Chapter 11 · Thermodynamics

11.3 The Second Law, Reversible Processes and the Carnot Engine

The first law allows any process that conserves energy; the second law says which of them can actually happen. No cyclic device turns heat completely into work, heat never flows by itself from cold to hot, and even a perfect reversible engine has efficiency $\eta=1-T_2/T_1$.

The oceans hold an enormous store of internal energy. Cooling the top hundred metres of the world's seas by just one-hundredth of a kelvin would release more energy than humanity uses in a year. So why does no ship sail by taking in warm sea water, extracting heat from it, and throwing it back as ice? Energy would be conserved — the first law has no objection. Yet no such machine has ever been built, and it never will be. In the same way, a cup of tea on a cold table never gets hotter by drawing heat from the table, although that too would conserve energy.

Something beyond energy conservation decides which way natural processes go. This section states that extra principle, the second law of thermodynamics, in two equivalent forms. It then separates reversible processes — the idealised, perfectly slow, frictionless kind that could run backwards — from the irreversible ones that make up the real world. Finally it builds the Carnot engine, a reversible cycle between a hot and a cold reservoir, and proves that its efficiency $1-\dfrac{T_2}{T_1}$ is the best that any engine can do.

The second law of thermodynamics

To turn heat into work continuously, a device must work in a cycle (Section 11.2): its working substance — steam, air, a refrigerant — must keep returning to its starting state, so that it can go round again. An isothermal expansion turns all the heat it absorbs into work, but it cannot go on for ever: the gas ends up at low pressure and must be compressed again, and the compression needs work. The question is how much of the heat absorbed in a cycle can come out as net work.

📘 Definition 11.9 — Heat reservoir

A heat reservoir (thermal reservoir) is a body so large that it can give out or take in any amount of heat without any change in its temperature — the ocean, the atmosphere, a large furnace kept at constant temperature. A reservoir at the higher temperature $T_1$ is called the source; one at the lower temperature $T_2$ is the sink.

A device working in a cycle that takes in heat $Q_1$ from a source, does net work $W$ and gives out heat $Q_2$ to a sink is a heat engine. Over a complete cycle $\Delta U=0$, so the first law gives $W=Q_1-Q_2$. Every engine ever built — steam turbine, petrol engine, jet engine — has $Q_2\gt0$: it throws some heat away. Every refrigerator needs work to be done on it to move heat from its cold interior into the warmer kitchen. These two facts of experience are the second law.

📐 Law 11.9 — Second law of thermodynamics

Kelvin–Planck statement. No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of this heat into work. (No engine working in a cycle can have $Q_2=0$.)

Clausius statement. No process is possible whose sole result is the transfer of heat from a colder body to a hotter body. (Heat can be moved from cold to hot only if work is done, as in a refrigerator.)

The two statements are equivalent: a device that broke either one could be combined with an ordinary engine or refrigerator to break the other.

The word sole matters. An isothermal expansion of an ideal gas does turn heat completely into work ($\Delta U=0$, $Q=W$), but that is not its only result — the gas is left expanded. A refrigerator does move heat from cold to hot, but not as the sole result — electrical work is used up and more heat comes out at the back than was taken from the inside.

Why the statements are equivalent. Suppose a "perfect refrigerator" broke the Clausius statement, moving heat $Q_2$ from the sink to the source with no work. Run an ordinary engine between the same reservoirs, taking $Q_1$ from the source, doing $W=Q_1-Q_2$ and dumping $Q_2$ in the sink. The perfect refrigerator returns that $Q_2$ to the source. The sink is left unchanged, and the combined device takes net heat $Q_1-Q_2$ from the source and turns all of it into work — breaking the Kelvin–Planck statement. The argument also runs the other way.

✏️ Example 1 — Checking three claims against the two laws
Which law, if any, does each claim break? (a) A ship's engine takes $5.0\ \text{MJ}$ of heat each second from sea water at $300\ \text{K}$, returns the water colder, and delivers $5.0\ \text{MW}$ of shaft power, with no other change. (b) An engine absorbs $800\ \text{J}$ per cycle from a hot reservoir, rejects $300\ \text{J}$ to a cold one, and does $600\ \text{J}$ of work. (c) A box with no power supply keeps its inside at $5\ ^\circ\text{C}$ in a room at $30\ ^\circ\text{C}$ by pumping heat out of the box into the room.
(a) Energy is conserved ($5.0\ \text{MJ}$ in, $5.0\ \text{MJ}$ of work out each second), so the first law is satisfied. But the sole result is heat taken from one reservoir and turned completely into work: it breaks the Kelvin–Planck statement.
(b) Over a cycle $W$ must equal $Q_1-Q_2=500\ \text{J}$, not $600\ \text{J}$: it creates $100\ \text{J}$ per cycle and breaks the first law.
(c) Heat would flow from the cold box to the warm room as the sole result, with no work done: it breaks the Clausius statement. (Insulation could only slow the inward leak; it cannot pump heat outward.)
Check A claim can break the first law, the second law, or neither — always check energy first ($W=Q_1-Q_2$), then the second law. Example 6 adds a stricter second-law test.

Reversible and irreversible processes

Film a pendulum swinging in a vacuum and play it backwards: nobody could tell. Film an egg breaking, a drop of ink spreading through water, a car braking to a stop with hot brakes, and the reversed film is obviously absurd. Most real processes have a direction.

📘 Definition 11.10 — Reversible and irreversible processes

A process is reversible if it can be run backwards so that both the system and its surroundings return exactly to their original states, with no change anywhere else in the universe. Any other process is irreversible.

A process can be reversible only if it is (i) quasi-static — the system passes through equilibrium states, with no finite differences of pressure or temperature inside it or between it and its surroundings — and (ii) free of dissipative effects such as friction, viscosity, electrical resistance and inelastic deformation, which turn work into internal energy.

Two features make most natural processes irreversible. The first is dissipation: friction turns the kinetic energy of a sliding block into internal energy of the block and the floor, and by the Kelvin–Planck statement there is no way to turn all of that back into work with no other change. The second is departure from equilibrium: heat flowing across a finite temperature difference, a gas rushing into a vacuum, two gases mixing, fuel burning. Reversing any of these would require heat to flow from cold to hot by itself, or a gas to collect itself back into one corner — forbidden by the second law.

A reversible process is therefore an idealisation, like a frictionless plane. A slow isothermal compression with a frictionless piston, in contact with a reservoir at the same temperature as the gas, comes close: run it backwards and the gas does the same work on the piston and returns the same heat to the reservoir. Such processes matter because, as the next part shows, they give the most work an engine can deliver.

✏️ Example 2 — Reversible or irreversible?
Classify each process and give the reason: (a) a gas is compressed very slowly by a frictionless piston, in contact with a large reservoir at its own temperature; (b) a hot iron block is dropped into a bucket of cold water; (c) a book slides across a table and stops; (d) a current flows through the heating coil of a kettle; (e) a drop of ink spreads through a glass of water.
(a) (Ideally) reversible: quasi-static, no friction, heat flows across a vanishingly small temperature difference.
(b) Irreversible: heat flows across a large, finite temperature difference; the water and iron will never sort themselves back into hot and cold.
(c) Irreversible: friction (dissipation) turns kinetic energy into internal energy.
(d) Irreversible: electrical resistance dissipates electrical work as internal energy (Joule heating).
(e) Irreversible: diffusion and mixing — a large departure from equilibrium, with no tendency to un-mix.
Check Ask "could a film of it run backwards with nothing else changing?" Only (a) passes.
✏️ Example 3 — Free expansion: nothing changes, yet it cannot be undone
A rigid insulated container is divided into two equal halves of $10\ \text{L}$ each. One half holds $1.0\ \text{mol}$ of an ideal gas at $300\ \text{K}$; the other is empty. The partition is broken. (a) Find $W$, $Q$ and $\Delta U$, and the final temperature and pressure. (b) Can this change be drawn on a $p$–$V$ diagram? (c) What would be needed to put the gas back into one half?
(a) The gas pushes against nothing, so $W=0$; the walls are insulated, so $Q=0$; hence $\Delta U=0$. For an ideal gas $U$ depends only on $T$, so $T$ stays $300\ \text{K}$. The volume doubles, so the pressure halves: from $\dfrac{nRT}{V}=2.49\times10^5\ \text{Pa}$ to $1.25\times10^5\ \text{Pa}$.
(b) No. While the gas rushes in, it has no uniform pressure or temperature — it is far from equilibrium. Only the initial and final states are points on the diagram; the process is not quasi-static.
(c) A compression, e.g. isothermal at $300\ \text{K}$: work $nRT\ln2=1.73\ \text{kJ}$ must be done on the gas and $1.73\ \text{kJ}$ of heat flows out to a reservoir. The gas is restored, but the surroundings have lost $1.73\ \text{kJ}$ of work and gained $1.73\ \text{kJ}$ of heat — they are changed. So free expansion is irreversible.
Check Compare Example 3 of Section 11.2: the same two end states reached by a slow isothermal expansion deliver $1.73\ \text{kJ}$ of work. Free expansion wastes that opportunity.

The Carnot engine and its efficiency

In 1824 the French engineer Sadi Carnot asked: given a source at $T_1$ and a sink at $T_2$, what is the greatest fraction of the heat taken from the source that any engine can turn into work? His insight was that every irreversible step wastes an opportunity to do work — heat flowing straight from a hot body into a cooler one, for example, could have driven an engine on the way. So the best engine must be reversible. It may take in heat only at $T_1$ and give it out only at $T_2$, and each heat exchange must happen across a vanishingly small temperature difference: the gas must be at $T_1$ while it touches the source and at $T_2$ while it touches the sink. In between, it must change temperature with no heat exchanged — adiabatically.

📘 Definition 11.11 — Carnot cycle and the efficiency of an engine

A Carnot engine takes a working substance (here an ideal gas) round a reversible cycle of four quasi-static steps (the Carnot cycle):

  1. 1 → 2, isothermal expansion at $T_1$ in contact with the source; the gas absorbs heat $Q_1$.
  2. 2 → 3, adiabatic expansion, insulated, until the gas has cooled to $T_2$.
  3. 3 → 4, isothermal compression at $T_2$ in contact with the sink; the gas gives out heat $Q_2$.
  4. 4 → 1, adiabatic compression, insulated, back to $T_1$ and the starting state.

The efficiency of the engine is the fraction of the heat absorbed that is turned into net work:

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

Finding $\eta$ for a Carnot engine. Take $n$ moles of ideal gas, with the four corners of the cycle at volumes $V_1$, $V_2$, $V_3$, $V_4$. On the two isotherms (Result 11.5):

$$Q_1=nRT_1\ln\frac{V_2}{V_1},\qquad Q_2=nRT_2\ln\frac{V_3}{V_4}.$$

On the two adiabats (Result 11.6, $TV^{\gamma-1}=$ const): $T_1V_2^{\gamma-1}=T_2V_3^{\gamma-1}$ and $T_1V_1^{\gamma-1}=T_2V_4^{\gamma-1}$. Dividing one by the other, $\left(\dfrac{V_2}{V_1}\right)^{\gamma-1}=\left(\dfrac{V_3}{V_4}\right)^{\gamma-1}$, so $\dfrac{V_2}{V_1}=\dfrac{V_3}{V_4}$ and the logarithms cancel:

$$\frac{Q_2}{Q_1}=\frac{T_2}{T_1}.$$

The work done by the gas in the adiabatic expansion, $nC_v(T_1-T_2)$, is exactly cancelled by the work done on it in the adiabatic compression, so the net work comes from the isotherms alone: $W=Q_1-Q_2$.

📐 Result 11.10 — Efficiency of a Carnot engine

A Carnot engine working between a source at absolute temperature $T_1$ and a sink at $T_2$ has efficiency

$$\eta=1-\frac{T_2}{T_1},\qquad\text{with}\qquad\frac{Q_2}{Q_1}=\frac{T_2}{T_1}.$$

It depends only on the two temperatures (in kelvin), not on the working substance or the size of the cycle. Since $T_2\gt0$, $\eta\lt1$: even a perfect engine must reject heat.

📐 Result 11.11 — Carnot's theorem

(a) No engine working between two given reservoirs can be more efficient than a Carnot (reversible) engine working between the same reservoirs. (b) All reversible engines working between the same two reservoirs have the same efficiency, $1-\dfrac{T_2}{T_1}$.

Reason. A reversible engine can be run backwards as a refrigerator. If some engine X were more efficient than a Carnot engine, X could drive the reversed Carnot engine: together they would take heat from the sink and deliver it to the source with no net work — or, adjusted, turn heat from one reservoir entirely into work. Both break the second law.

✏️ Example 4 — A Carnot engine in numbers
A Carnot engine works between a source at $500\ \text{K}$ and a sink at $300\ \text{K}$ and absorbs $1000\ \text{J}$ from the source in each cycle. Find (a) its efficiency, (b) the work done per cycle, (c) the heat rejected to the sink. (d) What would the efficiency be if you wrongly used $227\ ^\circ\text{C}$ and $27\ ^\circ\text{C}$?
(a) $\eta=1-\dfrac{300}{500}=0.40=40\%$.
(b) $W=\eta Q_1=0.40\times1000=400\ \text{J}$.
(c) $Q_2=Q_1-W=600\ \text{J}$; check $\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}=0.60$ ✓.
(d) $1-\dfrac{27}{227}=0.88$ — badly wrong. Temperatures in efficiency formulas are always absolute.
✏️ Example 5 — One full Carnot cycle, step by step
$1.0\ \text{mol}$ of a monatomic ideal gas is taken round a Carnot cycle between $T_1=500\ \text{K}$ and $T_2=300\ \text{K}$. The isothermal expansion goes from $V_1=10\ \text{L}$ to $V_2=20\ \text{L}$. Find (a) $Q_1$, (b) the volumes $V_3$ and $V_4$, (c) $Q_2$, the net work and the efficiency, (d) the work done in each adiabatic step.
(a) $Q_1=nRT_1\ln\dfrac{V_2}{V_1}=8.31\times500\times\ln2=2.88\times10^3\ \text{J}$.
(b) $\gamma-1=\tfrac23$: $\dfrac{V_3}{V_2}=\left(\dfrac{T_1}{T_2}\right)^{3/2}=\left(\dfrac53\right)^{1.5}=2.15$, so $V_3=43.0\ \text{L}$ and $V_4=2.15\times10=21.5\ \text{L}$.
(c) $Q_2=nRT_2\ln\dfrac{V_3}{V_4}=8.31\times300\times\ln2=1.73\times10^3\ \text{J}$; $W=2880-1728=1.15\times10^3\ \text{J}$; $\eta=\dfrac{1152}{2880}=0.40$ ✓ $=1-\dfrac{300}{500}$.
(d) Expansion 2 → 3: $W=nC_v(T_1-T_2)=\tfrac32(8.31)(200)=+2.49\times10^3\ \text{J}$. Compression 4 → 1: $-2.49\times10^3\ \text{J}$. They cancel.
Check Interactive 1 starts from this cycle: adding up strips $p\,\Delta V$ round the whole loop gives an enclosed area of $1152\ \text{J}=Q_1-Q_2$ ✓.
Interactive 1 — The Carnot Cycle

$1.0\ \text{mol}$ of monatomic ideal gas runs a Carnot cycle, expanding isothermally from $10\ \text{L}$ to $20\ \text{L}$ at the source temperature. Set the source and sink temperatures. Watch the cylinder: it touches the hot reservoir (red) on 1 → 2, is insulated on 2 → 3, touches the cold reservoir (blue) on 3 → 4 and is insulated again on 4 → 1. The enclosed area — the net work — is added up strip by strip and compared with $Q_1-Q_2$.

500 K300 K
heat absorbed $Q_1$—
heat rejected $Q_2$—
net work (area)—
efficiency $\eta$—

Carnot's result explains how real engines are designed. Power stations burn fuel to make the hottest steam their turbine blades can stand ($800$–$900\ \text{K}$) and cool the exhaust with river or sea water ($\approx300\ \text{K}$), for a Carnot limit near $65\%$. Friction, heat leaks and finite temperature differences bring the real figure to about $35$–$45\%$. The rejected heat is not a sign of bad engineering — the second law demands it.

✏️ Example 6 — An inventor's claim
An inventor claims that her engine, working between reservoirs at $400\ \text{K}$ and $300\ \text{K}$, absorbs $500\ \text{J}$ from the hot reservoir and rejects $300\ \text{J}$ to the cold one in each cycle. (a) Does the claim satisfy the first law? (b) Does it satisfy the second law? (c) What is the most work the engine could do per cycle, and the least heat it must reject?
(a) Yes, provided the work done is $W=Q_1-Q_2=200\ \text{J}$ per cycle.
(b) Claimed efficiency $\dfrac{200}{500}=40\%$. Carnot limit $1-\dfrac{300}{400}=25\%$. The claim beats a reversible engine, so by Carnot's theorem it breaks the second law — impossible.
(c) $W_\text{max}=0.25\times500=125\ \text{J}$, and then $Q_{2,\text{min}}=375\ \text{J}$.
Check Equivalent test: a reversible engine has $\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}$, i.e. $\dfrac{Q_2}{T_2}=\dfrac{Q_1}{T_1}$; any real engine must reject more, $\dfrac{Q_2}{T_2}\ge\dfrac{Q_1}{T_1}$. Here $\dfrac{300}{300}=1.0\lt\dfrac{500}{400}=1.25$ ✗. Interactive 2 tests any claim this way.
Interactive 2 — Is This Engine Possible?

An engine takes $Q_1=1000\ \text{J}$ per cycle from a source. Set the reservoir temperatures and the work the inventor claims. The energy-flow diagram must balance ($Q_2=Q_1-W$, first law), and the claimed efficiency is compared with the Carnot limit (second law). Find the boundary where a claim changes from "possible" to "impossible".

400 K300 K400 J
claimed $\eta=W/Q_1$—
Carnot limit $1-T_2/T_1$—
heat rejected $Q_2$—
verdict—
⚠️ Common mistakes

1. Celsius in $1-\dfrac{T_2}{T_1}$. The ratio $\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}$ holds only for absolute temperatures. Convert to kelvin first.

2. "The second law forbids converting heat completely into work." Only as the sole result of a cycle. A single isothermal expansion does convert heat completely into work — but leaves the gas changed.

3. "A reversible process is one that can go backwards." It must go back with the surroundings restored too. A gas can be pushed back after a free expansion, but the surroundings are left changed.

4. Efficiency as $\dfrac{Q_2}{Q_1}$. Efficiency is $\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1}$; $\dfrac{Q_2}{Q_1}$ is the fraction wasted.

5. Forgetting the adiabats. In a Carnot cycle no heat flows on the adiabats, but work is done on both; the two works cancel, which is why $W=Q_1-Q_2$ comes from the isotherms.

6. "A better design could reach 100%." No engine between $T_1$ and $T_2$ can beat $1-\dfrac{T_2}{T_1}$, and that equals $1$ only for a sink at absolute zero, which cannot be reached.

🚀 Beyond the board syllabus (JEE Advanced) — heat engines, refrigerators and heat pumps

For JEE Advanced only: its 2026 syllabus lists the Carnot engine and its efficiency, and general heat engines and refrigerators are set as applications of it. Not in the CBSE syllabus; the JEE Main and NEET 2026 syllabi stop at the second law and reversible and irreversible processes.

Any heat engine — petrol, diesel, steam turbine — takes $Q_1$ from a hot source, does work $W$ and rejects $Q_2$ in each cycle, with efficiency $\eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1}$ (Carnot's theorem: $\eta\le1-\dfrac{T_2}{T_1}$). Typical values: car petrol engine $25$–$30\%$, large diesel engine $40$–$50\%$, combined-cycle gas power station about $60\%$.

A refrigerator is an engine run backwards: work $W$ is done on the working substance (by the compressor), which takes heat $Q_2$ from the cold space at $T_2$ and delivers $Q_1=Q_2+W$ to the warmer surroundings at $T_1$. Its coefficient of performance is the heat removed per unit work,

$$\beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2},\qquad\beta_\text{Carnot}=\frac{T_2}{T_1-T_2}.$$

A heat pump is the same machine used to warm a building: what matters is the heat delivered, $\text{COP}_\text{hp}=\dfrac{Q_1}{W}=\beta+1$, at most $\dfrac{T_1}{T_1-T_2}$. Both can exceed 1 — that is not "free energy": the work only moves heat. For a reversible engine and refrigerator between the same reservoirs, $\beta=\dfrac{1-\eta}{\eta}$.

Example. A freezer at $-10\ ^\circ\text{C}$ ($263\ \text{K}$) in a room at $30\ ^\circ\text{C}$ ($303\ \text{K}$) freezes $1.0\ \text{kg}$ of water already at $0\ ^\circ\text{C}$ ($L=334\ \text{kJ/kg}$). The best possible $\beta=\dfrac{263}{40}=6.6$, so the work needed is at least $\dfrac{334}{6.58}=51\ \text{kJ}$, and at least $385\ \text{kJ}$ of heat goes into the room. Real freezers, with $\beta\approx2$–$3$, need about three times that work.

Exercises

Temperatures in the efficiency formula must be in kelvin ($T/\text{K}=t/^\circ\text{C}+273$). Take $R=8.31\ \text{J mol}^{-1}\text{K}^{-1}$. All engines work in cycles between two reservoirs.

I
The second law.
  • 1 State the Kelvin–Planck statement of the second law of thermodynamics.
  • 2 State the Clausius statement of the second law.
  • 3 Outline why a device that broke the Clausius statement could be used to break the Kelvin–Planck statement.
  • 4 A ship's engine is designed to run on heat taken from the sea, with no other reservoir. Which law forbids it? Does it break the first law?
  • 5 Can you cool a closed kitchen by leaving the refrigerator door open? Explain.
  • 6 Explain the difference between what the first law and the second law tell us about a proposed process.
  • 7 Can heat ever flow from a colder body to a hotter one? Explain, with an example.
1 No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work.
2 No process is possible whose sole result is the transfer of heat from a colder body to a hotter body.
3 Run an ordinary engine: it takes $Q_1$ from the source, does $W$ and dumps $Q_2$ in the sink. Use the Clausius-breaking device to carry $Q_2$ back to the source with no work. The sink is unchanged; the net result is $Q_1-Q_2$ taken from the source and turned entirely into work — breaking Kelvin–Planck.
4 The Kelvin–Planck statement: it would turn heat from a single reservoir completely into work in a cycle. It does not break the first law — energy is conserved.
5 No. The refrigerator takes heat $Q_2$ from inside and gives out $Q_2+W$ at the back into the same kitchen, so the kitchen gains the work $W$ each cycle and slowly warms up.
6 The first law checks the energy account ($\Delta U=Q-W$); it allows any process that balances. The second law decides which of those processes can actually occur, and in which direction — e.g. heat flows hot to cold, engines must reject heat.
7 Yes, but not as the sole result: work must be done, as in a refrigerator or air conditioner, which pumps heat from a cold room into the hot outdoors using electrical work.
II
Reversible and irreversible processes.
  • 1 Define a reversible process. State the two conditions it must satisfy.
  • 2 Classify as reversible (ideally) or irreversible: (a) slow frictionless adiabatic compression; (b) the explosion of a petrol–air mixture; (c) a falling stone hitting the ground; (d) ice melting very slowly in surroundings only infinitesimally above $0\ ^\circ\text{C}$; (e) two gases diffusing into each other.
  • 3 $2.0\ \text{mol}$ of an ideal gas at $350\ \text{K}$ in a $5.0\ \text{L}$ insulated vessel expands freely into an evacuated, insulated $10.0\ \text{L}$ vessel joined to it. Find $Q$, $W$, $\Delta U$, the final temperature, and the initial and final pressures.
  • 4 For the gas of part 3, how much work must be done to push it back slowly and isothermally into the $5.0\ \text{L}$ vessel? Why does this show that the free expansion is irreversible?
  • 5 Explain why all natural (spontaneous) processes are irreversible.
  • 6 $1000\ \text{J}$ of heat flows by conduction directly from a reservoir at $500\ \text{K}$ to one at $300\ \text{K}$. How much work could a reversible engine have produced from the same $1000\ \text{J}$ taken from the hotter reservoir?
1 One that can be reversed so that the system and its surroundings both return exactly to their initial states. It must be quasi-static and free of dissipative effects (friction, viscosity, resistance).
2 (a) reversible; (b) irreversible (combustion, far from equilibrium); (c) irreversible (inelastic impact — dissipation); (d) reversible (heat flows across an infinitesimal temperature difference); (e) irreversible (mixing).
3 $Q=0$ (insulated), $W=0$ (expands into vacuum), $\Delta U=0$, so $T$ stays $350\ \text{K}$. $p_1=\dfrac{2.0\times8.31\times350}{5.0\times10^{-3}}=1.16\times10^6\ \text{Pa}$; the volume triples to $15.0\ \text{L}$, so $p_2=3.88\times10^5\ \text{Pa}$.
4 $W_\text{on}=nRT\ln3=2.0\times8.31\times350\times1.099=6.39\times10^3\ \text{J}$, with $6.39\ \text{kJ}$ of heat given out to a reservoir. The gas is restored but the surroundings have lost work and gained heat, so the original process cannot be undone without leaving a change.
5 They involve finite differences (of temperature, pressure, concentration) or dissipation (friction, resistance); reversing them would need heat to flow from cold to hot, or dissipated energy to turn back into work, with no other change — which the second law forbids.
6 $W=\left(1-\dfrac{300}{500}\right)\times1000=400\ \text{J}$. Direct conduction wastes this opportunity.
III
The Carnot engine.
  • 1 Describe the four steps of a Carnot cycle for an ideal gas and sketch it on a $p$–$V$ diagram, marking where heat is absorbed and rejected.
  • 2 A Carnot engine works between $600\ \text{K}$ and $300\ \text{K}$ and absorbs $2000\ \text{J}$ per cycle. Find its efficiency, the work per cycle and the heat rejected.
  • 3 A Carnot engine of efficiency $30\%$ rejects $700\ \text{J}$ per cycle to a sink at $280\ \text{K}$. Find the heat absorbed, the work done and the source temperature.
  • 4 A steam power plant has its steam at $540\ ^\circ\text{C}$ and its condenser at $40\ ^\circ\text{C}$. Find the Carnot limit on its efficiency.
  • 5 A Carnot engine works between $500\ \text{K}$ and $300\ \text{K}$. Which raises its efficiency more: raising the source temperature by $50\ \text{K}$, or lowering the sink temperature by $50\ \text{K}$?
  • 6 Two Carnot engines work in series: the first between $800\ \text{K}$ and $T$, the second between $T$ and $200\ \text{K}$, the second taking in exactly the heat the first rejects. Find $T$ if (a) their efficiencies are equal, (b) they do equal work.
  • 7 Why can no engine have an efficiency of $100\%$?
  • 8 $1.0\ \text{mol}$ of a diatomic ideal gas runs a Carnot cycle between $400\ \text{K}$ and $300\ \text{K}$, expanding isothermally from $5.0\ \text{L}$ to $10.0\ \text{L}$. Find $Q_1$, $W$, $Q_2$ and the largest volume reached.
1 Isothermal expansion at $T_1$ (heat $Q_1$ absorbed from the source); adiabatic expansion cooling to $T_2$; isothermal compression at $T_2$ (heat $Q_2$ rejected to the sink); adiabatic compression back to $T_1$. On the $p$–$V$ diagram: two hyperbolic isotherms joined by two steeper adiabats, traversed clockwise.
2 $\eta=1-\tfrac{300}{600}=50\%$; $W=1000\ \text{J}$; $Q_2=1000\ \text{J}$.
3 $Q_2=0.70\,Q_1$: $Q_1=1000\ \text{J}$, $W=300\ \text{J}$; $\dfrac{T_2}{T_1}=0.70$, $T_1=\dfrac{280}{0.70}=400\ \text{K}$.
4 $T_1=813\ \text{K}$, $T_2=313\ \text{K}$: $\eta=1-\dfrac{313}{813}=0.615\approx62\%$ (real plants reach about $40\%$).
5 Raising $T_1$: $1-\tfrac{300}{550}=45.5\%$. Lowering $T_2$: $1-\tfrac{250}{500}=50.0\%$. Lowering the sink wins: $\eta$ changes by $\dfrac{\Delta T}{T_1}$ for the sink, but only by $\dfrac{T_2\,\Delta T}{T_1^2}$ (smaller, as $T_2\lt T_1$) for the source.
6 (a) $1-\dfrac{T}{800}=1-\dfrac{200}{T}$: $T^2=800\times200$, $T=400\ \text{K}$. (b) For reversible engines $\dfrac{Q}{T}$ is the same at every reservoir, so heat is proportional to temperature and equal works need $800-T=T-200$: $T=500\ \text{K}$.
7 Even a Carnot engine has $\eta=1-\dfrac{T_2}{T_1}$, which equals $1$ only if $T_2=0\ \text{K}$; absolute zero cannot be reached, and real engines do worse than Carnot. Equivalently, the Kelvin–Planck statement forbids $Q_2=0$.
8 $Q_1=8.31\times400\times\ln2=2.30\times10^3\ \text{J}$; $\eta=0.25$, $W=576\ \text{J}$; $Q_2=1.73\times10^3\ \text{J}$. Largest volume $V_3=10.0\times\left(\tfrac43\right)^{2.5}=20.5\ \text{L}$.
IV
Applications and inventors' claims.
  • 1 (Modelling) The burning gases in a car engine reach about $1100\ \text{K}$ and the exhaust leaves at about $400\ \text{K}$. Find the Carnot limit for these temperatures. The real engine turns $25\%$ of the fuel's energy into work; for $1.0\ \text{MJ}$ of fuel energy, find the work and the heat rejected, and what fraction of the Carnot limit the engine reaches.
  • 2 (Modelling) An ocean thermal power plant uses warm surface water at $25\ ^\circ\text{C}$ and cold deep water at $5\ ^\circ\text{C}$. Find its Carnot efficiency and the least heat it must take in each second to deliver $100\ \text{MW}$.
  • 3 (Modelling) A geothermal power plant takes heat from underground water at $450\ \text{K}$ and rejects heat at $300\ \text{K}$. It reaches $60\%$ of its Carnot efficiency and delivers $10\ \text{MW}$. Find the heat taken in and rejected per second.
  • 4 Show that for a Carnot cycle $\dfrac{Q_1}{T_1}=\dfrac{Q_2}{T_2}$.
  • 5 A Carnot engine does $200\ \text{J}$ of work per cycle and rejects $300\ \text{J}$ to a sink at $27\ ^\circ\text{C}$. Find the source temperature.
  • 6 An engine working between $500\ \text{K}$ and $300\ \text{K}$ takes in $1000\ \text{J}$ per cycle and rejects $650\ \text{J}$. Is it possible? Is it reversible?
  • 7 Another engine between the same reservoirs claims to take in $1000\ \text{J}$ and reject only $550\ \text{J}$ per cycle. Comment.
1 $\eta_\text{C}=1-\dfrac{400}{1100}=0.64$. $W=0.25\times1.0\ \text{MJ}=0.25\ \text{MJ}$; $Q_2=0.75\ \text{MJ}$; $\dfrac{0.25}{0.636}=0.39$ — about $40\%$ of the Carnot limit (friction, heat loss to the cylinder walls, and the fast, irreversible combustion).
2 $\eta_\text{C}=1-\dfrac{278}{298}=0.067=6.7\%$. $Q_1\ge\dfrac{100\ \text{MW}}{0.0671}=1.5\times10^9\ \text{W}$, i.e. $1.5\ \text{GJ}$ each second — hence the huge water pipes such plants need.
3 $\eta_\text{C}=1-\tfrac{300}{450}=\tfrac13$; actual $\eta=0.60\times\tfrac13=0.20$. $Q_1=\dfrac{10}{0.20}=50\ \text{MW}$ taken in; $Q_2=40\ \text{MW}$ rejected.
4 From Result 11.10, $1-\dfrac{Q_2}{Q_1}=1-\dfrac{T_2}{T_1}$, so $\dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}$, i.e. $\dfrac{Q_1}{T_1}=\dfrac{Q_2}{T_2}$.
5 $Q_1=500\ \text{J}$, $\eta=0.40$; $T_1=\dfrac{T_2}{1-\eta}=\dfrac{300}{0.60}=500\ \text{K}$ ($227\ ^\circ\text{C}$).
6 $W=350\ \text{J}$, $\eta=35\%$, below the Carnot limit $40\%$: possible, but not reversible (a reversible engine would reach exactly $40\%$).
7 $\eta=45\%\gt40\%$: it would beat a Carnot engine between the same reservoirs, contradicting Carnot's theorem and the second law. The claim is false.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks State the Kelvin–Planck and the Clausius statements of the second law of thermodynamics.

Kelvin–Planck No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of that heat into work. [1]
Clausius No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. [1]
Exam tip Do not leave out "sole result" — without it both statements are false (isothermal expansions and refrigerators exist).

E2. NEET JEE Main Which of these processes can be (very nearly) reversible? (a) A gas expanding freely into a vacuum (b) Heat flowing along a copper rod from boiling water to ice (c) An ideal gas compressed isothermally, very slowly, by a frictionless piston in a cylinder in contact with a large reservoir (d) Water churned by a paddle wheel driven by a falling weight

Idea A process is reversible only if it is quasi-static and free of dissipative effects (friction, viscosity) and of heat flow across a finite temperature difference.
Check (a) is not quasi-static (no equilibrium states on the way); (b) is heat flow across a large temperature difference; (d) turns work into internal energy by viscous friction. Only (c) meets every condition.
Answer (c).
Exam tip Every natural process is irreversible; a reversible process is an idealisation approached by going infinitely slowly with no friction and only infinitesimal temperature differences.

E3. JEE Advanced (Numerical value) A Carnot engine has an efficiency of $40\%$ with its sink at $300\ \text{K}$. By how many kelvin must the source temperature be raised, keeping the sink fixed, to make the efficiency $50\%$?

Now $1-\dfrac{300}{T_1}=0.40$: $T_1=500\ \text{K}$.
Then $1-\dfrac{300}{T_1'}=0.50$: $T_1'=600\ \text{K}$.
Answer $100$ (K).
Exam tip Each extra 10 percentage points costs more: going from $50\%$ to $60\%$ would need a further $150\ \text{K}$ ($T_1=750\ \text{K}$).

E4. JEE Advanced (One or more correct) $1$ mol of an ideal gas runs a Carnot cycle between $600\ \text{K}$ and $300\ \text{K}$; in the isothermal expansion its volume doubles. Which statements are correct? (A) The efficiency is $50\%$. (B) The heat absorbed per cycle is $600R\ln2$. (C) The net work per cycle is $300R\ln2$. (D) In the adiabatic expansion the volume also doubles.

(A) $1-\tfrac{300}{600}=0.50$ ✓.
(B) $Q_1=nRT_1\ln\dfrac{V_2}{V_1}=600R\ln2$ ✓.
(C) $W=\eta Q_1=300R\ln2$ ✓ (equivalently $Q_2=300R\ln2$, since $\dfrac{V_3}{V_4}=\dfrac{V_2}{V_1}=2$).
(D) On the adiabat $\dfrac{V_3}{V_2}=\left(\dfrac{T_1}{T_2}\right)^{1/(\gamma-1)}=2^{1/(\gamma-1)}$ — $2^{1.5}=2.83$ for a monatomic gas, $2^{2.5}=5.66$ for a diatomic one — never 2 ✗.
Answer (A), (B), (C).
Exam tip The two isothermal volume ratios are always equal ($V_2/V_1=V_3/V_4$); the adiabatic ratio depends on $\gamma$.

E5. JEE Advanced A Carnot refrigerator keeps its cold compartment at $-3\ ^\circ\text{C}$ in a room at $27\ ^\circ\text{C}$. Its coefficient of performance is (a) $9$ (b) $10$ (c) $0.9$ (d) $0.1$

Note The coefficient of performance of refrigerators is beyond the NCERT syllabus (see the 🚀 box above); this is JEE Advanced practice (JEE Main and NEET 2026 do not list it).
Idea $\beta=\dfrac{T_2}{T_1-T_2}=\dfrac{270}{300-270}=9$.
Answer (a) $9$.
Exam tip (b) $=\dfrac{T_1}{T_1-T_2}$ is the coefficient of performance of a heat pump; (d) $=\dfrac{30}{300}$ is the efficiency of the Carnot engine between the same temperatures, and $\beta=\dfrac{1-\eta}{\eta}=\dfrac{0.9}{0.1}=9$ ✓.

E6. Board · Case study An inventor claims to have built an engine that works between a geothermal source at $450\ \text{K}$ and a river at $300\ \text{K}$, takes in $1500\ \text{J}$ of heat per cycle and does $600\ \text{J}$ of work. (i) What is the greatest possible efficiency of any engine between these reservoirs? [1] (ii) Calculate the claimed efficiency and decide whether the claim can be true. [1] (iii) For the same heat input, find the most work possible per cycle and the least heat that must be rejected. [2]

(i) Carnot limit $\eta=1-\dfrac{300}{450}=\tfrac13\approx33\%$. [1]
(ii) Claimed $\eta=\dfrac{600}{1500}=40\%\gt33\%$: it would beat a Carnot engine, so the claim breaks the second law and is false (though it obeys the first law if $900\ \text{J}$ is rejected). [1]
(iii) $W_\text{max}=\tfrac13\times1500=500\ \text{J}$; $Q_{2,\text{min}}=1500-500=1000\ \text{J}$. [2]
Exam tip Quote Carnot's theorem by name in (ii); "it violates the second law" without a comparison of efficiencies earns no mark.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How close do the world's power stations come to the Carnot limit — and is the Carnot efficiency even the right benchmark?

Why it matters. Most of the world's electricity still comes from heat engines: coal, gas and nuclear stations, geothermal and solar-thermal plants. Each percentage point of efficiency means millions of tonnes of fuel and of carbon dioxide a year, and the rejected heat warms rivers and seas. Engineers in every country compare their plants with the thermodynamic limits you have just met.

What to do.

  1. Choose four kinds of plant: a coal-fired steam station, a nuclear (pressurised-water) station, a geothermal station and a combined-cycle gas station. From engineering websites or company and government reports, find for each a typical hottest working temperature, the cooling-water temperature and the actual efficiency (electrical output ÷ heat input).
  2. Convert the temperatures to kelvin and compute the Carnot limit $1-\dfrac{T_2}{T_1}$ for each plant.
  3. Compute the ratio (actual efficiency) ÷ (Carnot limit). Which kind of plant comes closest?
  4. A Carnot engine is infinitely slow, so it delivers zero power. Engineers have suggested that an engine run for maximum power has efficiency close to $1-\sqrt{T_2/T_1}$. Compute this too and compare it with the actual values.
  5. Explain why nuclear stations, which run their steam cooler than coal stations for safety, have lower efficiencies.

Tools: Spreadsheet; published plant data; a calculator for the square roots.

Hand in: A table of $T_1$, $T_2$, actual $\eta$, Carnot limit, maximum-power estimate and the two ratios; one paragraph on which benchmark predicts real plants better, and one on where the rejected heat goes.

Going further. Find out how a combined-cycle plant uses the hot exhaust of a gas turbine to raise steam for a second, steam turbine. Show that two Carnot engines in series between $T_1$, $T$ and $T_2$ have the same overall efficiency as one engine between $T_1$ and $T_2$.

Instructor note: with typical figures — coal steam $840\ \text{K}$/$300\ \text{K}$: Carnot $64\%$, $1-\sqrt{T_2/T_1}=40\%$, actual about $38\%$; nuclear $560\ \text{K}$/$300\ \text{K}$: Carnot $46\%$, maximum-power estimate $27\%$, actual about $33\%$; geothermal $450\ \text{K}$/$300\ \text{K}$: Carnot $33\%$, estimate $18\%$, actual $12$–$16\%$. The maximum-power estimate is usually much closer to reality than the Carnot limit — a nice discussion of idealisations. For engines in series, $\left(1-\dfrac{T}{T_1}\right)+\dfrac{T}{T_1}\left(1-\dfrac{T_2}{T}\right)=1-\dfrac{T_2}{T_1}$.

🧪 Self-check quiz

0/5 correct

Q1. The Kelvin–Planck statement of the second law forbids

Q2. A Carnot engine works between $400\ \text{K}$ and $300\ \text{K}$. Its efficiency is

Q3. Which of these processes is irreversible?

Q4. The efficiency of a Carnot engine depends on

Q5. A Carnot engine absorbs $800\ \text{J}$ per cycle from a source at $400\ \text{K}$ and rejects heat to a sink at $300\ \text{K}$. The heat rejected per cycle is

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