The Carnot engine
Derive Carnot efficiency, apply it as an upper bound, and interpret its practical limits.
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ContinueA Carnot engine is a reversible cycle exchanging heat with two fixed-temperature reservoirs. Its efficiency is a ceiling for heat engines under those boundary conditions, not a prediction of a real plant’s output. Energy conservation tells you how much heat must be rejected for a stated work output; the second law determines whether that output is possible.
Build on isothermal and adiabatic processes, and the second law. Use positive magnitudes: Qₕ enters from a hot reservoir, Qc leaves to a cold reservoir and Wₒᵤₜ is delivered over a complete cycle. Reservoir temperatures satisfy Tₕ > Tc > 0 in kelvin.
Follow a reversible cycle
For an ideal-gas working substance, the hot isothermal expansion absorbs heat and produces work. A reversible adiabatic expansion then lowers temperature without transferring heat. Cold isothermal compression rejects heat, and a reversible adiabatic compression restores the starting temperature and state. A complete return restores the working substance’s internal energy and entropy; it need not leave zero net heat or work transfers.
Heat exchange in the reversible model has an infinitesimal thermal difference, and the mechanical steps have no friction or finite driving pressure imbalance. Merely choosing four paths with the right names does not ensure these conditions. Reversing the ideal cycle gives a refrigerator; it changes the transfer directions, not the laws.
Derive the efficiency from energy and entropy
Because Δ U_cycle = 0, the first law gives Wₒᵤₜ = Qₕ-Qc and efficiency η = Wₒᵤₜ/Qₕ = 1-Qc/Qₕ.
The working substance returns to its starting state, while the hot reservoir loses entropy Qₕ/Tₕ and the cold reservoir gains Qc/Tc. Reversibility gives zero total entropy change: Qc/Tc-Qₕ/Tₕ = 0. Therefore Qc/Qₕ = Tc/Tₕ and ηC = 1-Tc/Tₕ. For an actual two-reservoir cyclic engine, total entropy generation is nonnegative. Substituting Qc = Qₕ-Wₒᵤₜ gives Wₒᵤₜ = Qₕ(1-Tc/Tₕ)-TcS_gen ≤ QₕηC. The reversible limit has S_gen = 0; positive generation reduces available work. The working substance’s cycle entropy change is still zero, distinct from entropy generated in the combined system and reservoirs. The MIT work-limit derivation develops this distinction.
Short worked examples
Maximum possible efficiency
Worked example 1
Efficiency ceiling
Problem
Show full solution
Apply the reversible bound
Method
Use eC = 1-Tc/Tₕ.Reason
The Carnot efficiency bounds every engine operating between these two reservoirs; the temperatures are already absolute.Working
Answer
eC = 1-300/900 = 2/3 ≈ 0.667.Express the ceiling as a percentage
Working
Answer
No engine between these reservoirs can exceed 66.7% efficiency.
Maximum work per cycle
Worked example 2
Work and rejected heat
Problem
Show full solution
Convert efficiency into work
Method
Multiply Qₕ by eC = 2/3.Reason
Efficiency is the fraction of absorbed heat delivered as work. Its maximum therefore gives the maximum output.Working
Answer
Wₘₐₓ = eCQₕ ≈ 0.667(1200) ≈ 8.0 × 10² J.Account for the remaining heat
Working
Answer
Q_(c, min) = Qₕ-Wₘₐₓ ≈ 400 J. Thus Wₘₐₓ = 800 J and Q_(c, min) = 400 J per cycle. The two sum to the 1200 J input.
Test a proposed engine
Guided practice 3
Check a claim that conserves energy
Problem
A cyclic engine exchanges heat only with reservoirs at 600 K and 300 K. It claims to absorb 1200 J and deliver 720 J of work per cycle, with no other transfers or net energy storage. Find the Carnot efficiency, the maximum work for that heat input and the total entropy change implied by the claim. Does the first-law balance alone establish feasibility?
Work through the model, then compare your reasoning
Hints
Hint 1
Check the model
The cycle stores no net energy or entropy. Calculate its rejected heat before testing the reservoir entropy changes.Show solution step by step
Find the reversible ceiling
Method
Use ηC = 1-Tc/Tₕ.Reason
Two fixed reservoirs impose a stronger condition than efficiency below one.Working
Answer
ηC = 1-300/600 = 0.50 and Wₘₐₓ = 1200(0.50) = 600 J. The claimed 720 J exceeds this ceiling despite being smaller than the 1200 J heat input.
Test the entropy account
Method
Use Qc = Qₕ-Wₒᵤₜ and Δ Sₜₒₜₐₗ = Qc/Tc-Qₕ/Tₕ.Reason
Energy conservation is necessary but does not enforce the second law.Working
Answer
The claim would reject Qc = 480 J, giving Δ Sₜₒₜₐₗ = 480/300-1200/600 = -0.40 J/K. That negative total is impossible for the stated combined engine and reservoirs, whose only external transfer is work. Report the failing assumption or measurement; do not approve the claim merely because its heat and work sum correctly.
Use the bound only for its stated surroundings
Convert Celsius to kelvin before forming the temperature ratio. Choose the actual heat-source and sink temperatures, not an arbitrary maximum working-fluid temperature. If there are additional reservoirs, noncyclic energy storage or other energy inputs, reconstruct the boundary balance instead of applying a two-reservoir formula to the wrong device.
As the reservoir temperatures approach one another, the efficiency ceiling tends to zero. The formal limit Tc → 0 gives a ceiling approaching one, but it does not establish an attainable zero-temperature sink or useful finite-power operation. Reversible heat exchange is a limiting process; a practical rate needs finite transfer mechanisms and their consequences.
A finite hot body cools as it supplies heat, so it is not one fixed-temperature reservoir. If a body of constant heat capacity C cools reversibly from Tᵢ to a fixed cold reservoir temperature Tc, take heat withdrawn in each small cooling step as -C dT. Its maximum differential work is (1-Tc/T)(-C dT). Integrate over the actual changing source temperature rather than applying the initial ceiling to every joule.
Change a reservoir into a finite source
Check your understanding 1
An idealised hot body of constant heat capacity 200 J/K cools from 600 K to 300 K while reversible engines extract work and reject heat to a fixed reservoir at 300 K. The engines complete cycles, and there are no other energy transfers. Find the total heat withdrawn, maximum total work and heat delivered to the cold reservoir in that limit. Compare the work with multiplying the entire withdrawn heat by the initial Carnot efficiency. Which model assumption makes those predictions differ? Check the total entropy change for the reversible extraction.
Show answer
The withdrawn heat is C(600-300) = 60000 J, but its source temperature falls. Integrating gives Wₘₐₓ = C∫₃₀₀⁶⁰⁰(1-300/T) dT = 200[300-300 ln 2] ≈ 18411 J. Multiplying all 60000 J by the initial efficiency 0.50 would give 30000 J and overstate the bound. That shortcut treats the cooling body as though it remained a 600 K reservoir.
The cold reservoir receives Qc = 60000-Wₘₐₓ ≈ 41589 J. The body’s entropy change is Δ S_body = 200 ln(300/600) ≈ -138.63 J/K, while the cold reservoir gains Qc/300 ≈ +138.63 J/K. These cancel in the reversible limit. Use unrounded values for that check.
The integral describes the supplied reversible, constant-capacity model; real extraction generates entropy and gives less work.
Practice set
Check your understanding 2
Find the Carnot efficiency for Tₕ = 600 K and Tc = 300 K.
Show hint
Use the colder temperature in the numerator.
Show answer
eC = 1-300/600 = 0.50 = 50%. Half the heat input can become work in the reversible limit.
Check your understanding 3
A plant operates between 1000 K and 320 K and receives 5.0 MJ per cycle. Find its maximum work output per cycle.
Show hint
Show answer
Check your understanding 4
Why must 200°C and 30°C be converted before using eC = 1-Tc/Tₕ?
Show hint
A temperature ratio depends on the scale’s zero.
Show answer
The formula requires absolute temperatures: Tₕ = 473.15 K and Tc = 303.15 K. Celsius has an offset zero, so 30/200 is not the required physical ratio.
Check your understanding 5
If Tc decreases while Tₕ stays fixed, what happens to Carnot efficiency?
Show hint
Show answer
It increases: decreasing Tc/Tₕ increases 1-Tc/Tₕ.
What to keep
A cyclic energy balance is necessary, and a two-reservoir entropy balance supplies the Carnot ceiling. Reversible engines attain the ideal bound; entropy generation lowers work for the same heat input and reservoir temperatures. A changing source needs a changing-temperature account.
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Syllabus and review details
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