Carnot engine

Every steam turbine, car engine and power-plant boiler is, at bottom, doing the same job: taking heat from somewhere hot and turning a slice of it into useful work, while dumping the rest somewhere cold. The question that occupied engineers in the 1820s was blunt and practical — how much of that heat can you ever turn into work, no matter how clever the machine? A young French engineer named Sadi Carnot answered it with a thought experiment so clean it still sets the ceiling on every real engine two centuries later. Carnot imagined a perfectly reversible engine — one with no friction, no wasted heat, no sudden jolts — cycling a gas between a hot reservoir and a cold reservoir through four idealised steps. He showed that such an engine's efficiency depends on nothing but the two absolute temperatures it runs between, $\eta = 1 - T_2/T_1$ , and that no real engine, however brilliantly engineered, can beat it. That single inequality is one of the deepest results in physics: it is the second law of thermodynamics wearing an engineer's hat. This article builds the Carnot cycle from the reversible-process and second-law ideas already in your Class 11 syllabus, derives the efficiency formula, and works two numerical examples in the board-exam style. A scope note up front: the CBSE 2025-26 Class 11 prescription stops at the qualitative second law and the classification of gas processes — it does not examine numerical Carnot-efficiency problems. Treat this section as an honest, self-contained preview of where your second-law study leads next, written so it does not contradict anything you are examined on this year.

Key takeaways

A Carnot engine is an idealised heat engine that carries a working substance through four fully reversible steps between a hot reservoir at $T_1$ and a cold reservoir at $T_2$ : isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression.
Because it is a closed cycle, $\Delta U_{cycle} = 0$ , so the first law gives the work output as the difference of the two heats exchanged, $W = Q_1 - Q_2$ .
Efficiency is defined for any heat engine as $\eta = W/Q_1 = 1 - Q_2/Q_1$ ; for the Carnot cycle specifically the heat ratio equals the temperature ratio, $Q_2/Q_1 = T_2/T_1$ .
That substitution gives the single most quoted formula in thermodynamics: $\eta_{Carnot} = 1 - T_2/T_1$ , using absolute (kelvin) temperatures only.
The Carnot efficiency is not just a good design target — it is the provable maximum for any engine running between $T_1$ and $T_2$ , a direct consequence of the Kelvin–Planck statement of the second law.
$\eta = 1$ would need $T_2 = 0\ \text{K}$ , an unreachable limit, so some heat rejection to the cold reservoir is unavoidable for every real engine.
Key idea: a Carnot engine's four reversible steps make its efficiency depend on the reservoir temperatures alone, $\eta = 1 - T_2/T_1$ , and this is the highest efficiency any engine — real or imagined — can reach between those two temperatures.

The core insight

A Carnot engine converts heat to work through four reversible steps between temperatures $T_1$ and $T_2$ , and the second law forces its efficiency $\eta = 1 - T_2/T_1$ to be the absolute ceiling that any real engine operating between the same two reservoirs can approach but never exceed or equal.

Where it fits in the syllabus

The CBSE Class 11 Physics 2025-26 syllabus builds thermal equilibrium and the zeroth law, the first law ( $\Delta Q = \Delta U + \Delta W$ ), the second law in its qualitative Kelvin–Planck and Clausius forms, and the classification of gas processes — isothermal ( $PV = \text{const}$ ), adiabatic ( $PV^\gamma = \text{const}$ ), reversible, irreversible, and cyclic. Heat engines, the Carnot cycle by name, a quantitative efficiency formula, and refrigerators/COP are not part of this year's prescribed Class 11 syllabus — they belong to Class 12 and beyond. This section exists because "Carnot engine" is a name every second-law discussion runs into, and understanding it deepens (without contradicting) the reversible-process ideas you are responsible for. Your Class 11 exam will not set a numerical Carnot-efficiency problem; read this as preview material that connects the dots ahead of time.

The physical picture

A heat engine is any device that absorbs heat from a hot source and delivers part of it as mechanical work, rejecting the remainder to a cold sink, while its working substance returns to its starting state at the end of every cycle. The Carnot engine is the theoretical limit of this idea: every one of its four steps is reversible, meaning it proceeds quasi-statically, through a sequence of equilibrium states, with only an infinitesimal driving temperature or pressure difference — exactly the reversibility condition you have already studied. The four steps of the Carnot cycle, carrying the working substance (often modelled as an ideal gas) between a hot reservoir at absolute temperature $T_1$ and a cold reservoir at absolute temperature $T_2$ , with $T_1 > T_2$ , are as follows. First, an isothermal expansion at $T_1$ , during which the gas absorbs heat $Q_1$ from the hot reservoir and does work as it expands at constant temperature, so its internal energy is unchanged and $Q_1$ equals the work done in that leg. Second, an adiabatic expansion, in which no heat is exchanged with either reservoir and the gas keeps expanding while its temperature falls from $T_1$ down to $T_2$ , purely at the expense of its own internal energy. Third, an isothermal compression at $T_2$ , during which the gas rejects heat $Q_2$ to the cold reservoir while being compressed at constant temperature. Fourth, an adiabatic compression, again with no heat exchange, which compresses the gas back to its original state and raises its temperature from $T_2$ back to $T_1$ , closing the cycle.

Figure

The four legs of the Carnot cycle drawn on a pressure versus volume diagram, with an isothermal expansion at the higher temperature, an adiabatic expansion, an isothermal compression at the lower temperature, and an adiabatic compression closing the loop back to the start

Because $\Delta U_{cycle} = 0$ over one complete circuit — the working substance ends exactly where it started — the first law applied to the whole cycle gives

W = Q_1 - Q_2

where: $W$ is the net work delivered by the engine in one cycle, $Q_1$ is the heat absorbed from the hot reservoir during the isothermal expansion, and $Q_2$ is the heat rejected to the cold reservoir during the isothermal compression. The physical reading is simple bookkeeping: whatever heat goes in but does not come back out as rejected heat must have left as work, since the gas itself carries away no net energy over a full cycle. The efficiency of any heat engine, Carnot or otherwise, is defined as the fraction of the absorbed heat that is converted into useful work,

\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}

where: $\eta$ is a pure number between $0$ and $1$ (often quoted as a percentage). The physical reading is that efficiency measures how little of the absorbed heat is "wasted" as rejected heat — the smaller $Q_2/Q_1$ , the closer $\eta$ is to $1$ , though the second law (below) guarantees it never actually reaches $1$ .

Derivation ACarnot efficiency from the reservoir temperatures

The goal of this derivation is to turn the general heat-engine efficiency $\eta = 1 - Q_2/Q_1$ , which involves two heat quantities that depend on the specific working substance and cycle details, into the far more useful $\eta_{Carnot} = 1 - T_2/T_1$ , which depends only on the two reservoir temperatures and nothing else about how the engine is built. The key fact borrowed here — proved formally with the ideal-gas isothermal and adiabatic relations in higher classes, and stated here as a quoted result consistent with your prescribed $PV = \text{const}$ and $PV^\gamma = \text{const}$ process relations — is that for a Carnot cycle the ratio of the two exchanged heats equals the ratio of the two absolute temperatures.

Step 1. Start from the general efficiency definition already established from the first law, $\eta = 1 - Q_2/Q_1$ , where $Q_1$ is heat absorbed at $T_1$ and $Q_2$ is heat rejected at $T_2$ . This holds for any heat engine, not just a Carnot one, and follows purely from $\Delta U_{cycle} = 0$ plus the definition of efficiency.

Step 2. Bring in the Carnot-specific result. Because every leg of the cycle is reversible and the two isothermal legs occur at the fixed temperatures $T_1$ and $T_2$ , it can be shown, using the ideal-gas isothermal relation $PV = \text{const}$ together with the adiabatic relation $PV^\gamma = \text{const}$ applied to the two adiabatic legs, that the volume ratios in the two isothermal legs are equal, which forces the heat ratio to equal the temperature ratio: $Q_2/Q_1 = T_2/T_1$ . This step is quoted rather than derived here — the full algebra with $PV^\gamma$ is beyond the Class 11 ceiling — but it is the single fact that makes the Carnot engine special.

Step 3. Substitute this ratio into the general efficiency formula from Step 1. Replacing $Q_2/Q_1$ with $T_2/T_1$ gives directly

\eta_{Carnot} = 1 - \frac{T_2}{T_1}

Step 4. Read the physical meaning of the result. Efficiency depends on nothing but the two absolute reservoir temperatures — not on the working substance, not on the size of the engine, not on how carefully any particular real machine is built. This is why the Carnot efficiency is called a limit: any actual heat engine running between the same two temperatures, having irreversibilities such as friction and finite-rate heat transfer that the Carnot engine is defined to exclude, can only do worse, so $\eta_{real} < \eta_{Carnot} = 1 - T_2/T_1$ always. The Kelvin–Planck statement of the second law — that no process can convert heat completely into work with nothing else changing — is exactly the statement that $\eta = 1$ is forbidden, and the Carnot formula shows precisely how close to $1$ you are allowed to get: only by making $T_2/T_1$ as small as possible, which means either a very hot source or a very cold sink, and never a true zero unless $T_2 = 0\ \text{K}$ , an unreachable limit.

Worked examples

Board level - efficiency and work output from given reservoir temperatures

A Carnot engine operates between a hot reservoir at $T_1 = 500\ \text{K}$ and a cold reservoir at $T_2 = 300\ \text{K}$ . Find its efficiency, and the work output if it absorbs $Q_1 = 1000\ \text{J}$ of heat per cycle from the hot reservoir.

Step 1. Compute the efficiency directly from the temperature ratio. $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.4$ . So the engine is 40% efficient — both temperatures are already in kelvin, so no unit conversion is needed here.

Step 2. Find the heat rejected using the Carnot heat-temperature ratio. $\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1} = 0.6 \Rightarrow Q_2 = 0.6 \times 1000\ \text{J} = 600\ \text{J}$ .

Step 3. Find the work output from the first law applied to the cycle. $W = Q_1 - Q_2 = 1000\ \text{J} - 600\ \text{J} = 400\ \text{J}$ .

Check. $\eta = W/Q_1 = 400/1000 = 0.4$ , matching Step 1 exactly — the two independent routes to $\eta$ agree, which is a useful self-check in an exam. The engine's efficiency is 40%, and its work output is 400 J per cycle, with the remaining 600 J unavoidably rejected to the cold reservoir as required by the second law, not lost to any engineering flaw.

Competitive level - minimum hot-reservoir temperature for a target efficiency

An engineer wants a Carnot engine, operating with a cold reservoir fixed at $T_2 = 290\ \text{K}$ (roughly room temperature), to reach an efficiency of at least 60%. Find the minimum hot-reservoir temperature $T_1$ required, in both kelvin and degrees Celsius.

Step 1. Write the efficiency condition as an inequality. $\eta = 1 - \dfrac{T_2}{T_1} \geq 0.60$ .

Step 2. Isolate the temperature ratio and invert it. $\dfrac{T_2}{T_1} \leq 0.40 \Rightarrow T_1 \geq \dfrac{T_2}{0.40} = \dfrac{290}{0.40} = 725\ \text{K}$ . Note the direction of the inequality flips correctly on inverting, since both sides are positive.

Step 3. Convert to Celsius for a physical feel: $T_1 \geq 725 - 273 = 452\,^\circ\text{C}$ .

Step 4. State the conclusion and its general lesson. The hot reservoir must be at least 725 K, about 452 degrees Celsius. For a fixed cold-reservoir temperature, raising efficiency always demands a hotter source — no amount of engineering cleverness in the moving parts can substitute for raising $T_1$ or lowering $T_2$ , because the Carnot limit depends on temperatures alone. Timing note: under two minutes. Rearrange $\eta \geq 1 - T_2/T_1$ straight to $T_1 \geq T_2/(1-\eta)$ in one line instead of re-deriving it each time — here $T_1 \geq 290/(1-0.6) = 290/0.4 = 725\ \text{K}$ . The one trap to watch is the inequality flip when you invert $T_2/T_1 \leq 0.40$ to $T_1 \geq T_2/0.40$ ; get the direction backwards and you would report a maximum instead of a minimum temperature.

Common traps

Using Celsius temperatures directly in $\eta = 1 - T_2/T_1$ — this formula requires absolute (kelvin) temperatures, since the ratio is tied to the thermodynamic (kelvin) temperature scale, not an arbitrary zero point like Celsius.
Believing a real engine can be built to match Carnot efficiency by using better materials or reducing friction to a very low but nonzero level — the Carnot limit is set by the temperatures alone together with the requirement of full reversibility at every step, which no real, finite-rate process can ever fully achieve.
Thinking $\eta = 1$ is achievable if $Q_2$ could somehow be made zero — that would require $T_2 = 0\ \text{K}$ (absolute zero), which is unattainable, so $\eta < 1$ always for any real pair of reservoirs.
Confusing the Carnot engine, an idealised standard used purely for comparison, with a specific real machine — no actual engine "is" a Carnot engine; every real engine operating between the same two temperatures is strictly less efficient than the Carnot value.
Forgetting the inequality direction when solving for a reservoir temperature from a target efficiency, as in the competitive example — dividing by a fraction less than one flips the sense of the inequality.

Common trapNo real engine can reach, let alone beat, the Carnot efficiency for its operating temperatures — this is a law of nature, not a current engineering limitation

It is tempting to think of $\eta_{Carnot} = 1 - T_2/T_1$ as merely a very good target that sufficiently advanced engineering might one day reach or exceed. It is not. It is a direct mathematical consequence of the second law of thermodynamics (the Kelvin–Planck statement), so no future material, no frictionless bearing, no exotic working fluid can ever push a real engine's efficiency to equal, let alone surpass, the Carnot value for the same two reservoir temperatures. Every real irreversibility — friction, a finite (non-infinitesimal) temperature difference driving heat flow, turbulence — only ever lowers efficiency below the Carnot ceiling; nothing can raise it above.

You should be able to…

List the four steps of the Carnot cycle in order and state that all four are reversible.
Explain why $W = Q_1 - Q_2$ follows from applying the first law to the closed Carnot cycle, where $\Delta U_{cycle} = 0$ .
State and correctly apply $\eta_{Carnot} = 1 - T_2/T_1$ , always using absolute (kelvin) temperatures, and rearrange it to solve for a target reservoir temperature.
Explain, qualitatively, why the Carnot efficiency is the maximum possible for any engine working between two given temperatures, connecting this to the Kelvin–Planck statement of the second law.
Recognise this as preview/extension material beyond the current CBSE Class 11 syllabus, built on the reversible-process and second-law ideas that genuinely are part of your Class 11 course.

Ask the tutor: Want to connect the Carnot cycle back to the reversible-process and second-law ideas from your Class 11 syllabus, or try a few more efficiency and reservoir-temperature calculations — ask the AI tutor to walk through them with you.

Why do we even study an engine that cannot really be built?

The Carnot engine is a limiting case, exactly the way "a frictionless surface" or "a massless string" is a limiting case in mechanics — you never meet one, but reasoning about it tells you the absolute best any real system can do. Once you know the ceiling $\eta = 1 - T_2/T_1$ , you immediately know that a claimed engine efficiency above that value, for its stated operating temperatures, must be a mistake or a fraud. It is a benchmark, not a blueprint.

My textbook uses 373 K and 300 K straight into the formula without converting from Celsius first — is that a shortcut I can copy?

That is not a shortcut, that is simply the temperatures already being given in kelvin, which is exactly what the formula needs. If a question instead states temperatures in degrees Celsius, such as 100 degrees Celsius and 27 degrees Celsius, you must add 273 to each before using $\eta = 1 - T_2/T_1$ , giving 373 K and 300 K. Skipping that conversion and plugging Celsius numbers straight in is one of the most common mark-losing errors on this topic.

Is this on my Class 11 exam, since my textbook does mention the second law?

No, not this year. The CBSE 2025-26 Class 11 syllabus covers the second law only in its qualitative Kelvin–Planck and Clausius forms and does not include the Carnot cycle by name or any numerical efficiency calculation — those appear in Class 12 and competitive-exam syllabi. This article exists so the idea does not feel unfamiliar when you meet it there; nothing here should conflict with what your Class 11 paper actually tests.

Why is it specifically the ratio of heats that equals the ratio of temperatures, and not something else?

That specific equality, $Q_2/Q_1 = T_2/T_1$ , is a consequence of combining the ideal-gas isothermal relation $PV = \text{const}$ with the adiabatic relation $PV^\gamma = \text{const}$ across all four legs of the cycle — the algebra shows the volume ratios in the two isothermal legs must match, which in turn forces the heat ratio to match the temperature ratio. The full derivation needs gas-law algebra a step beyond the Class 11 ceiling, so this section quotes the result rather than deriving it from scratch; you will derive it properly in Class 12.