Carnot engine

What is a Carnot engine?

A reversible heat engine operating between two temperatures is called a Carnot engine.

NCERT Physics Part II (Class 11), Section 11.11, page 237

In plain words: given a hot source and a cold sink, what is the most work you could possibly get from each joule of heat? Sadi Carnot, a French engineer, asked this in 1824 and found the answer: it depends only on the two temperatures. The hotter the source and the colder the sink, the better, but you can never reach 100%.

Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established.

NCERT Physics Part II (Class 11), Section 11.11, page 237

Why these four steps

The best engine must be reversible, because irreversibility lowers efficiency (page 237). For NCERT’s simple cycle with no internal heat-storage device, this gives the following steps:

• Heat must be taken in and given out with no finite temperature difference between the gas and the reservoir. So heat is absorbed from the hot reservoir isothermally at T₁, and given out to the cold reservoir isothermally at T₂.
• To change the gas’s temperature from T₁ to T₂ and back, using any process that exchanges heat, say an isochoric one, would need a whole series of reservoirs at temperatures in between. With only those contacts and no internal regenerator, these changes must be adiabatic, with no heat flow.

The Carnot cycle

With an ideal gas as the working substance (NCERT Fig. 11.9, pages 237–238), where μ is the fixed number of moles and γ = C_p/C_v is treated as constant:

1. Step 1 → 2: isothermal expansion at T₁, from (P₁, V₁) to (P₂, V₂). The gas takes in heat Q₁ = μRT₁ ln(V₂/V₁), equal to the work it does (NCERT Eq. 11.18).
2. Step 2 → 3: adiabatic expansion from T₁ to T₂. The gas does work μR(T₁ − T₂)/(γ − 1) (NCERT Eq. 11.19).
3. Step 3 → 4: isothermal compression at T₂. The gas gives out heat Q₂ = μRT₂ ln(V₃/V₄), equal to the work done on it (NCERT Eq. 11.20).
4. Step 4 → 1: adiabatic compression from T₂ back to T₁. Work μR(T₁ − T₂)/(γ − 1) is done on the gas (NCERT Eq. 11.21).

The two adiabatic works are equal and opposite, so they cancel. The net work per cycle is W = Q₁ − Q₂, the area enclosed by the cycle on a P–V graph.

Efficiency of a Carnot engine

The efficiency is η = W/Q₁ = 1 − Q₂/Q₁. Using the two isothermal heats,

(NCERT Eq. 11.23.)

For the adiabatic steps TV^γ−1 is constant, so T₁V₂^γ−1 = T₂V₃^γ−1 and T₁V₁^γ−1 = T₂V₄^γ−1 (the lines before NCERT Eqs. 11.24–11.25). Dividing the first by the second gives (V₂/V₁)^γ−1 = (V₃/V₄)^γ−1, so V₃/V₄ = V₂/V₁ (NCERT Eq. 11.26), so the logarithms cancel:

(In the printed NCERT Eq. 11.23 the natural logarithm ln appears as “In”, and in Eq. 11.25 the exponent should read 1/(γ − 1).)

Since Q₂/Q₁ = T₂/T₁, a Carnot engine always satisfies

Running it backwards. Every step of the Carnot cycle can be reversed. Then the engine takes heat Q₂ from the cold reservoir, has work W done on it, and gives Q₁ to the hot reservoir: a reversible refrigerator (page 238).

Indeed it is the only reversible engine possible that works between two reservoirs at different temperatures.

NCERT Physics Part II (Class 11), Section 11.11, page 238

NCERT’s “only reversible engine” wording refers to its simple construction. More elaborate ideal cycles with perfect internal heat storage, such as a regenerative Stirling cycle, can follow different paths while having the same Carnot efficiency. The universal result is the efficiency limit, not a unique mechanical design.

Carnot’s theorem

NCERT proves two results, sometimes called Carnot’s theorem (pages 238–239):

1. Working between two given temperatures, no engine can have a higher efficiency than a Carnot engine.
2. The efficiency of a Carnot engine does not depend on the working substance.

The idea of the proof of (1) (NCERT Fig. 11.10, page 239). Suppose an irreversible engine I were more efficient than a Carnot engine R. Let I take Q₁ from the source, give work W′ and reject Q₁ − W′ to the sink. Run R backwards as a refrigerator that returns the same Q₁ to the source. Because R is reversible, it then needs exactly the work W it would give as an engine taking in Q₁, and since η_R < η_I, W < W′. R takes Q₁ − W from the sink, so the pair takes (Q₁ − W) − (Q₁ − W′) = W′ − W from the sink and turns it all into work. The source is left unchanged. That breaks the Kelvin–Planck statement of the second law, so I cannot be more efficient than R. A similar argument shows that a reversible engine using one working substance cannot beat one using another.

Because of (2), we may calculate η with an ideal gas, whose equation of state is simple, and the result η = 1 − T₂/T₁ holds for every Carnot engine. NCERT’s Eq. 11.28 can therefore be used to define a truly universal thermodynamic temperature scale, independent of any particular substance. For an ideal gas it agrees with the ideal-gas (kelvin) temperature (page 239). NCERT’s text says that scale was “introduced in section 11.9”; it was introduced in Section 10.4 of the previous chapter, Thermal Properties of Matter.

Solved examples

Take R = 8.31 J mol^⁻¹ K^⁻¹.

Common mistakes to avoid

Quick revision summary

• A Carnot engine is a reversible engine working between two reservoirs at T₁ (hot) and T₂ (cold).
• Carnot cycle: isothermal expansion at T₁, adiabatic expansion, isothermal compression at T₂, adiabatic compression.
• Efficiency: η = 1 − T₂/T₁, with temperatures in kelvin; also Q₁/T₁ = Q₂/T₂.
• Carnot’s theorem: no engine between the same two temperatures is more efficient, and the efficiency does not depend on the working substance.
• Run backwards, a Carnot engine is a reversible refrigerator; Q₁/T₁ = Q₂/T₂ defines the thermodynamic temperature scale.

Questions students ask

Is the Carnot cycle in the JEE Main and NEET syllabus?

The Carnot cycle and Carnot engine are not named in the JEE Main 2027 or NEET (UG) 2026 Physics syllabuses: in both, Unit 8, Thermodynamics, ends with the second law of thermodynamics and reversible and irreversible processes. They are part of NCERT Class 11 Physics (Section 11.11), and the JEE Advanced 2026 syllabus does list the Carnot engine and its efficiency. The Carnot efficiency, 1 − T₂/T₁ with temperatures in kelvin, is the upper limit the second law sets for any heat engine working between two reservoirs, so it is still worth knowing for questions on engine efficiency.

Why is the Carnot cycle not practically possible?

The Carnot cycle cannot be built exactly because every step must be reversible: the isothermal steps need heat to flow across an almost zero temperature difference, so they must run infinitely slowly, and the adiabatic steps must also be quasi-static, with perfect insulation and a frictionless piston. Switching the cylinder between perfect thermal contact and perfect insulation in each cycle is also impossible in practice. Even if it could be done, an engine running infinitely slowly would deliver almost no power, so real engines run fast and accept a lower efficiency than 1 − T₂/T₁.

Is a Carnot engine 100% efficient because it is reversible?

No, a Carnot engine is reversible but not 100% efficient: its efficiency is 1 − T₂/T₁, which is less than 1 for any sink above 0 K. Being reversible means it wastes nothing beyond what the second law demands, yet in every cycle it must still give out heat Q₂ = Q₁T₂/T₁ to the cold reservoir. For example, between 800 K and 300 K a Carnot engine turns 1 − 300/800 = 62.5% of the heat it takes in into work and rejects the other 37.5%.

What is the difference between a Carnot engine and a Carnot refrigerator?

A Carnot refrigerator is a Carnot engine run backwards between the same two temperatures: it takes heat Q₂ from the cold reservoir, has work W done on it and gives Q₁ = Q₂ + W to the hot one. Its coefficient of performance is α = Q₂/W = T₂/(T₁ − T₂), which is linked to the engine’s efficiency η by α = (1 − η)/η. For example, between 350 K and 300 K the engine has η = 50/350 = 1/7 ≈ 14.3%, and the refrigerator has α = 300/50 = 6, which equals (6/7)/(1/7).

How is the Carnot cycle different from the Otto cycle of a petrol engine?

The Carnot cycle uses two isothermal and two adiabatic steps and takes in all its heat at the single hot temperature, while the Otto cycle, the idealised model of a petrol engine, uses two adiabatic and two constant-volume steps and takes in heat while its temperature rises. In the ideal-gas, constant-γ air-standard model, the Otto efficiency is 1 − 1/r^γ−1, where r = V_max/V_min is the compression ratio, and it is always lower than the Carnot efficiency between the highest and lowest temperatures of the same cycle. The Otto cycle is not part of NCERT Class 11 Physics.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why must a Carnot engine take in and give out heat isothermally?

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   To be reversible, heat must flow with no finite temperature difference between the gas and the reservoir. With fixed reservoirs at T₁ and T₂, the gas must be at T₁ while taking in heat and at T₂ while giving it out.

2. Q2. Find the efficiency of a Carnot engine working between 100 °C and 0 °C.

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   η = 1 − 273/373, about 0.268, that is about 27%.

3. Q3. Which is more effective for raising the efficiency of a Carnot engine: raising the source temperature by 10 K or lowering the sink temperature by 10 K?

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   Lowering the sink. η = 1 − T₂/T₁ changes by 10/T₁ when T₂ falls by 10 K, but by 10T_₂/[T_₁(T_₁ + 10)] when T_₁ rises by 10 K. The latter is smaller because T_₂ < T_₁. This compares mathematically allowed temperature changes, not their engineering cost; the sink must remain above 0 K.