PhysicsISC Class 11 · Thermodynamics

Heat Engines, Refrigerators and the Carnot Cycle: Second Law and Efficiency

Updated 17 October 2026 · Fact-checked

A heat engine takes heat Q₁ from a hot source, does work W and rejects Q₂ to a cold sink: η = W/Q₁ = 1 − Q₂/Q₁. For a Carnot engine η = 1 − T₂/T₁ (kelvin), the maximum possible between two temperatures. A refrigerator has coefficient of performance Q₂/W = T₂/(T₁ − T₂) when reversible.

Understand Heat Engines, Refrigerators and the Carnot Cycle

Heat engine. A device that converts heat to work in a cycle: it absorbs Q₁ from a source at T₁, delivers W, and rejects Q₂ to a sink at T₂, with W = Q₁ − Q₂ (since ΔU = 0 over the cycle). Efficiency η = W/Q₁ = 1 − Q₂/Q₁ is always less than 1.

Second law. Kelvin-Planck: no cyclic engine can take heat from a single reservoir and convert it entirely to work. Clausius: heat cannot flow by itself from a colder to a hotter body. Both statements say that some heat must be rejected to a cold sink and that heat flows spontaneously only downhill.

Reversible and irreversible processes. A reversible process can be reversed by an infinitesimal change in conditions, leaving no change elsewhere; it must be quasi-static and free of friction and dissipation. Real processes (free expansion, heat flow through a finite temperature difference, friction) are irreversible.

Carnot cycle. Four steps: isothermal expansion at T₁ (absorbs Q₁), adiabatic expansion (cools to T₂), isothermal compression at T₂ (rejects Q₂), adiabatic compression (back to T₁). Q₁/Q₂ = T₁/T₂ and η = 1 − T₂/T₁. No engine working between the same two temperatures can be more efficient than a Carnot engine; all reversible engines have the same efficiency. It is raised more by lowering T₂ than by raising T₁ by the same amount.

Refrigerator. A refrigerator works as a reversed engine: it takes Q₂ from the cold space, receives work W, and delivers Q₁ = Q₂ + W to the hotter surroundings. Coefficient of performance β = Q₂/W = T₂/(T₁ − T₂) for an ideal one.

Heat Engines, Refrigerators and the Carnot Cycle: formulas and key facts

Engine efficiency: $η = W / Q_{1} = 1 - Q_{2} / Q_{1}$; W = Q₁ − Q₂.
Carnot: $η = 1 - T_{2} / T_{1} ∣ Q_{1} / Q_{2} = T_{1} / T_{2}$; Temperatures in kelvin.
Refrigerator: $β = Q_{2} / W ∣ β = T_{2} / (T_{1} - T_{2})$; The second form is for an ideal (Carnot) refrigerator.
Energy balance: $Q_{1} = W + Q_{2}$; For a cyclic engine.

Heat Engines, Refrigerators and the Carnot Cycle in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

An ideal refrigerator

Suppose a kitchen refrigerator is idealised as a Carnot refrigerator, keeping the inside at 270 K with the room at 300 K, with an input of 100 W.

Find its coefficient of performance, the heat removed from the inside per second and the heat delivered to the room per second.

Try it first, then show the solutionSolution

β = T₂/(T₁ − T₂) = 270/30 = 9.
Heat removed per second = β × W = 9 × 100 = 900 J.
Heat delivered to the room = 900 + 100 = 1000 J every second.

Answer: β = 9; 900 W removed; 1000 W delivered to the room.

Memory hook: The room receives the heat removed plus the work done.

Tricks to remember Heat Engines, Refrigerators and the Carnot Cycle

Short memory aids. Say each one aloud once, then use it on a question the same day.

Lower the sink
For the same drop of 100 K, lowering T₂ raises efficiency more than raising T₁.
Use it when: Comparison questions on Carnot engines.
Kelvin always
η = 1 − T₂/T₁ needs absolute temperatures.
Use it when: Avoiding the °C trap.

Heat Engines, Refrigerators and the Carnot Cycle: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Efficiency of a Carnot engine · 1 mark · MCQBoard level

The efficiency of a Carnot engine depends on:
(a) the working substance
(b) the temperatures of the source and sink only
(c) the amount of heat absorbed
(d) the speed of the piston

Try it first, then show the solutionSolution

η = 1 − T₂/T₁, independent of the working substance.

Answer: (b) the temperatures of the source and sink only

Example 2: Efficiency from heats · 3 marks · Short answerBoard level

An engine absorbs 800 J of heat from the source and rejects 600 J to the sink in each cycle. Find the work done and the efficiency.

Try it first, then show the solutionSolution

W = Q₁ − Q₂ = 800 − 600 = 200 J. (1 mark)
η = W/Q₁ = 200/800. (1 mark)
= 0.25 = 25%. (1 mark)

Answer: W = 200 J; η = 25%.

Example 3: Carnot engine · 4 marks · Short answerHarder board level

A Carnot engine works between 500 K and 300 K and absorbs 2000 J from the source in each cycle. Find the efficiency, the work done and the heat rejected.

Try it first, then show the solutionSolution

η = 1 − 300/500 = 0.40. (1 mark)
W = ηQ₁ = 0.40 × 2000 = 800 J. (1 mark)
Q₂ = Q₁ − W = 1200 J. (1 mark)
Check: Q₂/Q₁ = T₂/T₁ = 0.6, so Q₂ = 1200 J. (1 mark)

Answer: η = 40%; W = 800 J; Q₂ = 1200 J.

Example 4: Raise the source or lower the sink? · 4 marks · Short answerHarder board level

A Carnot engine has a source at 400 K and a sink at 300 K. Compare the efficiency if (i) the source temperature is raised by 100 K, (ii) the sink temperature is lowered by 100 K.

Try it first, then show the solutionSolution

Original η = 1 − 300/400 = 25%. (1 mark)
(i) η = 1 − 300/500 = 40%. (1 mark)
(ii) η = 1 − 200/400 = 50%. (1 mark)
Lowering the sink is more effective. (1 mark)

Answer: 40% by raising the source; 50% by lowering the sink.

Example 5: Refrigerator performance · 5 marks · Long answerHOTS / competency

An ideal refrigerator keeps its inside at 270 K in a room at 300 K, with an input of 100 W. Find the coefficient of performance, the heat extracted per second, the heat delivered to the room per second, and the heat extracted per second if a real refrigerator works with a coefficient of performance of 4 on the same input.

Try it first, then show the solutionSolution

β = 270/(300 − 270) = 9. (1 mark)
Q₂ per second = 9 × 100 = 900 J. (1 mark)
Q₁ per second = 900 + 100 = 1000 J. (1 mark)
Real: Q₂ = 4 × 100 = 400 J per second. (1 mark)
The real machine extracts less than half of the ideal value because of irreversibilities. (1 mark)

Answer: β = 9; 900 J s⁻¹; 1000 J s⁻¹; 400 J s⁻¹ for the real machine.

Where students lose marks in Heat Engines, Refrigerators and the Carnot Cycle

The slips examiners see most often, and what to write instead.

Mistake: Using °C in the Carnot formula
Taking T₂/T₁ = 27/127.
Do this: Convert to kelvin: 300/400.
Usually costs: 2 marks.
Mistake: Calling efficiency 100% possible
Claiming an engine converts all the heat into work.
Do this: The second law forbids it; some heat Q₂ must be rejected.
Usually costs: 1 mark.
Mistake: Mixing up engine and refrigerator formulas
Using η for a refrigerator.
Do this: Refrigerator: β = Q₂/W = T₂/(T₁ − T₂).
Usually costs: 1 to 2 marks.

Exam tips

Draw the engine as a block diagram with Q₁, W, Q₂ and the temperatures.
State the Kelvin-Planck or Clausius statement exactly when asked.
Describe all four steps of the Carnot cycle in order.
Check W = Q₁ − Q₂ in every answer.

Practice questions on Heat Engines, Refrigerators and the Carnot Cycle

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
According to the Kelvin-Planck statement, it is impossible to build an engine that:
(a) works between two temperatures
(b) converts all the heat from a single reservoir into work in a cycle
(c) is reversible
(d) uses a gas
Hint
Some heat must be rejected to a cold sink.
Answer
(b) converts all the heat from a single reservoir into work in a cycle
Q2 · 2 marks · Short answerBoard level
A Carnot engine works between boiler steam at 177 °C and a condenser at 27 °C. What is its efficiency?
Hint
Convert to kelvin.
Answer
T₁ = 450 K, T₂ = 300 K; η = 1 − 300/450 = 33.3%.
Q3 · 3 marks · Short answerHarder board level
A Carnot engine with a source at 600 K has an efficiency of 50%. Find the sink temperature, and the new efficiency if the sink is raised to 360 K.
Hint
η = 1 − T₂/T₁.
Answer
T₂ = 600 × 0.5 = 300 K. New η = 1 − 360/600 = 40%.
Q4 · 3 marks · Short answerHarder board level
An engine takes 2000 J per cycle with an efficiency of 30% and makes 5 cycles each second. Find its power output and the rate of heat rejection.
Hint
W per cycle = ηQ₁.
Answer
W = 0.3 × 2000 = 600 J per cycle, so P = 3000 W. Rejected = 1400 J per cycle = 7000 W.
Q5 · 3 marks · Short answerHarder board level
A Carnot engine rejects heat to a river at 17 °C. Find the source temperature needed for an efficiency of 42%, and for 60%.
Hint
T₁ = T₂/(1 − η), in kelvin.
Answer
T₂ = 290 K. 42%: T₁ = 290/0.58 = 500 K = 227 °C. 60%: T₁ = 290/0.40 = 725 K = 452 °C.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a power plant takes in heat at a rate of 1000 MW from steam at 700 K and rejects heat to a river at 300 K. Its actual efficiency is 35%.
(i) Find the maximum possible efficiency. (1 mark)
(ii) Find the actual power output. (1 mark)
(iii) Find the rate of heat rejection. (1 mark)
(iv) Find the actual efficiency as a fraction of the ideal one. (1 mark)
Hint
The Carnot limit sets the ideal efficiency.
Answer
(i) 1 − 300/700 = 57.1%. (ii) 0.35 × 1000 = 350 MW. (iii) 1000 − 350 = 650 MW. (iv) 35/57.1 = 0.61, about 61% of the ideal value.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Efficiency of a Carnot engine?Show answer
1 − T₂/T₁.
2.Coefficient of performance of a refrigerator?Show answer
Q₂/W.
3.Name the four steps of the Carnot cycle.Show answer
Isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.
4.Can a real engine beat a Carnot engine between the same temperatures?Show answer
No.

Heat Engines, Refrigerators and the Carnot Cycle: frequently asked questions

Why can't efficiency be 100%?

The second law requires that some heat be rejected to a colder reservoir; only if the sink were at absolute zero would η = 1.

Why is the Carnot cycle important if it cannot be built?

It sets an upper limit on the efficiency of any engine working between two temperatures.

Is a refrigerator a violation of the second law?

No. Heat flows from cold to hot only because work is done on the system; the heat delivered to the room is more than the heat removed.