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The Second Law & Entropy

Source lecture(s): SC133 Lec 30

Intuition

Drop dye in water: it spreads. It never un-spreads. Coffee cools; it never re-heats by chilling the room. The first law would happily allow the reverse — energy balances either way. Something else forbids it: the second law, the only fundamental law of physics with a built-in arrow of time. Its currency is entropy — and the reason is ultimately just counting: disordered arrangements outnumber ordered ones so astronomically that "toward disorder" is simply "toward the overwhelmingly probable."

Classical statements (all equivalent)

  • Clausius: heat never flows spontaneously from cold to hot.
  • Kelvin–Planck: no cyclic engine can convert heat into work with no other effect (no perfect engine).
  • Entropy: for an isolated system,
\[\boxed{\,\Delta S \geq 0\,}\]

with equality only for idealized reversible processes.

Entropy, macroscopically

\[\Delta S = \int \frac{dQ_\text{rev}}{T} \qquad [\text{J/K}]\]

A state function, like \(E_\text{int}\). Canonical example — heat \(Q\) leaks from hot \(T_h\) to cold \(T_c\):

\[\Delta S = \frac{Q}{T_c} - \frac{Q}{T_h} > 0\]

The cold body gains more entropy than the hot one loses (dividing by a smaller \(T\)) — which is why heat flows that way and not backward.

Entropy, microscopically

Boltzmann's tombstone equation:

\[S = k_B \ln W\]

\(W\) = number of microscopic arrangements consistent with the macroscopic state. Gas spreading into a vacuum: each molecule doubles its options, so \(W \to 2^N W\) and \(\Delta S = Nk_B\ln 2\). With \(N \sim 10^{23}\), the probability of spontaneous un-spreading is \(2^{-10^{23}}\) — not forbidden, just never. The second law is statistics with teeth.

Heat engines and the Carnot ceiling

Any cyclic engine takes \(Q_h\) from a hot reservoir, dumps \(Q_c\) to a cold one, and delivers \(W = Q_h - Q_c\). The second law caps its efficiency at the Carnot limit:

\[e = \frac{W}{Q_h} \leq e_\text{Carnot} = 1 - \frac{T_c}{T_h}\]

Power plants at \(T_h \approx 800\) K, \(T_c \approx 300\) K: ceiling ~62%, real ~40%. The waste heat isn't engineering sloppiness — it is the entropy tax: dumping \(Q_c\) into the cold reservoir is what pays for the entropy removed from the hot one. Refrigerators run the cycle backward, using work to pump heat cold→hot — allowed, because something else (the power station) raises entropy on your behalf.

Worked example: the melting ice cube

50 g of ice melts at 0 °C in a 25 °C room. Entropy change of the universe?

Ice: \(\Delta S_\text{ice} = \frac{mL_f}{T} = \frac{0.05\times334000}{273} \approx +61.2\,\text{J/K}\). Room: \(\Delta S_\text{room} = \frac{-16\,700}{298} \approx -56.0\,\text{J/K}\). Total: \(+5.2\,\text{J/K} > 0\) ✓ — spontaneous, irreversible, and quantifiably so.

Common mistakes

  • "Entropy always increases." Only for isolated systems. Your freezer lowers entropy locally — the power plant more than compensates. Life itself is a local entropy-lowering enterprise funded by the Sun.
  • Entropy = messiness. It counts microstates, with precise units — a shuffled desk is a metaphor, not a calculation.
  • Blaming friction for all inefficiency. Even a perfect, frictionless engine hits the Carnot wall; the limit is thermodynamic, not mechanical.
  • Using °C in Carnot's formula — ratios of temperatures require kelvin.

Knowledge graph position

Prerequisites: First law, Ideal gas, Molecular speeds. Leads to: statistical mechanics, and the deep ends of every field — from shock physics to cosmology.

Quiz

Q1 (computational). A Carnot engine runs between 500 K and 300 K, drawing 1000 J per cycle. Maximum work?

Answer

\(e = 1 - 300/500 = 0.4 \Rightarrow W_\text{max} = 400\,\text{J}\), with 600 J necessarily dumped to the cold reservoir.

Q2 (conceptual). Does a growing crystal (highly ordered!) violate the second law?

Answer

No — freezing releases latent heat into the surroundings, whose entropy gain (\(Q/T_\text{surr}\)) exceeds the crystal's loss. Order can grow locally whenever it exports more than enough disorder.

Q3 (multiple choice). Which single change raises a Carnot engine's efficiency most, per kelvin? (a) raising \(T_h\) (b) lowering \(T_c\) (c) identical effect

Answer

(b). \(\partial e/\partial T_c = -1/T_h\) vs \(\partial e/\partial T_h = T_c/T_h^2\); since \(T_c < T_h\), one kelvin off the cold side beats one kelvin onto the hot side — though engineering usually finds \(T_h\) easier to raise.

Q4 (conceptual). Why is the second law the only law with a time direction?

Answer

Microscopic laws (Newton, Maxwell) are time-reversible; entropy's arrow is statistical: forward = toward the vastly more numerous microstates. Play the film backward and no collision breaks a law — but the sequence becomes astronomically improbable.