Entropy Balance Across a Carnot Cycle
1 · Predict
A Carnot engine exchanges heat with a hot reservoir during one leg and a cold reservoir during another. Does the hot reservoir's entropy loss exactly cancel the cold reservoir's entropy gain?
- Yes — for a reversible (Carnot) cycle, the two reservoirs' entropy changes exactly cancel, giving zero net entropy generation.
- The reservoirs' entropy changes don't cancel; there's a nonzero net entropy change overall.
- Reservoir entropy changes aren't related to the heat exchanged with the gas.
2 · Set Up
- Open the entropy-carnot preset and press Reset. The same 4-leg Carnot cycle as carnot-cycle runs between 600 K and 300 K reservoirs.
- Enable the per-leg heat readout.
- For each leg of the cycle (1 through 4), read off the heat exchanged and compute that leg's entropy contribution Q_leg/T_leg.
3 · Collect Data
| Leg number | Heat exchanged this leg Q (J) | Entropy contribution ΔS = Q/T (J/K) |
|---|---|---|
| 1 | ||
| 2 | ||
| 3 | ||
| 4 |
Sum your 4 legs' entropy contributions. Plot each leg's contribution as a bar chart, with legs 2 and 4 (adiabatic, Q = 0) showing zero.
4 · Analyze
- Confirm legs 2 and 4 (adiabatic) contribute zero entropy, since Q = 0 for those legs by definition. Sum the nonzero contributions from legs 1 (at Th = 600 K) and 3 (at Tc = 300 K).
- The heat absorbed by the gas during leg 1 becomes heat released by the hot reservoir; the heat released by the gas during leg 3 becomes heat absorbed by the cold reservoir. Explain why the reservoirs' combined entropy change over the full cycle comes out to zero.
5 · Extend
- A real engine has friction and finite-speed heat transfer, both irreversible. Explain why a real cycle's reservoir entropy changes would NOT cancel exactly, giving a positive net entropy generation — and why the second law requires that net generation to be ≥ 0.
- The Carnot cycle's zero net entropy generation is why it's called 'reversible' — you could run every leg backward and undo everything, including the reservoirs' states. Explain why a process with positive net entropy generation (like friction) can't be undone this way.
The Physics Behind This Experiment
Reservoir Entropy Balance
Each reservoir's entropy changes by −Q/T for heat it releases (or +Q/T for heat it absorbs) at its own fixed temperature. For a reversible cycle like Carnot's, the hot and cold reservoirs' entropy changes exactly cancel: ΔS_net = Qout/Tc − Qin/Th = 0.