Heat pumps and refrigerators
Model refrigerators and heat pumps as reversed heat engines and calculate COP in cooling and heating modes.
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ContinueA refrigerator, air conditioner and heating heat pump share an energy-transfer idea: work enables heat to move from a colder region to a warmer one. The useful benefit depends on which region you want to cool or heat. A large heat output per joule of work is not energy creation, but it still has a temperature-dependent upper bound.
Build on cyclic energy accounting and the Carnot engine. Use positive magnitudes Qc extracted from the cold side, Qₕ delivered to the hot side and Wᵢₙ supplied as work. Over a complete cycle with these as the only transfers, Qₕ = Qc + Wᵢₙ.
Match the numerator to the purpose
Cooling COP is Kc = Qc/Wᵢₙ; heating COP is Kₕ = Qₕ/Wᵢₙ. For the same complete energy account, Kₕ = Kc + 1. An engine efficiency asks how much work comes from absorbed heat, while COP asks how much heat is moved or delivered per work input. There is no common requirement that both ratios be below one.
Draw the boundary around the supplied device and state what is included in work input. Fan power, defrost, leaks, startup storage and useful heating losses can change an installation’s measured COP. Do not apply Kₕ-Kc = 1 to two reported ratings from different test conditions or to quantities that omit different channels. Rates can replace per-cycle energies in a consistent steady account.
Short worked examples
Refrigerator performance
Worked example 1
Cooling and room heating
Problem
Show full solution
Use extracted heat as the benefit
Method
Calculate Kc = Qc/Wᵢₙ.Reason
The desired effect is cooling the cold space, so delivered room heat is not the numerator.Working
Answer
Kc = Qc/Wᵢₙ = 2.4/0.8 = 3.0.Account for the work input
Working
Answer
The cooling COP is 3.0. The room receives both extracted heat and work: Qₕ = Qc + Wᵢₙ = 3.2 kJ. Thus the room gains more heat than the refrigerator removes from its interior.
Heating-mode performance
Worked example 2
Heating COP
Problem
Show full solution
Choose the heating numerator
Method
Divide Qₕ by Wᵢₙ.Reason
The heating benefit includes heat moved from the cold side plus energy supplied as work.Working
Answer
Kₕ = 5.0/1.25 = 4.0.Use the COP difference
Working
Answer
Kₕ = 5.0/1.25 = 4.0, Kc = Kₕ-1 = 3.0. Both are dimensionless; Kₕ-Kc = 1 accounts for the supplied work.
Derive the reversible temperature bound
For the combined cyclic pump and two fixed reservoirs, S_gen = Qₕ/Tₕ-Qc/Tc ≥ 0. Use Qc = Qₕ-Wᵢₙ to obtain Wᵢₙ ≥ Qₕ(Tₕ-Tc)/Tₕ, Kₕ ≤ Tₕ/(Tₕ-Tc). Similarly, Kc ≤ Tc/(Tₕ-Tc). These are reversible Carnot limits with Tₕ > Tc > 0 in kelvin. Real heat transfer, friction and other entropy generation increase the required input for the same useful transfer and reservoir temperatures.
A smaller temperature lift raises the ideal COP bound, but it does not guarantee the same improvement for every device: operating conditions, capacity and losses also matter. A formally divergent bound as Tₕ-Tc → 0 is not a promise of finite useful transfer with zero work in a practical device.
Guided practice 3
Test an attractive heat-pump rating
Problem
A cyclic heat pump exchanges heat only with reservoirs at 270 K and 300 K. A claimed cycle delivers 6000 J to the hot side using 300 J of work, with no other transfers or net energy storage. Find the claimed heating COP, its reversible upper bound and the minimum possible work input for that delivered heat. Does an internally consistent energy balance make the claim feasible?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
The first law gives the extracted cold-side heat. The second law limits how little work can produce that transfer.Show solution step by step
Calculate the claim and ceiling
Method
Compare Qₕ/Wᵢₙ with Tₕ/(Tₕ-Tc).Reason
COP above one is allowed, but COP above the reversible bound is not.Working
Answer
The claimed COP is 6000/300 = 20. The reversible bound is 300/(300-270) = 10, so the minimum possible input is 6000/10 = 600 J. The claim asks for half that minimum.
Check the complete transfer account
Method
Use Qc = Qₕ-Wᵢₙ.Reason
An energy balance can close while total entropy changes in a forbidden direction.Working
Answer
The claim extracts 5700 J, so energy conservation closes: 6000 = 5700 + 300. But S_gen = 6000/300-5700/270 = -1.11… J/K is negative. The claim is incompatible with the stated two-reservoir cycle. Recheck the rating conditions, quantities and boundary before drawing a device conclusion.
Trace where the rejected heat goes
A refrigerator warms its surroundings with the heat extracted from its interior plus the work input. It can cool a separate space only when its heat rejection is routed somewhere else. For a larger boundary containing both cold and hot sides, those heat transfers are internal and cancel; the external work input still adds energy to the combined system.
This boundary reasoning also prevents double-counting a pump’s useful outputs. A heating calculation uses heat delivered to the intended space, with any delivery losses accounted for. A cooling calculation uses heat actually removed from the intended cold space, not the hot-side total.
Choose the boundary for an open refrigerator
Check your understanding 1
An open refrigerator operates inside a thermally insulated room. Over an interval, its cyclic mechanism extracts 2400 J from room air, rejects 3200 J back into the room and receives 800 J of electrical work from outside. Neglect changes in macroscopic kinetic and potential energy and any storage change in the refrigerator itself. Find net external heat and work into the room and refrigerator together, and their combined internal-energy change. A colleague expects room cooling because the device has cooling COP 3: identify the incorrect boundary argument.
Show answer
For the combined room–refrigerator boundary, the 2400 J and 3200 J transfers are internal. Net external heat is zero and electrical work input is 800 J, so the combined internal energy increases by 800 J. Equivalently, room air loses 2400 J but receives 3200 J, a net gain of 800 J. COP 3 describes removal from the cold side, but both sides are inside the same room. It does not imply net room cooling; in this model the room gains energy. A closed refrigerated compartment and heat rejection to a separate surrounding space answer a different boundary question.
Practice set
Check your understanding 2
A heat pump delivers 6.0 kJ and extracts 4.5 kJ per cycle. Find its work input, heating COP and cooling COP.
Show hint
Show answer
Check your understanding 3
A refrigerator has cooling COP 2.5 and uses 400 J per cycle. Find the heat extracted per cycle.
Show hint
Show answer
Check your understanding 4
Does a heating COP of 1.8 violate energy conservation? Explain.
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Show answer
Check your understanding 5
With indoor temperature fixed, what usually happens to a heat pump’s heating COP when the colder outside air warms closer to indoor temperature?
Show hint
Show answer
Heating COP usually increases: the smaller temperature lift requires less work per unit of delivered heat. Actual performance also depends on the device and operating conditions.
What to keep
Heating and cooling COP use different useful heat transfers. A complete cyclic balance gives their difference; the reservoir temperatures impose stronger Carnot bounds. Follow every transfer across the chosen boundary before interpreting a rating or a room-heating claim.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.