SCIENCE · INVESTIGATE AND EXPLAIN

Use reservoir temperatures to bound engine efficiency

Derive the Carnot limit and distinguish a reversible bound from a practical power rating.

Learning outcome: Apply eta ≤ 1 − Tc/Th using kelvin, connect the equality case to reversible operation and reject unsupported claims of attainable power.

Starting recap

For the two-reservoir cyclic proposal, total entropy change is Qout/Tc − Qin/Th. It must be nonnegative.

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Try

Would hotter source conditions make a previously rejected 50% efficiency proposal pass the same thermodynamic screen?

Open the experiment

Predict

Keep the proposed efficiency at 50%. Predict its status with source temperatures 400 K, 600 K and 800 K while the sink stays at 300 K.

Your prediction and reason

Experiment

Compare the three source temperatures and the maximum-work marker. Record one rejected proposal, one equality case and one positive-entropy case. Explain why the same proposed work can have different thermodynamic status.

Screen a two-reservoir engine proposal

Worked teaching model · synthetic values · A proposed cyclic engine absorbs 1000 J per cycle from a fixed-temperature hot reservoir and rejects heat only to a 300 K reservoir. No mass exchange, other heat contacts, auxiliary inputs or state accumulation; work goes to ideal mechanical storage. The engine returns to its initial energy and entropy. Energy balance is enforced by construction; the entropy screen may reject a proposal. Passing is a necessary thermodynamic check, not a device design or measured performance. Reservoir temperatures stay fixed; no rate or power is predicted.

Hot reservoir absolute temperature: 600 K Proposed thermal efficiency: 40 %

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Passes entropy screen · Compatible with this entropy balance; no machine performance has been demonstrated.

Entropy accounting for one proposed cycle
Part	Heat into part (J)	Temperature (K)	Entropy change (J/K)
Hot reservoir	−1000	600	-1.666667
Cold reservoir	600	300	2.000000
Cyclic engine	Net 400	Not specified along cycle	0 (returns to initial state)
Combined including ideal work storage	No external heat	No single temperature	0.333333

Proposed net work 400 J + rejected heat 600 J = input 1000 J. Minimum permitted rejection for these reservoirs is 500.000 J. The entropy of ideal mechanical work storage is unchanged; work is not assigned a heat-over-temperature entropy transfer.

Passes entropy screen. Proposed work 400 J and heat rejection 600 J per cycle.

Required total entropy change 0.333333 J/K; Carnot limit 50.000%. The screen does not establish a realizable finite-power machine.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Turn the entropy screen into a limit

Keep the heat input Qin at 1000 J per cycle and the cold reservoir at Tc = 300 K. The hot reservoir temperature is Th. Qin and Qout are positive heat magnitudes; net work by the engine is W = Qin − Qout. Efficiency eta is W/Qin.

For a cyclic engine exchanging heat only with these two reservoirs, the combined entropy change is Qout/Tc − Qin/Th ≥ 0. Multiplying by the positive temperatures and rearranging gives Qout/Qin ≥ Tc/Th. Substitution into eta = 1 − Qout/Qin gives the bound:

eta ≤ 1 − Tc/Th.

Read this as: the work fraction cannot exceed one minus the absolute cold-to-hot temperature ratio. This is the Carnot limit for the specified reservoirs. It does not depend on our earlier monatomic-gas heat capacity or on a particular piston material. The cyclic two-reservoir boundary is essential.

Compare three source temperatures

Set the proposed efficiency to 50%, corresponding to 500 J of work and 500 J of rejected heat. At 400 K, the maximum fraction is 1 − 300/400 = 25%. Maximum reversible work is 250 J, so the proposal is rejected; its combined entropy change is 500/300 − 1000/400 = −0.833333 J/K.

At 600 K, the limit is 50% and maximum reversible work is 500 J. The proposal lies exactly at the reversible boundary: the reservoir changes cancel.

At 800 K, the limit is 62.5% and maximum reversible work is 625 J. The same 500 J work proposal passes the screen, with positive combined entropy change 500/300 − 1000/800 ≈ 0.416667 J/K. Changing the source temperature changed the available thermodynamic limit even though the proposed energy split stayed fixed.

Capture these cases. Use the dashed maximum-work marker and the numerical ledger together; an energy bar extending beyond the marker represents a rejected claim, not extra usable performance.

What would equality require?

Equality requires no total entropy generation. A reversible Carnot cycle can connect the reservoirs through two isothermal heat-exchange stages and two reversible adiabatic stages. In an ideal-gas example, it expands isothermally while receiving heat at Th, expands adiabatically to Tc, compresses isothermally while rejecting heat at Tc, then compresses adiabatically back to the initial state. The OpenStax Carnot-cycle reference develops that ideal construction.

The model here screens the cycle’s total transfers; it does not construct or animate those four paths. Reversible heat exchange is a limiting process with vanishing temperature difference. Ordinary friction and heat transfer across finite differences generate entropy. A displayed equality does not establish attainable finite power, cycle rate or an engineering design.

Our previous rectangular gas cycle has heat exchange while the gas temperature changes. Its hottest and coldest gas states are not automatically the two fixed reservoirs assumed here. Do not transplant this formula without checking the thermal contacts and full system boundary.

Use kelvin, and distinguish a bound from a guarantee

A source at 500 K and sink at 300 K give a 40% limit. In Celsius these are 226.85°C and 26.85°C. Substituting those Celsius numbers into the ratio would give about 88.2%, a wrong result: Celsius has an offset zero and cannot be used in this temperature ratio.

Increasing Th or lowering Tc raises this ideal bound when Th > Tc > 0. It does not guarantee the same improvement in a real machine. Nor does it make 100% thermal efficiency attainable between finite positive reservoir temperatures. Material choices, losses, rates and auxiliary inputs would need their own evidence.

Transfer to a new pair of reservoirs

An engine receives 2400 J per cycle from a 750 K reservoir and rejects heat only to 300 K. Find the maximum reversible work and minimum rejected heat. A second claim proposes 1600 J of work; screen it independently using entropy.

Check the Carnot transfer

The limit is 1−300/750 = 0.6, so maximum work is 1440 J and minimum rejected heat is 960 J. The 1600 J proposal would reject only 800 J. Its combined entropy change would be 800/300 − 2400/750 ≈ −0.533333 J/K, so it is rejected despite conserving energy.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

State the system boundaries needed for the Carnot limit, why temperatures must be absolute, and why equality is an ideal reversible limit rather than proof of a finite-power device.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Check entropy as well as energy Next: Move heat from cold to warm with work

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