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The first law of thermodynamics tells you that energy is conserved — heat in plus work in equals change in internal energy. But the first law alone is too permissive: it would allow a cup of cold coffee to spontaneously heat up by absorbing thermal energy from the air, with the air simultaneously cooling. That would conserve energy. It just never happens. The second law of thermodynamics is the principle that captures this asymmetry of nature. It says, in two equivalent forms, that heat does not spontaneously flow from cold to hot, and that no heat engine can convert all the heat it absorbs into work. Together with the introduction of entropy — a state function whose total is increased by every irreversible process — the second law sets fundamental ceilings on what any engine, refrigerator or biological organism can achieve. The Carnot cycle, designed by Sadi Carnot in 1824, is the theoretically ideal heat engine; its efficiency η_C = 1 − T_C/T_H is an absolute upper bound that no real engine operating between the same two temperatures can ever exceed.
Spec mapping (AQA 7408 §3.11.2 — Thermodynamics, option C): This lesson covers the Kelvin-Planck and Clausius statements of the second law; entropy S as a state function with dS = dQ_rev/T; the principle that the total entropy of the universe increases for any irreversible process (and is constant for any reversible process); the four-stage Carnot cycle (isothermal expansion at T_H, adiabatic expansion, isothermal compression at T_C, adiabatic compression); and the Carnot efficiency formula η_C = 1 − T_C/T_H as the upper bound on any heat engine operating between two reservoirs. (Refer to the official AQA specification document for exact wording.)
Synoptic links: The second law sits above the four-process algebra of Lesson 2 and the engine cycles of Lesson 3 — it tells you what those cycles cannot do, however cleverly designed. It connects forward to Lesson 5 (refrigerators and heat pumps) by setting the COP ceiling, and outward to statistical mechanics (entropy is a measure of microscopic disorder, S = k ln W in Boltzmann's formulation — out of A-Level scope but the underlying meaning of entropy). It also touches cosmology (the heat death of the universe is the unbounded-time limit of the second law), and information theory (Shannon entropy is structurally identical to Boltzmann entropy and underlies modern data compression and cryptography).
The second law has two everyday formulations that turn out to be logically equivalent.
Kelvin-Planck statement. No process is possible whose sole result is the absorption of heat from a single reservoir and the complete conversion of that heat into work.
In other words, you cannot build a heat engine that operates with only one thermal reservoir — there must be a hot reservoir to take heat from and a cold reservoir to dump heat to. Every real heat engine has a hot side (combustion chamber, boiler, nuclear core) and a cold side (radiator, cooling tower, atmosphere). The cold-side heat rejection is not a design flaw — it is a fundamental requirement.
Clausius statement. No process is possible whose sole result is the transfer of heat from a colder body to a hotter body.
Heat does not spontaneously flow uphill in temperature. Refrigerators and heat pumps do move heat from cold to hot, but they require an external work input — that work is the "non-sole result" that exempts them from the Clausius prohibition.
The two statements look different but are equivalent. If you could violate Kelvin-Planck (converting heat from a single reservoir entirely into work), you could couple that work to a refrigerator and end up moving heat from cold to hot for free — violating Clausius. Conversely, a Clausius-violation could be coupled to a heat engine to produce free work from a single reservoir — violating Kelvin-Planck. They stand or fall together.
Exam Tip: AQA expects you to be able to state both forms and to explain why they are equivalent (in words, not formally). Specimen mark schemes typically award one mark for each correct statement and one mark for the equivalence argument.
For a reversible process at absolute temperature T, the change in entropy of the system is defined by:
dS = dQ_rev / T
Over a finite reversible process from state A to state B: ΔS = ∫_A^B dQ_rev / T.
The units of entropy are J K⁻¹. Entropy is a state function — its value at a given thermodynamic state is fixed, regardless of how the system reached that state. The change in entropy ΔS depends only on the initial and final states, not on the path taken. This is exactly analogous to internal energy U, and exactly unlike heat Q and work W (both of which are path-dependent).
For a reversible cycle, ∮dQ_rev/T = 0 — the entropy returns to its starting value after one cycle, as it must for any state function.
For an irreversible process, the entropy of the universe (system + surroundings) increases:
ΔS_universe > 0 for any irreversible process.
For a reversible process: ΔS_universe = 0.
The entropy form of the second law: The total entropy of an isolated system never decreases. It increases for any irreversible process and is constant for any reversible process.
Some example irreversible processes that increase universe entropy:
All natural processes that occur spontaneously are accompanied by an increase in the entropy of the universe. The second law makes spontaneous reversal impossible.
Question: 1000 J of heat flows directly from a thermal reservoir at 500 K to a reservoir at 300 K (e.g. by conduction through a metal bar). Calculate the change in entropy of (a) the hot reservoir, (b) the cold reservoir, (c) the universe.
Solution:
(a) ΔS_hot = − Q/T_H = − 1000/500 = −2.00 J K⁻¹.
(b) ΔS_cold = + Q/T_C = + 1000/300 = +3.33 J K⁻¹.
(c) ΔS_universe = −2.00 + 3.33 = +1.33 J K⁻¹.
The universe gained 1.33 J K⁻¹ of entropy — confirmation that direct heat transfer across a finite ΔT is irreversible.
Sadi Carnot, working in the 1820s, asked: what is the maximum possible efficiency of a heat engine operating between a hot reservoir at T_H and a cold reservoir at T_C? His answer, the Carnot cycle, consists of four reversible stages:
graph LR
S1["1: V₁, p₁<br/>at T_H"] -->|"1→2 isothermal<br/>Q_H in at T_H"| S2["2: V₂, p₂<br/>at T_H"]
S2 -->|"2→3 adiabatic<br/>cool to T_C"| S3["3: V₃, p₃<br/>at T_C"]
S3 -->|"3→4 isothermal<br/>Q_C out at T_C"| S4["4: V₄, p₄<br/>at T_C"]
S4 -->|"4→1 adiabatic<br/>heat to T_H"| S1
style S1 fill:#e74c3c,color:#fff
style S3 fill:#3498db,color:#fff
On a pV diagram, the Carnot cycle is bounded by two isotherms (1-2 at T_H, 3-4 at T_C, both hyperbolic curves) and two adiabats (2-3 and 4-1, both steeper than the isotherms). The enclosed area is the net work output per cycle.
The clean way to derive η_C uses entropy. Both adiabatic legs are reversible and adiabatic, so dQ_rev = 0 on those stages, and ΔS = 0 for each of them. The only entropy exchanges are on the isothermal legs:
Around the closed reversible cycle, the gas returns to its starting state so ΔS_cycle = 0. Therefore:
Q_H/T_H − Q_C/T_C = 0 ⇒ Q_C/Q_H = T_C/T_H
Efficiency: η = W_net/Q_H = (Q_H − Q_C)/Q_H = 1 − Q_C/Q_H = 1 − T_C/T_H.
Carnot efficiency: η_C = 1 − T_C/T_H (T in kelvin)
This is a function of the two reservoir temperatures only. It does not depend on the working fluid, the size of the engine, the speed of the cycle, the masses involved — nothing else. Any reversible heat engine operating between the same two reservoirs has the same efficiency η_C. Any irreversible engine between the same two reservoirs has lower efficiency.
Question: A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. Per cycle, it absorbs 4000 J from the hot reservoir. Calculate (a) the Carnot efficiency, (b) the work output per cycle, (c) the heat rejected per cycle, (d) the entropy changes of the two reservoirs.
Solution:
(a) η_C = 1 − T_C/T_H = 1 − 300/600 = 0.500 or 50.0%.
(b) W_net = η × Q_H = 0.500 × 4000 = 2000 J.
(c) Q_C = Q_H − W_net = 4000 − 2000 = 2000 J.
(d) ΔS_hot = − Q_H/T_H = − 4000/600 = − 6.67 J K⁻¹. ΔS_cold = + Q_C/T_C = + 2000/300 = + 6.67 J K⁻¹. ΔS_universe = 0. ✓ Reversible cycle.
The entropy change of the hot reservoir exactly cancels that of the cold reservoir — the cycle is reversible and the universe's entropy is conserved. Real engines are irreversible: they reject more than 2000 J at T_C, the cold-side entropy gain exceeds the hot-side entropy loss, and the universe's entropy increases.
Suppose a hypothetical engine had η_X > η_C between the same two reservoirs. Then for the same Q_H input, W_X > W_C. Use that extra work to drive a Carnot refrigerator in reverse (running the Carnot cycle anticlockwise, using work to pump heat from T_C to T_H). The Carnot refrigerator, by reversibility, exactly cancels its own outputs when run in reverse: it requires W_C of work to move Q_C from cold to hot and Q_H back to hot. Net result of the coupled (X + reverse Carnot) system: extra work W_X − W_C is produced from nothing, with no net heat flow from the hot reservoir. That violates Kelvin-Planck. So η_X ≤ η_C — the Carnot efficiency is an absolute upper bound.
Exam Tip: The Carnot efficiency formula η = 1 − T_C/T_H requires temperatures in kelvin, not Celsius. Using Celsius gives nonsense (and often a negative efficiency). This is the most common 2-mark loss on Carnot questions.
A Carnot engine is a theoretical construct — it requires fully reversible processes, which in turn require all heat exchanges to occur at zero temperature difference (which makes them infinitely slow) and all mechanical processes to occur without friction. No real engine satisfies these conditions. Real engines therefore have efficiencies strictly below the Carnot bound for their reservoir temperatures.
Realistic examples:
| Engine | T_H (K) | T_C (K) | η_Carnot | Real η (typical) |
|---|---|---|---|---|
| Petrol engine (peak cycle T) | ~2500 | ~300 | ~0.88 | 0.25–0.30 |
| Diesel engine | ~2700 | ~300 | ~0.89 | 0.35–0.40 |
| Coal-fired steam (subcritical) | ~810 | ~310 | ~0.62 | 0.33–0.40 |
| Supercritical / CCGT | ~1700 | ~300 | ~0.82 | 0.55–0.60 |
| Nuclear (PWR steam cycle) | ~600 | ~310 | ~0.48 | 0.30–0.35 |
Two patterns to notice:
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