Heat engines and heat pumps
Build the core model of heat engines and heat pumps, including sign conventions and the cycle energy balance.
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ContinueA working substance can return to its starting state while transferring energy between its surroundings. That is the purpose of a heat-engine or heat-pump cycle: the device is reused, while heat and work have a net effect outside it. Begin by choosing the boundary and confirming a complete cycle before cancelling its energy-storage term.
Use the first law: Δ U = Q + Wₒₙ, with heat and work positive into a closed system. Assume negligible changes in bulk kinetic and potential energy. In this lesson, Qₕ,Qc,Wₒᵤₜ and Wᵢₙ are instead positive transfer magnitudes; their signs in the first law follow their directions.
Translate the arrows into one energy balance
A two-reservoir engine absorbs Qₕ from the hot reservoir, rejects Qc to the cold reservoir and supplies work. For the working substance, Δ U = Qₕ-Qc-Wₒᵤₜ. A complete cycle restores its internal energy, so Wₒᵤₜ = Qₕ-Qc. This is net work over the whole cycle, including any compression steps that consume work; expansion work alone is not the output.
A refrigerator or heat pump extracts Qc from the cold side, receives work and delivers Qₕ to the hot side: Δ U = Qc + Wᵢₙ-Qₕ. For its complete cycle, Qₕ = Qc + Wᵢₙ. A heating pump benefits from heat delivered to the hot space; a refrigerator benefits from heat removed from the cold space. They need not be physically different machines.
The working substance returns to its state, but the reservoirs and work receiver generally do not. “Cyclic” describes the device, not an absence of change everywhere. The second law also constrains these transfers: energy balance alone does not prove a proposed device possible.
Short worked examples
Heat engine work output
Worked example 1
Engine energy balance
Problem
Show full solution
Balance a complete cycle
Method
Subtract the rejected heat from the absorbed heat.Reason
The working substance returns to its starting internal energy, so the net heat input equals the work output.Working
Answer
Wₒᵤₜ = Qₕ - Qc = 900 - 540 = 360 J.State the output
Working
Answer
The engine delivers 360 J of work per cycle. Together with the 540 J rejected heat, this accounts for the 900 J input.
Heat pump work input
Worked example 2
Heating a room
Problem
Show full solution
Identify the extra energy supplied
Method
Use Wᵢₙ = Qₕ-Qc.Reason
The delivered heat contains both the extracted outdoor energy and the supplied work.Working
Answer
Wᵢₙ = Qₕ - Qc = 1400 - 1100 = 300 J.State the required input
Working
Answer
The pump requires 300 J of work per cycle. It moves existing thermal energy as well as receiving work, so the delivered heat exceeds the work input.
Check the interval before declaring a cycle
A repeated control signal or moving piston is not enough to show that the working substance has returned to its complete state. During startup, energy can accumulate. Retain Δ U for that interval, rather than labelling every difference between heat rates as exported work. Even Δ U = 0 alone is insufficient to prove a cycle: different states can have the same internal energy.
Guided practice 3
Test the storage term before using a cycle formula
Problem
During a startup interval, a closed working device absorbs 1000 J of heat from a hot source, rejects 600 J to a cooler sink and delivers 250 J of work. These are its only energy transfers, and bulk energy changes are negligible. Find signed net heat, the internal-energy change and the work that the same heat transfers would support over a complete cycle. Is this interval a cycle?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
Use the stated 250 J output for the actual interval; do not replace it with the result of assuming a cycle.Show solution step by step
Close the actual interval balance
Method
Use Δ U = Qₕ-Qc-Wₒᵤₜ.Reason
Startup can change the working device’s state.Working
Answer
Q = 1000-600 = 400 J and Δ U = 400-250 = 150 J. The output is only 250 J because 150 J remains as increased internal energy. A nonzero internal-energy change rules out a complete cycle.
Compare with the cycle condition
Method
Set Δ U = 0 only for the hypothetical complete cycle.Reason
A cycle formula is a consequence of the state return, not a definition of measured work.Working
Answer
With the same heat transfers in a complete cycle, Wₒᵤₜ = 1000-600 = 400 J. Applying that answer to the actual startup interval would discard its storage term. After startup, consistently measured steady rates can satisfy the cycle-average balance, but the averaging interval and boundary still matter.
Combine devices by cancelling internal transfers
For an engine driving a pump, the engine’s work output can be the pump’s work input. Both appear in the separate device balances; they cancel across a boundary containing both. The same is true for heat exchanged internally. Count only transfers that cross the chosen boundary, and distinguish a device’s output from a larger installation’s output after auxiliary loads.
When there are additional heat sources, electrical inputs, mass flows or other work channels, the simple two-reservoir formula is no longer the whole account. Write those channels explicitly. A heat-driven refrigeration installation, for example, can receive energy from a third heat source; absence of shaft work does not mean absence of every driving input.
Follow the net effect of an engine driving a pump
Check your understanding 1
An engine absorbs 1000 J from a 600 K reservoir, rejects 600 J to a 300 K reservoir and supplies 400 J of work. That work entirely drives a cyclic pump which extracts 400 J from the same cold reservoir and delivers 800 J to the same hot reservoir. Both devices return to their initial states and have no other transfers. Find the combined work export and each reservoir’s net heat change. What is the combined external effect?
Show answer
Across the combined boundary, the engine’s 400 J work transfer to the pump is internal, so external work is zero. The hot reservoir loses 1000 J to the engine and receives 800 J from the pump: it loses 200 J overall. The cold reservoir receives 600 J and supplies 400 J: it gains 200 J overall. The sole combined effect is 200 J transferred from hot to cold, with no net device storage or exported work. Adding the engine output to the pump’s delivered heat would count an internal transfer as another useful external output.
Practice set
Check your understanding 2
An engine absorbs 1200 J and rejects 800 J per cycle. Find its work output.
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Show answer
Check your understanding 3
A refrigerator extracts 600 J from a cold space and rejects 900 J to the room per cycle. Find its work input.
Show hint
Show answer
Check your understanding 4
A device absorbs 500 J and rejects 520 J per cycle. Can it supply net work as a heat engine?
Show hint
Show answer
Check your understanding 5
Why must a cyclic engine operating between two reservoirs reject some heat rather than convert all its absorbed heat to work?
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What to keep
Choose the boundary, translate each transfer direction and retain storage until a complete cycle is established. Engines supply net work; pumps receive work and move heat. In a combined installation, internal transfers cancel and additional inputs must be included.
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Syllabus and review details
No official syllabus alignment is listed for this lesson.