Carnot EngineJEE Advanced

Concept

A Carnot engine runs the most efficient possible cycle between two temperatures: two isothermal steps (heat exchange) and two adiabatic steps (no heat exchange). Its efficiency depends only on the reservoir temperatures — never on the working substance.

Key formula

Derivation

Heat absorbed during isothermal expansion at $T_h$: $Q_h = nRT_h\ln(V_2/V_1)$.

Heat rejected during isothermal compression at $T_c$: $Q_c = nRT_c\ln(V_3/V_4)$.

The two adiabats give $V_2/V_1 = V_3/V_4$, so $\dfrac{Q_c}{Q_h} = \dfrac{T_c}{T_h}$, hence $\eta = 1 - T_c/T_h$.

Scenarios to explore

• Carnot Engine — The most efficient cycle — η = 1 − T_c/Tₕ on a P–V loop.
• A steam engine — 800 K source, 400 K sink: even a perfect engine gives only 50%.
• Close to the limit — A 1200 K source and 250 K sink approach 80% — the best any engine could do.

Real-world applications

• Upper bound on every real heat engine's efficiency.
• Refrigerators & heat pumps run the Carnot cycle in reverse.
• Defining the thermodynamic (Kelvin) temperature scale.

JEE exam tips

• Efficiency rises faster by lowering $T_c$ than by raising $T_h$ by the same amount.
• For a Carnot refrigerator, COP $= T_c/(T_h - T_c)$.

Common mistakes

• Using Celsius instead of Kelvin in $1 - T_c/T_h$.
• Thinking 100% efficiency is reachable — it needs $T_c = 0$ K.
• Forgetting the adiabats carry no heat, only work.

Exam traps to avoid

• No engine between the same two reservoirs can beat the Carnot efficiency (2nd law).
• Net work equals the area enclosed by the loop on the P–V diagram.