The Second Law

The question

The ocean holds an almost unimaginable amount of internal energy. The first law would happily let a ship extract heat from the sea, turn all of it into work, and sail on forever leaving a trail of slightly cooler water. No ship does — and every power plant ever built dumps more than half its fuel energy into a river or the sky. Why must every power plant throw heat away?

What the first law doesn't say

The first law is a law of bookkeeping: energy is neither created nor destroyed. But it is silent about direction. It would not object if a cold cup of coffee spontaneously got hotter by cooling the room around it, or if a puddle of water climbed back into a glass. Our experience says loudly that these things never happen. Processes have a direction; they run one way and not the other.

The captures that direction. It has two classical statements, and they are equivalent — violate one and you can build a machine that violates the other.

The Clausius statement. It is impossible for any system to operate so that its sole result is heat transfer from a cooler body to a hotter one. Heat flows downhill on its own; pushing it uphill always costs something — which is why your refrigerator has a plug.

The Kelvin–Planck statement. It is impossible for any system to operate in a thermodynamic cycle and deliver a net amount of work while exchanging heat with a single reservoir. To get work out of heat in a cycle, you need a hot reservoir to take heat from and a colder one to reject heat to. That is the answer to the question: the condenser and the cooling tower are not engineering failures; they are demanded by the second law.

Irreversibility: the real enemy

A process is if, after it happens, the system and its surroundings can both be returned exactly to their initial states. No real process is. The usual culprits, which Moran and Shapiro call irreversibilities, are:

heat transfer across a finite temperature difference — the flame at 1900 K heating steam at 800 K;
friction, of solids or fluids;
unrestrained expansion — gas rushing through a valve, or into a vacuum;
mixing of different substances or temperatures;
electrical resistance, inelastic deformation, chemical reactions running freely.

Reversible processes are idealisations — the limit of doing things infinitely slowly, with infinitely small temperature differences and no friction. But they are enormously useful, because they set the best any device could possibly do.

Carnot's engine and the ultimate efficiency

In 1824, a young French engineer, Sadi Carnot, asked what limits the efficiency of steam engines. He imagined an engine in which every process is reversible: heat taken in at a single high temperature $T_H$ , rejected at a single low temperature $T_C$ , with reversible adiabatic processes linking them.

ExploreCarnot’s engine, forwards and backwards

Hot reservoir T_H600 K · 327 °C

Cold reservoir T_C300 K · 27 °C

Carnot efficiency50.0%1 − T_C / T_H

Heat in at T_H136 kJ/kg

Heat rejected at T_C68 kJ/kg

Challenges0/3

Make a Carnot engine more than 70% efficient.
Run a refrigerator between a −18 °C freezer and a 25 °C kitchen. What is its best possible COP?
Find conditions where an ideal heat pump delivers 10 units of heat per unit of work.

An ideal gas runs the Carnot cycle: two reversible isothermal processes linked by two reversible adiabatic ones. Left, the pressure–volume diagram (log scales); right, the temperature–entropy diagram, where the cycle is a rectangle. Run it backwards and the same machine becomes a refrigerator or a heat pump.

Carnot reached two conclusions — the Carnot corollaries:

No engine operating between two reservoirs can be more efficient than a reversible one between the same reservoirs.
All reversible engines between the same two reservoirs have the same efficiency, whatever their working fluid or design.

The second corollary is remarkable: the maximum efficiency depends only on the two temperatures. Kelvin realised this lets you define temperature by it, with no reference to mercury or gas — the Kelvin scale, $Q_C/Q_H = T_C/T_H$ for a reversible cycle. It agrees exactly with the ideal-gas scale of Lesson 1. The result is the most famous formula in engineering:

$\eta_\text{max} = 1 - \frac{T_C}{T_H}.$

Temperatures in kelvin, always. For reversed cycles, the same reasoning gives the best possible coefficients of performance:

$\beta_\text{max} = \frac{T_C}{T_H - T_C}\ \ \text{(refrigerator)}, \qquad \gamma_\text{max} = \frac{T_H}{T_H - T_C}\ \ \text{(heat pump)}.$

Estimate first

A modern steam plant takes heat at about 600 °C and rejects it at about 30 °C. What is the Carnot efficiency, in percent?

110100

Drag along the line to place your guess — each tick is ten times the last.

Judging claims

The Carnot limit is the engineer's lie detector. Inventors, start-ups and patent applicants regularly claim performance that no machine can achieve. Test yourself:

ChallengeJudge the inventors

A geothermal start-up: “Our plant draws on 150 °C brine, rejects heat to a 20 °C river, and converts 25% of the heat into electricity.”

A car-engine designer: “Combustion gases peak at 2300 K, exhaust ambient is 300 K. Our engine reaches 45% efficiency.”

An ocean-energy company: “Our OTEC plant runs between 27 °C surface water and 5 °C deep water at 12% efficiency.”

A refrigerator maker: “Our freezer holds −20 °C in a 25 °C room with a coefficient of performance of 4.”

A heat-pump installer: “On a 0 °C day, our heat pump delivers 35 °C water with a heating COP of 9.5.”

A perpetual-motion enthusiast: “Our engine takes 100 kJ of heat from the ocean at 15 °C and delivers 100 kJ of work, with no other reservoir.”

Your engine: hot source800 K

Cold sink300 K

Claimed efficiency40.0%

Carnot limit62.5%

Verdictpossible64% of the limit

Your score0 / 6

Challenges0/2

Judge all six claims correctly.
Set up your own engine exactly at the Carnot limit.

Every claim below has been made, in some form, to a patent office or an investor. For each, decide: impossible (it beats the Carnot limit for its temperatures), possible (below the limit), or the reversible limit itself. Temperatures must be in kelvin.

Check your intuition

A heat pump heats a house at 20 °C from outdoor air at −10 °C. Which COP could it possibly achieve?

Going deeper
Going deeper: why the two statements are the same law

Suppose you had a refrigerator that violated Clausius — moving heat $Q$ from cold to hot with no work. Pair it with an ordinary engine that takes $Q_H$ from the hot reservoir, rejects $Q$ to the cold one, and produces $W = Q_H - Q$ . The refrigerator returns that $Q$ straight back to the hot reservoir. Net result: the cold reservoir is unchanged, the hot one gave up $Q_H - Q$ , and you got exactly that much work. A single-reservoir engine — a Kelvin–Planck violation. The argument also runs the other way. Two statements, one law.

The same style of argument proves the Carnot corollaries: if any engine beat a reversible one, run the reversible one backwards with it and you get a machine that violates Kelvin–Planck.

Recap

What to carry forward

The second law gives processes a direction. Clausius: heat doesn't flow from cold to hot on its own. Kelvin–Planck: no cycle makes work from a single reservoir.
So every power plant must reject heat to a colder sink.
Irreversibilities — friction, heat transfer across finite temperature differences, unrestrained expansion, mixing — make real devices fall short of reversible ones.
Carnot: the best efficiency between $T_H$ and $T_C$ is $1 - T_C/T_H$ , the same for every reversible engine. Reversed cycles: $\beta_\text{max} = T_C/(T_H - T_C)$ , $\gamma_\text{max} = T_H/(T_H - T_C)$ .
The Kelvin scale is defined by reversible cycles and is independent of any substance.

Next: the second law as an inequality is powerful but clumsy. Can we turn it into a property we can calculate — something that grows whenever something irreversible happens?