JEE Mains Physics · Thermodynamics

Heat Engines, Carnot Cycle and Refrigerators

A heat engine turns part of the heat it takes from a hot reservoir into work and rejects the rest; no engine between two temperatures beats the Carnot efficiency 1 − T₂/T₁, and a refrigerator runs the cycle backwards.

All Thermodynamics notes JEE Mains Physics strategy

Why this matters

Twenty-three PYQs, five of them numerical, and none yet from 2026. Ten use the Carnot efficiency, and six of those change one reservoir temperature and write the efficiency twice. Eight move between heat, work and temperature: four find a heat or the work, three find a reservoir temperature from the heats, and one is a refrigerator. Five go beyond one engine: three put engines in series and two ask about entropy.

Concept 1 of 3: Carnot efficiency η = 1 − T₂/T₁

A heat engine takes heat from a hot reservoir, turns part of it into work and rejects the rest to a cold one. No engine working between two temperatures can beat a reversible (Carnot) engine, whose efficiency depends only on the two temperatures in kelvin. Most questions change one temperature and give the efficiency before and after: write the formula twice and solve.

Definition

$\eta = 1 - \dfrac{T_2}{T_1}$ , with $T_1$ the source and $T_2$ the sink, both in kelvin.
Any real engine between the same two temperatures has a lower efficiency than the Carnot engine.
A hotter source or a colder sink raises η. A temperature change has the same size in °C and in K.
Two conditions: write $T_2 = (1 - \eta)T_1$ for each and compare.
"Efficiency increases by 100%" means it doubles. "Increases by 30%" is usually read as 1.3 times the old value; check that your reading gives an option.
η = 1 would need a sink at 0 K, which cannot be reached.

Carnot efficiency

\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}

Worked example

A Carnot engine has an efficiency of 40% with its sink at 87 °C. By how much must the source temperature rise to make the efficiency 50%?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 1 February 2023 · Q98Moderate

Example 1 · Thermodynamics · Heat Engines, Carnot Cycle and Refrigerators

A Carnot engine operating between two reservoirs has efficiency

\frac{1}{3}

. When the temperature of cold reservoir raised by

x

, its efficiency decreases to

\frac{1}{6}

. The value of

x

, if the temperature of hot reservoir is

99^{\circ}C

, will be:

Kelvin, not Celsius

For an engine between 327 °C and 27 °C, η = 1 − 300/600, not 1 − 27/327. Only temperature differences may stay in °C.

Per cent or percentage points?

"Efficiency increases by 30%" from 50% could mean 65% or 80%. Try the relative reading, 1.3 times, first, and check it against the options.

Drill 9 more on carnot efficiency η = 1 − t₂/t₁

Concept 2 of 3: Heat, work and temperature in engines and refrigerators

Energy is conserved round an engine: heat in = work out + heat rejected. For a Carnot engine the heats are also in the ratio of the temperatures, Q₂/Q₁ = T₂/T₁. With these two equations, any two of Q₁, Q₂, W, T₁ and T₂ give the rest. A refrigerator runs the cycle backwards: work put in carries heat out of the cold side.

Definition

$Q_1 = W + Q_2$ and $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$ .
Carnot engine: $\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1}$ , so $Q_1 = \dfrac{W}{\eta}$ .
A real engine that takes $Q_1$ and rejects $Q_2$ cannot beat Carnot, so the minimum source temperature is $T_1 = T_2\dfrac{Q_1}{Q_2}$ .
Refrigerator: $\text{COP} = \dfrac{Q_2}{W} = \dfrac{T_2}{T_1 - T_2}$ for a Carnot refrigerator. Heat removed each second = COP × power.
$1\ \text{kcal} \approx 4.2 \times 10^{3}\ \text{J}$ .

Engine and refrigerator

Q_1 = W + Q_2 \qquad \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \qquad \text{COP} = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}

Worked example

A Carnot engine works between 600 K and 450 K and rejects 1500 J of heat each cycle. Find the heat it takes in and the work it does each cycle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q102Moderate

Example 2 · Thermodynamics · Heat Engines, Carnot Cycle and Refrigerators

Work done by a Carnot engine operating between temperatures

127^{\circ}C

and

27^{\circ}C

2\text{ }kJ

. The amount of heat transferred to the engine by the reservoir is:

Efficiency divides by the heat taken in

η = W/Q₁, with Q₁ the heat from the hot reservoir. Dividing the work by the rejected heat Q₂ gives a number that is too large.

COP is not an efficiency

A refrigerator's COP = Q₂/W = T₂/(T₁ − T₂) is usually larger than 1. The engine formula (T₁ − T₂)/T₁ is the wrong one for a refrigerator.

Drill 7 more on heat, work and temperature in engines and refrigerators

Concept 3 of 3: Engines in series and entropy

Two Carnot engines in series, the second taking all the heat the first rejects, act as one engine between the outer two temperatures. So the pair's efficiency is 1 − T₃/T₁, which is less than the sum of the two efficiencies. Entropy measures heat divided by the temperature at which it flows: reversible heating adds up dQ/T step by step, and an irreversible change makes the total entropy rise.

Definition

Carnot engines in series between $T_1 \to T_2 \to T_3$ : $\eta = 1 - \dfrac{T_3}{T_1} = \eta_1 + \eta_2 - \eta_1\eta_2$ , so $\eta < \eta_1 + \eta_2$ .
If the two engines do equal work, the middle temperature is $T_2 = \dfrac{T_1 + T_3}{2}$ .
Entropy: $\Delta S = \int \dfrac{dQ}{T}$ . At a fixed temperature $\Delta S = \dfrac{Q}{T}$ . Heating a mass m of specific heat s: $\Delta S = ms\ln\dfrac{T_2}{T_1}$ , with m in kg when s is per kg.
Entropy is extensive: for two parts already in equilibrium, $S = S_1 + S_2$ . Any irreversible process raises the total entropy.
Second law (Kelvin): no engine turns all the heat it takes in into work.

Engines in series and entropy

\eta = \eta_1 + \eta_2 - \eta_1\eta_2 = 1 - \frac{T_3}{T_1} \qquad \Delta S = ms\ln\frac{T_2}{T_1}

Worked example

Two Carnot engines run in series between 800 K, 400 K and 200 K, the second taking all the heat the first rejects. Find each efficiency and the efficiency of the pair.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q7Moderate

Example 3 · Thermodynamics · Heat Engines, Carnot Cycle and Refrigerators

A Carnot engine (E) is working between two temperatures 473 K and 273 K . In a new system two engines - engine

E_{1}

works between 473 K to 373 K and engine

E_{2}

works between 373 K to 273 K . If

\eta_{12}

\eta_{1}

and

\eta_{2}

are the efficiencies of the engines

E,E_{1}

and

E_{2}

, respectively, then

Efficiencies in series do not add

Two engines in series have η = η₁ + η₂ − η₁η₂, less than η₁ + η₂. The pair does no better than one Carnot engine across the whole range.

Mass units in ΔS

With s in J kg⁻¹ K⁻¹, the mass must be in kg. A mass given in grams brings a factor of 10⁻³.

Drill 4 more on engines in series and entropy

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 (3)

Carnot efficiency η = 1 − T₂/T₁
Carnot efficiency
$\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}$
Heat, work and temperature in engines and refrigerators
Engine and refrigerator
$Q_1 = W + Q_2 \qquad \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \qquad \text{COP} = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$
Engines in series and entropy
Engines in series and entropy
$\eta = \eta_1 + \eta_2 - \eta_1\eta_2 = 1 - \frac{T_3}{T_1} \qquad \Delta S = ms\ln\frac{T_2}{T_1}$

Watch out for (6)

Kelvin, not Celsius
→ Carnot efficiency η = 1 − T₂/T₁
Per cent or percentage points?
→ Carnot efficiency η = 1 − T₂/T₁
Efficiency divides by the heat taken in
→ Heat, work and temperature in engines and refrigerators
COP is not an efficiency
→ Heat, work and temperature in engines and refrigerators
Efficiencies in series do not add
→ Engines in series and entropy
Mass units in ΔS
→ Engines in series and entropy

Test yourself on Thermodynamics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Thermodynamics chapter test Drill all 23 questions