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MHT-CET Physics · Thermodynamics

The Carnot Engine and the Refrigerator

A Carnot engine runs between a hot source at T₁ and a cold sink at T₂ through two isothermal and two adiabatic steps, and its efficiency 1 − T₂/T₁ is the most any engine between those temperatures can reach; run backwards it is a refrigerator with coefficient of performance T₂/(T₁ − T₂).

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Why this matters

11 PYQs, 3 of them HARD. The favourite gives the efficiency before and after the sink is lowered by some kelvins and asks for both temperatures. Others ask for the work from a given heat, the source temperature for a higher efficiency, a refrigerator's room temperature, the link between efficiency and COP, and the order of the cycle's steps. One card.

Concept 1 of 1: Efficiency and Coefficient of Performance

The cycle is isothermal expansion at T₁, adiabatic expansion, isothermal compression at T₂, adiabatic compression. Its efficiency η = W/Q₁ = 1 − T₂/T₁, temperatures in kelvin. Lowering the sink by a known amount changes η, and the two equations fix both temperatures. For a fixed sink, a higher efficiency needs a hotter source: T₁ = T₂/(1 − η). A refrigerator moves heat from the cold side; its coefficient of performance β = T₂/(T₁ − T₂) = (1 − η)/η, so η = 1/(β + 1). With a diatomic working gas, the adiabatic volume ratio fixes T₂/T₁ = (V₁/V₂)^0.4.

Definition

  • η=WQ1=1−T2T1\eta = \dfrac{W}{Q_1} = 1 - \dfrac{T_2}{T_1} (227 °C and 27 °C ⇒ 40%; 50 kJ in ⇒ 20 kJ out).
  • New source for a new η, same sink: T1′=T21−η′T_1' = \dfrac{T_2}{1 - \eta'} (50% at 600 K → 70% ⇒ 1000 K).
  • Sink lowered by x: 1−T2−xT1=η′1 - \dfrac{T_2 - x}{T_1} = \eta' together with 1−T2T1=η1 - \dfrac{T_2}{T_1} = \eta.
  • Refrigerator: β=T2T1−T2=1−ηη\beta = \dfrac{T_2}{T_1 - T_2} = \dfrac{1 - \eta}{\eta}, η=1β+1\eta = \dfrac{1}{\beta + 1}.
  • First step of the cycle: isothermal expansion.

Carnot

η=1−T2T1,β=T2T1−T2\eta = 1 - \frac{T_2}{T_1}, \qquad \beta = \frac{T_2}{T_1 - T_2}

Worked example

An engine's efficiency is 1/4. Lowering its sink by 50 K raises it to 1/2. Source and sink temperatures?
Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q21Hard

Example 1 · Thermodynamics · Carnot Engine, Efficiency, and Refrigerator

A Carnot engine has efficiency 16\frac{1}{6}. It becomes 13\frac{1}{3}, when the temperature of sink is lowered by 57 K . The temperature of the source is

Using degrees Celsius in the efficiency

227 °C and 27 °C give 1 − 300/500 = 40%, not 1 − 27/227. Convert to kelvin first.

Changing the source when the sink moves

'Sink lowered by 57 K' leaves T₁ alone. Write the two efficiencies with the same T₁ and solve.
Drill 10 more on efficiency and coefficient of performance

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Watch out for (2)

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