SCIENCE · INVESTIGATE AND EXPLAIN

Check entropy as well as energy

Find engine proposals that conserve energy but require an impossible total entropy decrease.

Learning outcome: Calculate reservoir entropy changes, distinguish a subsystem decrease from a total decrease and screen a cyclic engine with explicit boundaries.

Starting recap

A cyclic engine returns to its initial state, so both internal-energy and entropy changes are zero over a cycle. Heat and work transfers need not vanish.

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Try

An engine claims 800 J of work from 1000 J of heat, rejecting the other 200 J. Its energy account balances. Is that sufficient?

Open the experiment

Predict

With a 600 K source and 300 K sink, predict whether a proposed 40% engine and a proposed 80% engine pass the entropy check.

Your prediction and reason

Experiment

Capture the 40% proposal, then raise the proposed efficiency to 80%. Check both energy splits and add the reservoir entropy changes. Locate the 50% boundary and test zero work. State which numbers are rejected proposals rather than achievable output.

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 %

Reset keeps your prediction, captures and practice history.

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

An energy balance is necessary, but not sufficient

A proposal takes 1000 J from a 600 K reservoir, delivers 800 J as work and sends 200 J to a 300 K reservoir. The numbers conserve energy: 1000 = 800 + 200. Before accepting the claim, inspect what happens to the two reservoirs as well as the cyclic engine.

Our boundary includes the engine, both ideal reservoirs and ideal mechanical work storage. There is no matter exchange, hidden heat source or auxiliary work input. The reservoirs remain at their stated temperatures. The engine returns to the same state after each cycle; its internal energy and entropy therefore have zero net change. The reservoirs do not return to their original energy contents.

Calculate entropy changes with the right signs

Thermodynamic entropy S is a state property measured in joules per kelvin (J/K). For an ideal fixed-temperature reservoir receiving heat Q, its change is ΔS = Q/T, using absolute temperature T in kelvin. Q is negative when heat leaves that reservoir. This reservoir calculation remains valid even if the overall engine process is irreversible.

For a general system undergoing an irreversible process, do not divide its actual heat by an arbitrary temperature to obtain its entropy change. Entropy change can instead be calculated along a reversible path between the same states; with varying temperature, the contributions must be summed appropriately. See the OpenStax entropy reference for this distinction.

In the proposal, the hot reservoir loses 1000 J, giving ΔSh = −1000/600 = −1.666667 J/K. The cold reservoir receives 200 J, giving ΔSc = 200/300 = +0.666667 J/K. The engine’s change is zero. Ideal mechanical storage receives work without an entropy transfer or generation in this model. The combined change would therefore be −1 J/K.

The second law forbids that total decrease for the specified combined system. The display labels the proposal as rejected. It has not simulated an engine producing 800 J; it has calculated the consequences that make the claim impossible under its assumptions.

A subsystem may lose entropy

Reset to the default proposed efficiency of 40%, still at 600 K. Work is 400 J and rejected heat is 600 J. The hot reservoir still loses 1.666667 J/K, but the cold one gains 2 J/K. The combined change is +0.333333 J/K, so this proposal passes the entropy screen.

The hot reservoir’s decrease is allowed because the surroundings gain enough entropy. “Entropy never decreases” needs a boundary: it applies to the total of an isolated combined system, not necessarily to every part. Also, the engine’s zero change is not zero entropy production for the complete process. Entropy generated during a real cyclic process is exported rather than accumulated indefinitely in the working substance.

Passing this balance does not supply a working mechanism, material limits or a speed. It only means this particular thermodynamic constraint has not rejected the proposal.

Locate the boundary and a zero-work case

Set 50% at 600 K. The hot change is −1.666667 J/K and the cold change is +1.666667 J/K. Their sum is zero: the reversible boundary under these assumptions.

Set 0%. The proposal delivers no work and rejects all 1000 J. The total change is 1000/300 − 1000/600 = +1.666667 J/K. Direct heat transfer from hot to cold has this accounting. Zero work does not mean zero entropy generation, and energy has not disappeared.

Transfer: a reversed heat-flow claim

An isolated pair of ideal reservoirs is claimed to transfer 600 J from 300 K to 600 K without any work or other change. Calculate the two entropy changes and decide whether energy conservation is enough to support the claim.

Check the entropy transfer

The colder reservoir loses 600/300 = 2 J/K, while the hotter one gains 600/600 = 1 J/K. The combined change would be −1 J/K. Energy balances, but the second law rejects the proposed transfer without another compensating change. A refrigerator uses work input and has a different full energy and entropy account.

These are macroscopic entropy balances, not a count of molecular arrangements or a probability lesson. Describing entropy merely as visual “disorder” would not tell you which boundary, temperature or heat sign to use here.

Apply

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

Explain why the hot reservoir may lose entropy while the total may not. Distinguish engine entropy change from entropy generated in the engine-plus-surroundings process.

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: Audit a heat engine’s efficiency Next: Use reservoir temperatures to bound engine efficiency

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