Thermodynamics
Reversible Carnot cycle — entropy
A reversible Carnot cycle exchanges entropy with reservoirs — net ΔS_univ ≈ 0.
Reversible Carnot cycle — entropy — interactive Thermodynamics simulation. A reversible Carnot cycle exchanges entropy with reservoirs — net ΔS_univ ≈ 0. Free browser-based virtual physics lab with live SI measurements and a guided tutorial.
Reversible entropy
A reversible Carnot cycle exchanges entropy with reservoirs — net ΔS_univ ≈ 0.
For a full reversible cycle the system's entropy returns to its starting value, while the hot reservoir loses entropy Q_h/T_h and the cold reservoir gains Q_c/T_c with equal magnitude. The universe's total entropy change is zero—defining maximum efficiency.
- ΔS_univ = 0 (reversible)
- Entropy exchange
Investigation brief
Plan the question before you open the lab
The brief mirrors the prerendered page: driving question, competing predictions, variable roles, governing laws, setup, analysis and extension prompts remain visible and in this order.
Driving question
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?
Predictions to weigh
- The reservoirs' entropy changes don't cancel; there's a nonzero net entropy change overall.
- Yes — for a reversible (Carnot) cycle, the two reservoirs' entropy changes exactly cancel, giving zero net entropy generation.
- Reservoir entropy changes aren't related to the heat exchanged with the gas.
Variable roles
What you set:
- Leg number
What you measure:
- Heat exchanged this leg Q (J)
- Entropy contribution ΔS = Q/T (J/K)
How the investigation runs
- 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.
Governing equation
Reservoir Entropy Balance — ΔS = Qout/Tc − Qin/Th
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.
What the printable worksheet asks students to work out
- 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.
Where this shows up beyond the lab
- 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.
- AP Physics 2 — Unit 9: Thermodynamics
- IB Physics — B.4 Thermodynamics
- General High School Physics — Heat, temperature & gas laws
- NGSS High School Physics — Thermal energy transfer
- Welcome to Entropy Carnot
- Select the chamber
- Press Play
- Entropy in a reversible cycle
- Open the Properties panel
- You did it!
Open the interactive simulation to build the scene, press Play, and explore with live measurements and a guided tutorial.