Second law and Carnot cycle

No engine, however cleverly built, converts all the heat it absorbs into useful work — this is the hard limit the second law of thermodynamics places on every heat engine ever imagined, from a coal-fired power plant to a car engine. The first law only tracks energy bookkeeping; it says nothing about which processes actually happen or how efficient an engine can possibly be. The second law fills that gap, and the Carnot cycle names the best any engine could ever do between two given temperatures. This section builds the idea of reversible and irreversible processes, states the second law in Kelvin's and Clausius's forms, walks through the four steps of the idealised Carnot cycle on a $p$ – $V$ diagram, and applies the efficiency formula $\eta = 1 - T_2/T_1$ to numericals — exactly as prescribed for this syllabus, with no formal proof required.

The core insight

No cyclic heat engine can be 100% efficient because some heat must always be rejected to a colder sink, and the Carnot cycle — an idealised reversible engine of isothermal and adiabatic legs — sets the theoretical ceiling $\eta = 1 - T_2/T_1$ on that efficiency.

The core idea

Reversible vs irreversible. A reversible process is an idealised process that can be exactly retraced, restoring both the system and its surroundings to their original states, passing through a continuous sequence of equilibrium states with no friction or dissipation. An irreversible process — essentially every real process — leaves some permanent change behind once it runs (friction, uncontrolled expansion, heat flowing across a finite temperature difference). Real engines are always irreversible to some degree; the Carnot cycle is the reversible limit they can only approach.

The second law, two ways. The first law of thermodynamics allows a "perfect" engine (100% efficient, converting all absorbed heat to work) and a "perfect" refrigerator (moving heat from cold to hot with no work input) — yet neither is ever observed. The second law forbids both:

Kelvin's statement: it is impossible to construct a heat engine that, operating in a cycle, converts heat completely into work with no other effect.
Clausius's statement: it is impossible to construct a device that, operating in a cycle, produces no effect other than transferring heat from a colder body to a hotter body.

The heat engine. A heat engine absorbs heat $Q_1$ from a hot reservoir (source) at temperature $T_1$ , converts part of it into work $W$ , and rejects the rest, $Q_2$ , to a cold reservoir (sink) at temperature $T_2 < T_1$ , returning to its initial state each cycle. Its efficiency is

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

where: $W$ is the net work output per cycle, $Q_1$ is the heat absorbed from the source, and $Q_2$ is the heat rejected to the sink. Since $Q_2$ can never be zero (Kelvin's statement), $\eta$ can never reach 1.

The Carnot cycle. The Carnot cycle is an idealised, reversible cycle giving the most efficient heat engine possible between two fixed temperatures. On a $p$ – $V$ diagram it has four legs: isothermal expansion at $T_1$ (absorbing $Q_1$ from a source of infinite thermal capacity, so its temperature stays fixed), adiabatic expansion cooling the gas from $T_1$ to $T_2$ (requiring perfect thermal insulation, no heat exchange), isothermal compression at $T_2$ (rejecting $Q_2$ to the sink), and adiabatic compression returning the gas to $T_1$ and its starting state.

Figure

The four steps of the Carnot cycle on a pressure volume diagram forming a closed loop of two isothermal curves and two steeper adiabatic curves with the enclosed area equal to the net work done

For a Carnot engine the two heat quantities are proportional to the two absolute temperatures, $Q_2/Q_1 = T_2/T_1$ , so its efficiency depends only on $T_1$ and $T_2$ , not on the working substance.

Derivation ACarnot efficiency from the temperature ratio

The goal is to turn the Carnot-engine heat relation $Q_2/Q_1 = T_2/T_1$ — the defining property of the reversible Carnot cycle, given here as a starting fact rather than proved from the four $p$ – $V$ legs — into the working formula $\eta = 1 - T_2/T_1$ that the syllabus asks you to apply directly in numericals, showing exactly which substitution collapses heat ratios into a temperature ratio.

Step 1. Start from the general definition of heat-engine efficiency, which holds for any cyclic engine, Carnot or not: $\eta = W/Q_1$ . Using energy conservation over one complete cycle (the first law applied to a cycle, where the internal energy returns to its starting value), $W = Q_1 - Q_2$ , so $\eta = (Q_1 - Q_2)/Q_1 = 1 - Q_2/Q_1$ .

Step 2. Bring in the Carnot-specific fact. For a reversible Carnot cycle operating between a source at $T_1$ and a sink at $T_2$ , the heat exchanged on the two isothermal legs obeys $Q_2/Q_1 = T_2/T_1$ (stated here as the prescribed result; the syllabus does not require its derivation from the cycle's $p$ – $V$ legs).

Step 3. Substitute this ratio into the Step 1 result. Replacing $Q_2/Q_1$ with $T_2/T_1$ gives $\eta = 1 - T_2/T_1$ , with $T_1$ and $T_2$ both in kelvin. This is the Carnot efficiency formula, and it shows the efficiency depends only on the two reservoir temperatures — nothing about the gas, its amount, or the engine's design enters.

Step 4. Read off the physical limit. Since $T_2 > 0$ K always (absolute zero is unattainable), $T_2/T_1$ is always strictly positive, so $\eta < 1$ always — no engine, not even this ideal one, reaches 100% efficiency. This is Kelvin's statement wearing numbers.

Worked example

A Carnot engine operates between a source at $227^\circ\text{C}$ and a sink at $27^\circ\text{C}$ , absorbing 2000 J of heat from the source per cycle. Find the efficiency, the work output per cycle, and the heat rejected per cycle.

Step 1. Convert both temperatures to kelvin, since the formula requires absolute temperatures: $T_1 = 227 + 273 = 500\ \text{K}$ , $T_2 = 27 + 273 = 300\ \text{K}$ .

Step 2. Apply the Carnot efficiency formula: $\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{500} = 1 - 0.6 = 0.4$ , so the efficiency is 40%.

Step 3. Find the work output using $\eta = W/Q_1$ : $W = \eta \times Q_1 = 0.4 \times 2000 = 800\ \text{J}$ .

Step 4. Find the heat rejected using $Q_2 = Q_1 - W$ : $Q_2 = 2000 - 800 = 1200\ \text{J}$ . Check: $Q_2/Q_1 = 1200/2000 = 0.6 = T_2/T_1 = 300/500$ — consistent.

Final answer: efficiency $\eta = 40\%$ , work output $W = 800\ \text{J}$ per cycle, and heat rejected $Q_2 = 1200\ \text{J}$ per cycle — even this ideal engine cannot convert all 2000 J into work, exactly as Kelvin's statement demands.

Competitive stretch

A Carnot engine has efficiency 30% with its sink at 27°C. Keeping the sink fixed, how much must the source temperature rise to reach 50% efficiency? Sink: $T_2 = 300\ \text{K}$ throughout. First case: $1 - 300/T_{1a} = 0.30 \Rightarrow T_{1a} = 300/0.70 \approx 428.6\ \text{K}$ . Second case: $1 - 300/T_{1b} = 0.50 \Rightarrow T_{1b} = 300/0.50 = 600\ \text{K}$ . Required rise: $\Delta T_1 = 600 - 428.6 \approx \textbf{171.4 K}$ .

Common traps

Using Celsius temperatures directly in $\eta = 1 - T_2/T_1$ — the formula requires absolute (kelvin) temperatures; Celsius values give a wrong, sometimes negative, answer.
Believing efficiency can reach 100% by making $T_2$ very small but nonzero — $\eta$ approaches but never reaches 1 unless $T_2 = 0$ K, which is physically unattainable.
Mixing up Kelvin's statement (about a perfect engine converting heat completely to work) with Clausius's statement (about a perfect refrigerator moving heat cold-to-hot with no work input) — they describe different devices, though they express the same law.
Trying to derive $\eta = 1 - T_2/T_1$ from scratch in an exam — the syllabus expects you to apply the formula, not re-derive it from the four $p$ – $V$ legs.

Ask the tutor: want to see how the Carnot cycle's four legs trace out on an actual $p$ – $V$ graph with numbers, or work through more efficiency-and-heat numericals?

Why can Q2/Q1 = T2/T1 be used without deriving it — isn't that just assuming the answer?

It looks that way, but the syllabus deliberately keeps the full derivation (which needs the isothermal and adiabatic work integrals for an ideal gas) out of scope at this level — it belongs to a more advanced treatment. What you are expected to master here is the physical picture of the four-step cycle and the ability to apply $\eta = 1 - T_2/T_1$ correctly, including unit conversion and interpreting the result. Treat the ratio as a given property of the reversible Carnot cycle, the same way you'd treat a standard formula handed to you on a formula sheet.

If the Carnot cycle is only an idealisation, why bother studying it?

Because it sets the theoretical ceiling. No real engine — however well engineered — can beat the Carnot efficiency for the same two operating temperatures, since real cycles involve friction, finite-time heat transfer, and other irreversibilities that the Carnot cycle assumes away. Knowing $\eta_{\text{Carnot}} = 1 - T_2/T_1$ tells engineers the best possible payoff from raising a furnace's temperature or lowering a condenser's temperature, even though no actual power plant reaches that number.

Does raising the source temperature or lowering the sink temperature help efficiency more?

Both increase $\eta = 1 - T_2/T_1$ , but they are not symmetric in practice. Lowering $T_2$ toward absolute zero is far harder and more expensive than raising $T_1$ , since most engines already reject heat near ambient temperature. That is why real power-plant design usually focuses on running the source (boiler/furnace) hotter rather than chasing an impractically cold sink — though both routes push efficiency toward, never to, 100%.

Second law and Carnot cycle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads