Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.12
Entropy and Disorder.
A hot cup of tea left on the table goes cold, but a cold cup never warms itself up. Energy is not lost in either direction, yet things only run one way. Why? The answer is one quantity: entropy.
1. Entropy: a measure of disorder
The measure of how disordered a system is, is called entropy, and in nature disorder keeps growing. The book says it in one line: “Entropy is a measure of disorderliness”. Clausius's definition takes the same idea from the other side: the thermal property of a body that stays constant in an adiabatic process is entropy.
Entropy is a measurable physical property of a body. It shows which way heat flows and helps fix the body's thermodynamic state, which is why it is called thermal inertia. Like temperature, volume and pressure it is a natural quantity: its value depends on the present state of the body, not on the path taken to reach it or on its history. Absorb or reject heat and it changes. Unlike temperature or pressure, though, you cannot feel it.
Eight cards in order. Shuffle them, then see whether they ever tidy themselves again.
From order to disorder
The cards are a model; 8 cards can be laid out in 8! = 40320 ways.
When entropy increases, a body passes from an ordered state to a disordered state.
2. dS = dQ/T and its sign
Let a system absorb or reject heat dQ at absolute temperature T. The change of entropy is dS = dQ/T (1.28). T is in kelvin and dQ in joules, so the SI unit of entropy is the joule per kelvin, JK⁻¹.
Heat absorbed (dQ > 0) raises entropy; heat rejected (dQ < 0) lowers it. So melting ice gives a positive change and freezing water a negative one. For each process below ask one thing only: is heat absorbed, rejected, or neither?
Which way entropy goes
3. The second law in terms of entropy
In a reversible adiabatic process no heat is absorbed or rejected, so dQ = 0 and the change of entropy is zero, dS = 0. That is exactly what Clausius's definition said.
Clausius's first law: the total energy of the universe is constant (conservation of energy). Second law: the entropy of the universe keeps increasing (the law of increase of entropy).
The mathematical form of the second law: SB − SA = ∫A→B dQ/T (1.29). For two nearby states dS = dQ/T, that is dQ = T dS (1.30).
In an irreversible process entropy increases; in a reversible process it stays constant. Most processes in the universe are irreversible, so the entropy of the universe goes on rising.
4. Entropy round the Carnot cycle
In the Carnot cycle, along AB (isothermal expansion) heat Q₁ is absorbed and the entropy change is Q₁/T₁. Along CD heat Q₂ is rejected, the change is Q₂/T₂ and entropy falls. Along the adiabatic BC and DA the change is zero. In the Carnot cycle Q₁/T₁ = Q₂/T₂, so the total dS = 0.
An irreversible engine has efficiency η′ below that of a reversible one, η (Carnot's theorem). From this we get T₂/T₁ < Q₂/Q₁, that is Q₂/T₂ − Q₁/T₁ > 0. So entropy increases in an irreversible process.
Tap the four steps: entropy rises in the first, stays constant in the second and fourth, and falls in the third.
Entropy over four steps
| step | heat | entropy |
|---|---|---|
| A→B isothermal expansion | Q₁ absorbed | rises, +Q₁/T₁ |
| B→C adiabatic expansion | none | constant |
| C→D isothermal compression | Q₂ rejected | falls, −Q₂/T₂ |
| D→A adiabatic compression | none | constant |
| total: Q₁/T₁ − Q₂/T₂ = 0 | ||
5. Heat death
Every spontaneous change moves toward equilibrium and raises entropy. When entropy reaches its maximum everything will be at one temperature, and heat energy can no longer turn into mechanical energy. This is the heat death of the universe. Step through the chain.
Heat death
6. Entropy sums
In every sum the temperature is in kelvin: °C + 273. The first is the book's example 1.8(1): heating 5 kg of water from 10°C to 100°C gives dS = ∫dQ/T = ms ln(T₂/T₁) = 5 × 4200 × ln(373/283) = 21000 × 0.2761 ≈ 5.799 × 103 JK⁻¹. A calculator gives ln = 0.2763, so the answer is close to 5.80 × 103.
Example 1.8(2): 4 kg of water at 100°C turning to steam (L = 2.26 × 106 J kg⁻¹) happens at a fixed temperature, so dS = mL/T = 4 × 2.26 × 106/373 ≈ 2.42 × 104 JK⁻¹. Example 1.8(3): the same problem with L = 2.268 × 106 gives dS = 4 × 2.268 × 106/373; the book also prints 2.42 × 104 JK⁻¹, and the exact value is about 2.43 × 104.
Melting 1 kg of ice at 0°C (L = 3.36 × 105 J kg⁻¹) happens at a fixed temperature: dS = mL/T = 3.36 × 105/273 ≈ 1230.8 ≈ 1.23 × 103 JK⁻¹. And heating 0.01 kg of water from 0°C to 10°C gives dS = ms ln(T₂/T₁) = 0.01 × 4200 × ln(283/273) ≈ 1.5 JK⁻¹.
A two-stage sum is just two terms added. Heating 20 kg of water from 20°C to 100°C and then vaporising it: 20 × 4200 × ln(373/293) + 20 × 2.26 × 106/373 ≈ 2.03 × 104 + 1.212 × 105 ≈ 1.41 × 105 JK⁻¹. In the other direction, 1 kg of water at 100°C turning into ice at 0°C (L = 336000, s = 4200) gives dS = 4200 ln(273/373) − 336000/273 ≈ −1311 − 1231 ≈ −2541.6 JK⁻¹. It is negative because heat is rejected.
The calculator below carries these six examples, and you can type your own numbers. Nothing you enter goes anywhere. The notebook is yours.
Entropy sums
The number is a model, not a measurement of you.
7. Check yourself
Three questions in the exam's style. Start with the first.
Question 1
Which of the following is the mathematical form of the second law of thermodynamics?
In the second, watch the sign and the direction.
Question 2
What is the change of entropy in an irreversible process?
The last one may make you think.
Question 3
About entropy:
i. it increases in natural changes
ii. it stays constant in a reversible adiabatic process
iii. its net change is zero in a reversible process
The whole note on eight cards.
Eight cards
Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.12.