heat gives zero internal-energy change.
Inputs and output in one cooling cycle
Lengths encode energy, not power. Cooling COP is 3; heating COP is 4. Reservoir temperatures still need a separate feasibility check.
The two performance ratios answer different questions about this same account. Cooling COP is 600 divided by 200, or three. Heating COP is 800 divided by 200, or four. The extra unit in heating COP comes from the work input that eventually contributes to rejected heat. It does not imply a second supply of extracted environmental heat. A ratio above one is reasonable because its numerator includes energy moved across the cold boundary, rather than only energy supplied as work.
Now enclose both reservoirs and the device within a larger boundary, leaving only the work source outside. The 600 J and 800 J heat transfers become internal exchanges. The larger system gains 200 J, exactly the entering work. This resolves the open-refrigerator question without guessing how cold air circulates. If the cold region and hot region are parts of the same otherwise isolated room, moving energy between them cannot reduce the room's total energy. Work input increases it after the device completes its cycle.
Move the boundary around both heat transfers
The dashed outer boundary receives only work. Heat exchanged among its parts cancels in the combined energy account. Geometry and arrow lengths are schematic.
A finite cold object differs from a reservoir. Suppose an object has whole-sample heat capacity 300 J/K, does negligible work, and remains within one phase. Removing 600 J lowers its temperature by two kelvin. Removing the same energy from an object with twice that capacity lowers temperature by only one kelvin. The capacity is for the whole object, not per kilogram or per mole. COP alone does not determine either temperature change because it says nothing about the cooled object's capacity.
Before checking temperatures, practice moving the boundary on the 600 J example. For the cold reservoir, heat input is −600 J. For the hot reservoir, heat input is +800 J. For the device, net heat input is −200 J and work output is −200 J. For everything except the work source, the energy increase is +200 J. Each internal transfer appears once with each sign. Next check whether the proposed transfers satisfy the entropy condition.
Use the reversible temperature bound
To avoid decreasing the entropy of the two reservoirs, sufficient heat must be rejected at the hot temperature. At the reversible limit, cooling COP is cold temperature divided by the difference between hot and cold temperatures. This temperature difference is the lift. Keep all temperatures in kelvin, even though a difference of one kelvin equals a difference of one Celsius degree.
With a hot reservoir at 300 K and a cold one at 270 K, the ideal cooling COP is nine. For each 100 J of work, the device extracts 900 J and rejects 1,000 J. A smaller 10 K lift gives a higher ideal COP of 29; a 60 K lift gives four. The graph compares these bounds at a fixed 300 K hot reservoir and fixed work input.
Derive the work requirement from reservoir entropy
The cold reservoir loses entropy when energy leaves it. If its fixed temperature is Tc, removing Qc changes its entropy by −Qc/Tc. The hot reservoir gains Qh/Th. The device returns to its initial entropy after a full cycle. With no other changes, adding these reservoir contributions must give a nonnegative result. The temperatures belong to the reservoirs where the heat enters or leaves. They must be absolute temperatures because entropy per joule depends on inverse kelvin temperature, not inverse Celsius reading.
Apply this test to the 600 J extraction and 200 J work example with reservoirs at 250 K and 300 K. The cold reservoir changes by −600/250 = −2.4 J/K. The hot reservoir gains 800/300, about 2.667 J/K. Their sum is about +0.267 J/K, so this account passes the entropy test. Energy conservation alone would also allow 600 J extraction with only 50 J work and 650 J rejection. That alternative gives about −0.233 J/K overall and fails.
Compare three proposals at 250 K and 300 K
Work input
Hot rejection
Cold ΔS
Hot ΔS
Total ΔS
50 J
650 J
−2.400 J/K
+2.167 J/K
−0.233 J/K: forbidden
120 J
720 J
−2.400 J/K
+2.400 J/K
0 J/K: reversible limit
200 J
800 J
−2.400 J/K
+2.667 J/K
+0.267 J/K: permitted account
All rows extract 600 J. Reservoir entropy changes use signed heat into each reservoir; the cyclic device has zero net entropy change. Values are rounded.
To find the smallest allowed work, replace the hot heat Qh by Qc plus Win in the entropy inequality. Multiplying by the positive hot temperature gives Qc plus Win at least Qc times Th/Tc. Subtract Qc from each side. The result is Win at least Qc times the temperature lift divided by Tc. This is the reversible minimum. Dividing Qc by that minimum work gives the maximum cooling COP, Tc divided by Th minus Tc.
For 600 J extraction between 250 K and 300 K, the lift is 50 K. Minimum work is 600 times 50/250, or 120 J. At that limit the hot heat is 720 J. Its entropy gain is 720/300 = 2.4 J/K, exactly matching the cold loss. The earlier 200 J input exceeds the minimum by 80 J. At fixed extraction and reservoir temperatures, that excess adds 80/300 J/K to the reservoir entropy total. The result matches the positive total calculated directly.
Heating performance has its own useful form. At the reversible limit, heating COP is Th divided by the lift. Between 250 K and 300 K it is six, while cooling COP is five. Suppose instead that the required task is delivering 600 J to the hot reservoir. The minimum work is then 600 divided by six, or 100 J, and extraction is 500 J. Fixing delivered heat is a different comparison from fixing extracted heat, even when both comparisons use the same reservoirs.
Choose which useful heat is fixed
Gold marks work; blue marks heat. The first task fixes extraction at 600 J, while the second fixes delivery at 600 J. Both are reversible limiting accounts, not device predictions.
Use these relations to test an unfamiliar account on paper. Keep reservoirs at 250 K and 300 K, require 1,000 J extraction, and propose 150 J work. Minimum work is 1,000 times 50/250 = 200 J, so the proposal fails before any device mechanism is considered. Its rejection would be 1,150 J, giving 1,150/300 minus 1,000/250, about −0.167 J/K. A consistent energy sum has therefore supplied a counterexample to the claim that energy conservation alone guarantees feasible cooling.
Work through the mechanism
Cooling coefficient of performance
Win is a positive work-input magnitude. Qc is heat removed from the cold reservoir. The bound applies to a complete cycle between two fixed-temperature reservoirs.
Heating and cooling count different outputs
Qh = Qc + Win, so dividing by Win gives this identity for the same cycle.
Compare temperature lifts at a fixed hot side
Cold reservoir 270 K. Cooling COP 9; heating COP 10. Each 100 J of work extracts 900 J and rejects 1,000 J.
Case
Work input
Cold heat extracted
Hot heat delivered
30 K reference
100 J
900 J
1,000 J
Selected
100 J
900 J
1,000 J
The curve is a reversible cooling-COP bound with hot temperature fixed at 300 K. The outlined ring marks the 30 K reference; the filled point marks the selection. Work input is 100 J per cycle in both cases. Controls replace only the lift, without a time or real-appliance model.
Temperature lift, hot side 300 K
Ideal cooling COP
Qc for 100 J work
Qh
10 K
29
2,900 J
3,000 J
30 K
9
900 J
1,000 J
60 K
4
400 J
500 J
These are reversible bounds, not measured appliance ratings.
Explore lift without changing the work budget
The comparison fixes the hot reservoir at 300 K and work at 100 J. Changing lift changes the cold temperature and the COP denominator. Begin with the reference lift of 30 K. The cold reservoir is 270 K, cooling COP is nine, and the extracted heat is 900 J. Rejection is 1,000 J. These are reversible accounts calculated from temperatures, not observations of a working refrigerator.
Before selecting the small-lift preset, predict whether extraction will double, triple, or change by another factor. At 10 K lift the cold reservoir is 290 K, so COP is 290/10 = 29. Extracted heat becomes 2,900 J, which is 29/9 times the reference, about 3.22 times as much. The factor is not exactly three because the numerator also changes. Rejection becomes 3,000 J. The hot-side entropy gain is ten J/K, exactly balancing the cold-side loss of 2,900/290 = ten J/K.
Select the large-lift preset next. A 60 K lift puts the cold reservoir at 240 K. COP is four, extraction is 400 J, and rejection is 500 J. Doubling the reference lift has reduced extraction to four ninths, not one half. The hot temperature stayed fixed, so the colder numerator decreased from 270 K to 240 K. At an intermediate slider setting of 20 K, predict COP fourteen and extraction 1,400 J, then compare the displayed result with that arithmetic.
A second comparison separates task size from work budget. Keep extraction fixed at 900 J on paper instead of holding work fixed. At the reference lift, minimum work is 100 J. At 10 K it is 900/29, about 31.03 J; at 60 K it is 900/4 = 225 J. These reversible COP values answer a different question, not an additional interface control. The work bars show why a COP increase means less minimum work for a fixed cooling task.
Hold extraction fixed instead of work
These are paper comparisons for the same 900 J task. In the existing interactive model, work is fixed instead and extraction changes. Bar lengths start at zero.
The reservoir temperatures are not necessarily the temperatures of a device's working fluid during heat exchange. To absorb heat at a finite rate from a 270 K region, an ordinary heat exchanger needs a colder receiving surface. To reject heat into a 300 K region, it needs a hotter donating surface. Illustrative surface temperatures of 265 K and 305 K would create a 40 K internal lift. A reversible cycle evaluated between those surfaces has cooling COP 265/40 = 6.625, below the reservoir-only bound of nine.
Finite heat-transfer differences increase the working lift
Vertical position encodes temperature on a linear scale. Horizontal positions are schematic locations, not distance or time. Heat enters the colder surface and leaves the hotter surface; the work-driven connection is not a passive temperature profile.
The surface comparison explains a mechanism for lower performance without supplying a performance prediction. Finite temperature differences generate entropy in heat transfer, and friction or other losses can add more. The interactive graph has no surface-temperature or heat-transfer-area controls, and its curve does not include these losses. After each preset, state the fixed work, cold temperature, extraction, and rejection aloud or in writing. Then identify which missing information would be needed to infer a cooling rate. No cycle time appears in any of these energy accounts.
Do not treat the bound as a product specification
Real equipment requires temperature differences between its working fluid and the surrounding air or water to transfer heat at useful rates. Fans, pumps, defrost cycles, compressor losses, and changing outdoor temperatures also affect measured performance. A seasonal heating COP is not the same quantity as one ideal cycle evaluated at two nominal air temperatures.
As the temperature lift approaches zero, the reversible COP formula grows without bound. This is not a promise of infinite cooling power. Reversible transfer approaches vanishing driving differences, and the ideal ratio contains no heat-exchanger area or time model. Exactly zero lift is excluded from this formula. Compare devices with the same useful task, energy inputs, operating temperatures, and averaging period.
Separate a large ratio from a large cooling rate
The reversible formula becomes large near zero lift because very little work is needed to move energy between almost equal temperatures. For a fixed 300 K hot reservoir and 900 J extraction, a 10 K lift needs about 31.03 J of work. A 1 K lift would need 900/299, about 3.01 J. These paper examples extend below the slider range. Their work requirements do not specify energy transfer per second.
Small lift reduces minimum work for a fixed task
This additional comparison extends below the interface range. The open endpoint is a limit, not a finite COP or a prediction of cooling rate.
At exactly equal temperatures, the distinction between a cold and hot reservoir disappears. The formula divides by zero and cannot assign a finite COP. Reversible heat exchange also requires temperature differences to approach zero, while ordinary heat-transfer rates fall as their driving differences decrease for fixed conductance. Consequently, the limiting ratio cannot establish infinite useful cooling power. One statement compares cycle energies. The other concerns energy transferred per unit time through a specified transport mechanism.
For example, two hypothetical accounts can each extract 900 J using 100 J of work, so both have cooling COP nine. If one account describes a cycle every second, its cooling rate is 900 W and its work-input rate is 100 W. If the other describes a cycle every ten seconds, those rates are 90 W and ten W. Equal ratios hide a tenfold difference in rates. These durations are stipulated, not predicted by the COP model.
The same COP can accompany different rates
Account
Extraction per cycle
Work per cycle
Cycle duration
Cooling rate
Work rate
A: COP 9
900 J
100 J
1 s
900 W
100 W
B: COP 9
900 J
100 J
10 s
90 W
10 W
Cycle times are stipulated for comparison. Energy per cycle divided by seconds per cycle gives watts; neither time follows from the COP formula.
Likewise, the time needed to cool a finite object requires more than an initial COP. A constant-capacity object of 300 J/K must lose 3,000 J to cool by ten kelvin without phase change or work. A stipulated constant extraction rate of 100 W would remove that energy in 30 seconds. But maintaining that rate as the object's temperature falls is an additional assumption. With a fixed hot side, the lift usually increases during cooling, so a fixed initial reversible COP is not a complete account of the interval.
Compare boundaries before comparing reported heating ratios. A ratio that counts only work entering the cyclic working substance differs from one that includes every auxiliary work input. If an abstract account delivers 1,000 J while receiving 100 J at the cyclic device and another 25 J across the larger chosen boundary, its first ratio is ten. If the same delivered heat is the useful output of the larger boundary, its ratio is eight. Neither number alone identifies the complete physical account.
Finish by reversing the usual question. Suppose a proposed two-reservoir cycle passes both balances but its rate is unknown. What has been established? The reported energies and temperatures are compatible with conservation and nondecreasing total entropy under the stated cyclic assumptions. Nothing in that result establishes a working mechanism, a transition time, or measured performance. Conversely, a negative total entropy change is enough to reject that complete account. A warmer store cooling down or another energy source could change the conclusion only by changing the stated boundary and process.
Exercises
Reconcile a COP above one
A heat pump delivers 1,200 J to a room using 300 J of work per cycle. Find the extracted heat, heating COP, and cooling COP.
Hint
Subtract work from delivered heat before calculating cooling COP.
Worked answer
Quantity
Value
Heat taken from cold side
900 J
Work input
300 J
Heat delivered
1,200 J
Heating / cooling COP
4 / 3
Check temperatures separately before accepting the performance claim.
It extracts 900 J from the cold side. Heating COP is 1,200/300 = 4; cooling COP is 900/300 = 3. The delivered heat includes both transferred environmental heat and the work input. This energy account alone does not prove feasibility at unspecified temperatures.
Test a refrigerator specification
A refrigerator operates between 250 K and 300 K. It claims to extract 800 J for 100 J of work each cycle, with no other transfers. Is that possible?
Hint
Compare its claimed cooling COP with the reversible bound.
Worked answer
Check
Result
Claimed cooling COP
8
Maximum cooling COP
5
Total reservoir entropy change
−0.2 J/K
A plausible energy sum can still violate the entropy condition.
The claimed COP is eight. The bound is 250 divided by the 50 K lift, or five. The claim is impossible for the stated two-reservoir cycle. Its 900 J rejection would give a reservoir entropy change of 900/300 minus 800/250 = −0.2 J/K.
Sources and further reading
These optional references support the stated laws and model assumptions. The numerical examples and diagrams are authored teaching models, not experimental measurements or engineering specifications. Selected relevant passages were reviewed, not entire textbooks or a comprehensive research literature.
Selected energy-flow definitions and cooling versus heating coefficients of performance. The refrigerator paragraph says work done by the working substance, but its positive denominator Qh minus Qc is work input. Our equations explicitly use positive Win.