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The Gold Standard of Thermodynamics | Why an Average COP Can Reverse a Heat Pump Comparison

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Choose a route or read from the beginning. All teaching chapters stay open; the grouped index below is expandable.

Define performance

Chapters 1–2. Name the service, electrical boundary and point ratios.

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Follow the reversal

Chapters 3–5. Combine energy records and derive the correct weights.

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Make the comparison fair

Chapters 6–7. Match duty and distinguish a separate boundary change.

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Inspect the record

Chapters 8–10. Keep coverage, operating modes and comparison purpose visible.

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Practise and continue

Chapters 11–12. Check worked answers and follow the physics and metric guides.

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Full chapter index · Practice and worked checks · Thermodynamics subject root

Heat pump A has a higher coefficient of performance than heat pump B in each of two stated operating conditions. Yet B’s overall reported ratio is higher. That can happen without an arithmetic error when the two systems spend different shares of their work in those conditions.

This guide builds an original numerical example of that reversal. We separate useful heat from electrical input, combine periods correctly, compare a common duty and examine how auxiliaries or missing meter intervals change the meaning of the result. The purpose is to assess a performance claim before turning it into a judgement about equipment.

All systems, energies and condition labels below are fictional. They are not product measurements, predicted savings or certified seasonal ratings. The Thermodynamics subject root explains the underlying energy account; this article develops the narrower problem of combining and comparing ratios.

Chapter index

Define performance · Chapters 1–2

1. Define the useful service and the electrical boundary
2. Calculate two point ratios without averaging them yet

Follow the reversal · Chapters 3–5

3. Combine system A's energy records
4. Build the reversal with a different operating mix
5. Derive the correct weights

Make the comparison fair · Chapters 6–7

6. Compare a common duty before judging the point models
7. A boundary change can create another reversal

Inspect the record · Chapters 8–10

8. Match meter intervals and keep missing periods visible
9. Standby, defrost and auxiliary service need a declared treatment
10. Transfer the result to a performance report

Practise and continue · Chapters 11–12

11. Practice checks with worked answers
12. Questions readers ask, sources and next reading

CHAPTER 1 OF 12 · Define performance

1. Define the useful service and the electrical boundary

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For the heating examples here, COP = useful heat delivered/electrical energy supplied over the same interval and stated boundary. Both energies use kilowatt-hours, so the numerical ratio is dimensionless.

Cooling COP uses a different useful output: heat removed from the cooled region. A heating figure should not quietly become a cooling figure. Similarly, a compressor-only electrical denominator differs from one that also includes specified pumps, fans, controls, standby or auxiliary heaters.

COP above one does not create energy. In a simplified heat-pump energy account, heat delivered to the warm side includes energy extracted from the source as well as work supplied. OpenStax’s refrigerator and heat-pump chapter explains that relationship.

For our first comparison, assume the same heating-service and electrical boundaries in every row. That lets us isolate the effect of differing condition mixes. Later we will deliberately change the boundary and observe the separate problem it creates.

Write these definitions before ranking any ratios. A higher number attached to another service or a narrower denominator is not automatically better performance on the task the reader cares about.

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CHAPTER 2 OF 12 · Define performance

2. Calculate two point ratios without averaging them yet

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Suppose the exercise defines two operating bins, called mild and demanding. The labels simply distinguish supplied conditions. They do not specify actual outdoor temperatures, refrigerant states or a universal performance curve.

Under the fictional inputs, system A has COP six in the mild bin and two in the demanding bin. System B has COP five and 1.5 respectively. A therefore has the higher point COP in both bins under the stated common boundaries.

Operating binSystem A point COPSystem B point COP
Mild65
Demanding21.5
Fictional point COPs under common stated heating and electrical boundaries.

An arithmetic mean gives A four and B 3.25. Those numbers are averages of two ratio values. We have not yet supplied the energies, durations or duty mix needed to describe either system’s actual combined service.

A point COP answers a conditional question: how much useful heat per electrical input under the specified condition and boundary? It does not alone tell us how much work occurs in that condition.

The refrigeration cycles and COP guide develops cycle physics, temperature lift and part-load context. Here we retain the supplied point values and ask how their weights affect the comparison.

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CHAPTER 3 OF 12 · Follow the reversal

3. Combine system A’s energy records

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System A delivers twelve kilowatt-hours of useful heat in the mild bin and uses two kilowatt-hours of electricity. Its mild-bin COP is 12/2 = six.

In the demanding bin it delivers sixty kilowatt-hours and uses thirty kilowatt-hours of electricity. Its COP there is 60/30 = two. Across both bins it delivers 72 kilowatt-hours of useful heat using 32 kilowatt-hours of electricity.

The combined ratio is therefore 72/32 = 2.25. It is not the arithmetic mean of six and two, which was four.

System A recordUseful heatElectrical inputRatio
Mild bin12 kWh2 kWh6
Demanding bin60 kWh30 kWh2
Combined72 kWh32 kWh2.25
System A energy totals and combined ratio in the invented record.

The demanding condition accounts for most of this system’s electrical input. Its lower point ratio therefore has a large influence on the combined result.

The account is easy to check: add the energy columns first, then divide. A calculation that averages the two displayed COPs treats the rows as equally weighted observations, even though their input-energy weights are very different.

The combined result is still useful. It describes A’s supplied operating record. The error would be presenting it as a condition-free property that can be compared with any other installation’s ratio without examining its duty.

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CHAPTER 4 OF 12 · Follow the reversal

4. Build the reversal with a different operating mix

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System B delivers one hundred kilowatt-hours of useful heat in the mild bin using twenty kilowatt-hours of electricity, giving COP five. In the demanding bin it delivers three kilowatt-hours using two, giving COP 1.5.

Its combined heat is 103 kilowatt-hours and combined electricity is 22. The overall ratio is 103/22 ≈ 4.682, higher than A’s 2.25 despite B having the lower point COP in both bins.

System B recordUseful heatElectrical inputRatio
Mild bin100 kWh20 kWh5
Demanding bin3 kWh2 kWh1.5
Combined103 kWh22 kWhApproximately 4.682
System B energy totals under a different operating-condition mix.

Nothing in the table contradicts the point comparisons. B’s supplied record is dominated by the easier bin; A’s by the more demanding one. The aggregate is answering a different question from a matched condition-by-condition comparison.

This kind of aggregation reversal is often discussed under Simpson’s paradox. The important work is identifying the changed weights and comparison question, rather than treating the name as an explanation.

We cannot conclude from these totals alone that B’s equipment would use less electricity on A’s duty. Nor should we erase B’s actual aggregate because it seems inconvenient. Report what each quantity describes and compare like conditions for the judgement being attempted.

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CHAPTER 5 OF 12 · Follow the reversal

5. Derive the correct weights

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For each bin i, useful heat Qi = COPi × Ei, where Ei is electrical input. Adding the heat and dividing by total electricity gives:

COPcombined = Σ(COPi Ei)/ΣEi.

Thus the combined ratio is an electrical-input-weighted average of the bin ratios. For A, the weights are 2/32 and 30/32. The result is 6(2/32) + 2(30/32) = 2.25.

For B, the weights are 20/22 and 2/22. The result is 5(20/22) + 1.5(2/22) = 103/22. The input weights expose the reversal directly.

An unweighted mean would be appropriate for this particular aggregation only when the bins contribute equal electrical energies, apart from coincidental cases. Equal durations do not guarantee equal energies if input power differs.

There is a second useful form when heat duties are supplied: COPcombined = ΣQi/Σ(Qi/COPi). This is a heat-duty-weighted harmonic relationship, not an arithmetic heat-weighted mean of COPs.

Rather than memorising two different averaging rules, return to the energy columns. Useful heat adds to useful heat; electricity adds to electricity. Divide the relevant totals after aligning their boundaries and intervals.

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CHAPTER 6 OF 12 · Make the comparison fair

6. Compare a common duty before judging the point models

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To isolate the supplied point-performance difference, give each system the same hypothetical duty: sixty kilowatt-hours of useful heat in each bin. Assume the stipulated point COPs remain applicable throughout those duties and that capacity is sufficient.

A needs 60/6 = ten kilowatt-hours of electricity in the mild bin and 60/2 = thirty in the demanding bin. Total input is forty for 120 useful, giving combined COP three.

B needs 60/5 = twelve kilowatt-hours in the mild bin and 60/1.5 = forty in the demanding bin. Total input is 52 for the same 120 useful, giving combined COP 120/52 ≈ 2.308.

Common-duty modelMild-bin electricityDemanding-bin electricityTotal electricityCombined ratio
A: 60 kWh useful heat in each bin10 kWh30 kWh40 kWh3
B: 60 kWh useful heat in each bin12 kWh40 kWh52 kWhApproximately 2.308
A common hypothetical heating duty using the supplied point COP models.

The ranking now follows the higher point COPs because the service mix is matched. This does not overturn the earlier measured-record arithmetic; it answers a different counterfactual question under explicit assumptions.

Actual equipment comparison also needs output capacity, controls, changing conditions and a validated performance map. A model cannot deliver heat beyond a device’s capability merely because dividing by COP is convenient. Keep that physical requirement beside the energy calculation.

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CHAPTER 7 OF 12 · Make the comparison fair

7. A boundary change can create another reversal

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Use a separate pair of fictional units, C and D, to avoid mixing this issue with the operating-bin example. Each delivers sixty kilowatt-hours of useful heat during the comparison interval.

Unit C uses ten kilowatt-hours at the compressor and five in the other electrical loads included by the stipulated system boundary. Its compressor-only ratio is six; its whole-system ratio is 60/(10+5) = four.

Unit D uses twelve compressor kilowatt-hours and one for the other included loads. Its compressor-only ratio is five, but its whole-system ratio is 60/13 ≈ 4.615.

C leads on the narrower denominator, while D leads on the specified broader denominator. This is not the earlier condition-mix effect. The amount of electrical input included in the ratio has changed.

The useful output is assumed measured at the same defined service boundary in these rows. Do not add or subtract auxiliary energy from that output casually; the physical heat pathways determine what belongs in the numerator.

The geothermal heat pump engineering guide develops whole-installation questions, including loop and distribution requirements. It offers professional context for a denominator that a compressor data point cannot supply on its own.

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CHAPTER 8 OF 12 · Inspect the record

8. Match meter intervals and keep missing periods visible

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A ratio needs matched heat and electricity coverage. Suppose interval one has twelve kilowatt-hours of useful heat and three of electricity; interval two has six and three. Their combined ratio is 18/6 = three.

If heat data for interval two is missing, twelve divided by the full six electrical kilowatt-hours gives two. That number treats missing heat as zero. If electricity for interval two is missing but both heat values remain, eighteen divided by three gives six. That silently treats missing input as zero.

Neither result describes the complete two-interval performance. Retain the missing-data status and identify a valid common coverage period. On interval one alone, COP is four, but that narrower interval does not establish the missing interval’s performance.

IntervalUseful heatElectricityValid interval COP
112 kWh3 kWh4
26 kWh3 kWh2
Both complete intervals18 kWh6 kWh3
A matched two-interval energy record; missing readings must retain their status.

NREL’s field-validation dataset description illustrates recording operating modes alongside heat-pump measurements. We use it as a primary example of monitoring context, not as evidence for our fictional values or a particular product ranking.

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CHAPTER 9 OF 12 · Inspect the record

9. Standby, defrost and auxiliary service need a declared treatment

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A complete period may include time when no useful heating is delivered but electricity is still consumed. A heating-only compressor record and a full service-period energy account may therefore have different denominators and coverage.

In a fictional extension of A’s matched duty, include two additional electrical kilowatt-hours of stipulated standby consumption without additional useful heat. The combined ratio becomes 120/(40+2) ≈ 2.857 rather than three.

Defrost can involve changing operating modes and heat flows. Auxiliary heating can add useful heat and electrical input through another pathway. Their treatment needs the chosen measurement boundary and the relevant method; it cannot be inferred simply from a device being labelled a heat pump.

Formal seasonal ratings have defined procedures, climates and component treatments. An analyst’s ratio of site totals should not automatically be called a certified SCOP, HSPF or another regulated rating. Read the method and unit conventions.

The existing COP, EER, SEER and SCOP guide keeps these labels distinct. The practical lesson here is to preserve the periods and energy paths that the stated ratio actually includes.

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CHAPTER 10 OF 12 · Inspect the record

10. Transfer the result to a performance report

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Imagine a report that says “B is more efficient: its annual ratio is higher.” Ask which comparison the report intends. Is it describing delivered service in two actual installations, comparing equipment on a common duty, or predicting an intervention in one building?

Actual-installation totals can be valid descriptions while reflecting different weather, demand, schedules, setpoints and system boundaries. A common-duty calculation can help isolate a specified difference while relying on an adequate model of both systems.

If conditions overlap poorly, a simple adjustment may not be supported. Do not extend a performance map into an untested region merely to force a comparison. State where the supplied evidence ends.

The same reasoning applies beyond heat pumps. Comparing average travel speeds across different mixes of roads or service times across different mixes of cases can create a related weighting problem. The domain quantities and physical mechanisms change, but the need to preserve numerator, denominator and case mix remains.

A useful report gives the total energies, condition coverage, boundaries and comparison purpose before the conclusion. That lets another reader reproduce the ratio and judge whether it answers the intended question.

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CHAPTER 11 OF 12 · Practise and continue

11. Practice checks with worked answers

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Equal input weights Two intervals each use two electrical kilowatt-hours, with COPs three and five. Their useful heat totals six plus ten, so combined COP is 16/4 = four. Here the arithmetic mean works because the electrical inputs are equal.

Equal useful heat duties Each interval delivers twelve useful kilowatt-hours, with COPs three and six. Inputs are four and two, so combined COP is 24/6 = four. The arithmetic mean, 4.5, answers another question.

Equal time is not enough Two one-hour intervals use one and three electrical kilowatt-hours, with COPs six and two. Useful heat is six in each. Combined COP is 12/4 = three, rather than the unweighted mean four.

A missing interval If one meter record is absent, identify matching valid coverage. Do not insert zero simply to make the spreadsheet formula return a number. Any estimation method needs an explicit basis and appropriate uncertainty.

A ratio above the component values With positive electrical inputs, a combined ratio under the same boundary must lie between the smallest and largest component COPs. A result outside that range signals incompatible definitions, a calculation error or a more complex signed-flow account than the one assumed here.

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CHAPTER 12 OF 12 · Practise and continue

12. Questions readers ask, sources and next reading

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Is an average COP always wrong? No. The weighting and intended quantity matter. For a combined useful-heat-to-electricity ratio, add matched energy totals or use the corresponding electrical-input weights.

Does a higher annual ratio prove better equipment? It describes performance over its stated record. Equipment comparison also needs comparable duty, conditions, capacity and system boundaries.

Can this method calculate a certified seasonal rating? Only when the relevant prescribed method is followed. Our fictional aggregation is an educational energy account, not a certification calculation.

Sources and scope OpenStax supports the heat-pump energy and COP framework. The DOE field-validation dataset description supplies monitoring context. All systems, duties, reversals, missing-data examples and practice calculations in this article are original fictional constructions.

Follow the physics and the metric Return to The Gold Standard of Thermodynamics, continue into refrigeration-cycle mathematics, distinguish HVAC efficiency labels, or explore geothermal heat pump engineering for system-level context.

Connect the learning Learning That Transfers into Life explains why a changed comparison can reveal hidden assumptions. The How X Works hub connects the wider library.

When two performance claims seem to contradict each other, ask what work each ratio combines. The weights, boundaries and covered intervals may explain the difference before any device has changed.

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Sources checked 21 October 2026. Worked examples are educational models with the assumptions stated above.