No Engine Can Ever Turn All Its Heat Into Work

The first law says energy cannot be created or destroyed. It would allow an engine that turns every joule of heat into work. Yet no such engine exists — every car engine and power station throws away a large share of its heat.

The reason is the second law of thermodynamics, which sets a firm limit on how much work heat can give.

This part covers the two statements of the second law, reversible and irreversible processes, the Carnot cycle, and Carnot efficiency.

Kelvin-Planck: no process is possible whose sole result is absorbing heat from a reservoir and converting it completely into work. Clausius: no process is possible whose sole result is transferring heat from a colder body to a hotter body.

Heat engine. It absorbs $Q_1$ from a hot source, does work $W$ and rejects $Q_2$ to a cold sink:

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

Refrigerator. Work $W$ removes $Q_2$ from a cold space and releases $Q_1=Q_2+W$; its coefficient of performance is $\alpha =\frac{Q_2}{W}$.

Worked example 1 — an engine. It absorbs $1000$ J and rejects $600$ J:

$$W=400\text{ J}\eta =\frac{400}{1000}=0.4=40\%$$

Worked example 2 — a refrigerator. It removes $300$ J from inside using $100$ J of work: $\alpha =3$.

An everyday example. A car's radiator and exhaust get hot — that is the heat $Q_2$ every engine must reject to its surroundings.

The substance. **Kelvin-Planck forbids $\eta =1$ and Clausius forbids a refrigerator with $W=0$** — neither is ruled out by the first law alone.

A reversible process can be run backwards so that both the system and its surroundings return exactly to their original states; any process involving friction, viscosity, sudden changes or heat flow across a finite temperature difference is irreversible.

Causes of irreversibility:

- Friction and viscous drag, turning work into heat
- Heat flowing from a hot to a cold body across a temperature gap
- Free expansion of a gas into a vacuum
- Mixing of different substances

Worked example — free expansion versus slow expansion. $1$ mol of ideal gas at $300$ K doubles its volume.

- Free expansion into vacuum: $W=0$, $Q=0$, so $\Delta U=0$ — no work is obtained
- Slow, reversible isothermal expansion: $W=8.314\times 300\times \ln 2\approx 1.73\times 10^3$ J

Same start and end states, but the irreversible path **wastes the chance to get $1.73\times 10^3$ J of work.

An everyday example. A drop of ink spreading through a glass of water never gathers itself back into a drop.

The substance. Every real process is irreversible**; the reversible process is an ideal that gives the most work possible.

**The Carnot cycle is a reversible cycle of four steps — isothermal expansion at the source temperature $T_1$, adiabatic expansion down to $T_2$, isothermal compression at the sink temperature $T_2$, and adiabatic compression back to $T_1$.

- Step 1 — isothermal expansion** at $T_1$: absorbs heat $Q_1$ and does work
- Step 2 — adiabatic expansion: temperature falls from $T_1$ to $T_2$ with no heat exchange
- Step 3 — isothermal compression at $T_2$: rejects heat $Q_2$
- Step 4 — adiabatic compression: temperature rises back to $T_1$

On a P-V diagram, the cycle is a closed loop bounded by two isotherms and two steeper adiabats; the enclosed area is the net work $W=Q_1-Q_2$.

Worked example. $1$ mol of ideal gas runs between $600$ K and $300$ K, doubling its volume in step 1. Because the two adiabats give equal volume ratios, step 3 halves the volume:

$$Q_1=8.314\times 600\times \ln 2\approx 3.46\times 10^3\text{ J}{Q}_2=8.314\times 300\times \ln 2\approx 1.73\times 10^3\text{ J}$$

$$W\approx 1.73\times 10^3\text{ J}\eta =\frac{1.73}{3.46}=50\%$$

An everyday example. A thermal power station takes in heat at a high temperature in its boiler and dumps heat at a low temperature in its cooling towers — the same source-and-sink pattern.

The substance. No real engine runs a true Carnot cycle, but it is the benchmark every engine is compared with.

**For a Carnot engine, $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$, so its efficiency is $\eta =1-\frac{T_2}{T_1}$; Carnot's theorem states that no engine working between the same two temperatures can be more efficient than a reversible one.

Derivation outline.** From the isotherms, $Q_1=\mu R{T}_1\ln \frac{V_2}{V_1}$ and $Q_2=\mu R{T}_2\ln \frac{V_3}{V_4}$. The adiabats, with $T{V}^{\gamma -1}$ constant, give $\frac{V_3}{V_4}=\frac{V_2}{V_1}$, so

$$\frac{Q_2}{Q_1}=\frac{T_2}{T_1}\Rightarrow \eta =1-\frac{T_2}{T_1}$$

Worked example 1. Steam at $800$ K with a sink at $300$ K:

$${\eta }_{max}=1-\frac{300}{800}=0.625=62.5\%$$

Worked example 2 — which change helps more?

- **Lower the sink to $250$ K**: $\eta =1-\frac{250}{800}\approx 68.8\%$
- **Raise the source to $900$ K**: $\eta =1-\frac{300}{900}\approx 66.7\%$

Why it is the limit. If some engine beat a Carnot engine, it could drive the Carnot engine backwards as a refrigerator and together they would move heat from cold to hot with no work — breaking the Clausius statement.

An everyday example. Cooling towers at a power plant keep the sink as cold as possible, because a lower $T_2$ raises the possible efficiency.

The substance. **Efficiency could reach $100\%$ only with a sink at $0$ K**, which cannot be reached — and it does not depend on the working gas.