Carnot engine: the ceiling on efficiency
"Carnot engine: the ceiling on efficiency" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the lesson "Carnot engine" in Thermodynamics. Set the two reservoir temperatures. The arrows are drawn to scale: heat in from the top, work out to the right, heat rejected to the bottom. A Carnot engine rejects Q₂ = Q₁T₂/T₁, so η = 1 − T₂/T₁ with kelvin temperatures. It reaches 1 only if T₂ = 0 K, and every real engine between the same two temperatures does worse. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Carnot engine in Thermodynamics. Change the values and watch what happens.
In JEE Main
42 shifts analysed
48 questions from Thermodynamics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §11.11
- Question posed by Sadi Carnot, a French engineer, in 1824: with a hot reservoir at T₁ and a cold one at T₂, what is the largest possible efficiency of a heat engine, and what cycle gives it? He found the right answer before the basic ideas of heat were firmly settled.
- The ideal engine must be reversible. Heat exchange with a finite temperature difference is not quasi-static, so heat must be taken in isothermally at T₁ and given out isothermally at T₂.
- To move the working substance between T₁ and T₂ without other reservoirs, only reversible adiabatic steps will do; any other process, e.g. isochoric, would need a whole series of reservoirs between T₂ and T₁.
- A reversible engine working between two temperatures is a Carnot engine. The Carnot cycle, with an ideal gas: 1→2 isothermal expansion at T₁, absorbing Q₁ = W₁₂ = μRT₁ ln(V₂/V₁); 2→3 adiabatic expansion from T₁ to T₂, work by the gas μR(T₁ − T₂)/(γ − 1).
- 3→4 isothermal compression at T₂, releasing Q₂ = W₃₄ = μRT₂ ln(V₃/V₄) (work done on the gas); 4→1 adiabatic compression from T₂ back to T₁, work on the gas μR(T₁ − T₂)/(γ − 1).
- The two adiabatic works cancel. Net work W = μRT₁ ln(V₂/V₁) − μRT₂ ln(V₃/V₄), and the adiabatic steps (TV^(γ−1) = constant) give V₃/V₄ = V₂/V₁.
- Efficiency: η = W/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁, with temperatures in kelvin. For example, between 500 K and 300 K, η = 1 − 300/500 = 0.40 = 40%.
- Every step can be reversed. Run backwards, the cycle takes Q₂ from the cold reservoir, has work W done on it, and delivers Q₁ to the hot reservoir: a reversible refrigerator.
- Carnot's theorem: (a) no engine working between two given temperatures is more efficient than a Carnot engine; (b) the Carnot efficiency does not depend on the working substance.
- Proof of (a): couple an irreversible engine I, used as an engine, to a Carnot engine R, run as a refrigerator that returns the same Q₁ to the source. If I were more efficient (W′ > W), the pair would take W′ − W of heat from the cold reservoir and turn it wholly into work with no other change, which the Kelvin–Planck statement forbids.
- Since Q₂/Q₁ = T₂/T₁ holds for every Carnot engine, whatever the substance, it can define a universal thermodynamic temperature scale; with an ideal gas as the working substance this scale is the same as the ideal-gas temperature.
Questions
3 answers
- What does the "Carnot engine: the ceiling on efficiency" simulation show?
- Set the two reservoir temperatures. The arrows are drawn to scale: heat in from the top, work out to the right, heat rejected to the bottom.
- Does JEE Main ask questions from Thermodynamics?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Carnot engine" in Thermodynamics (Class 11 Physics).
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