The big idea: Stir a spoon of sugar into your coffee and it spreads out and dissolves; it never gathers itself back into a neat cube. Entropy (symbol S) is the measure behind that one-way spreading.
What entropy counts: Entropy is the number of microscopic ways (microstates) the particles can be arranged while still looking the same on the outside. More ways to arrange the particles ⇒ higher entropy.
Low entropy
- A tidy, ordered state
High entropy
- A spread-out, jumbled state
Microstates in plain words: Imagine gas atoms in a box. There are vastly more ways for them to be spread evenly throughout the box than to be bunched in one corner. So the spread-out state has far more microstates — it is the high-entropy state, and it is the one you actually find.
Entropy is measured in joules per kelvin (J K⁻¹). It is a property of the whole system, like temperature or internal energy.
Counting microstates
- entropy (J K⁻¹)
- Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹
- number of microstates (no unit)
How S follows Ω: Because S depends on ln Ω, squaring Ω doubles S, and taking the square root of Ω halves S. Three coins with two heads and one tail have 3 microstates (HHT, HTH, THH), so S = kB ln 3.
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When an amount of heat ΔQ flows into or out of a body at constant temperature T, the body's entropy changes by a definite amount. The temperature must always be in kelvin.
- entropy change (J K⁻¹)
- heat transferred (J): positive in, negative out
- absolute temperature (K)
600 J of heat flows into a large reservoir held at a constant temperature of 300 K. Find the change in entropy of the reservoir.
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A change of state happens at constant temperature, so ΔS = ΔQ/T works directly, with ΔQ = mL.
20 g of water boils away at 100 °C. The specific latent heat of vaporization of water is 2.3 × 10⁶ J kg⁻¹. Calculate the change in entropy of the water.
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The second law: The second law of thermodynamics states:
The entropy of an isolated system never decreases.
What that means
- It stays the same for an ideal (reversible) process.
- It increases for every real (irreversible) process.
- The entropy of the universe as a whole therefore always goes up.
Look at the whole, isolated system: One part can lose entropy — but only if another part gains more. The law applies to the total entropy of an isolated system (one where neither mass nor energy crosses the boundary).
The second law can be stated in two other, equivalent ways. Each forbids a process that would lower the total entropy.
Clausius form
- Heat cannot flow from a colder body to a hotter body on its own.
- A fridge can do it only because work is done on it.
Kelvin form
- No engine working in a cycle can turn heat from a hot reservoir completely into work.
- Some heat must always be given out to a colder reservoir.
Entropy can fall locally: When water freezes, it gives out latent heat, so its entropy falls (ΔS = −mL/T). The colder surroundings, such as the ice around it, take in that heat at a lower temperature and gain more entropy, so the total still rises.
A working fridge with its door open in a sealed room cools its inside, but its motor and back plate give out more heat than it removes: the entropy of the room increases.
1200 J of heat flows from a hot body at 400 K to a cold body at 300 K. Treating the two bodies together as an isolated system, find the net change in entropy.
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Why real processes only run one way: The second law gives time a direction ('time's arrow'). A process happens spontaneously only if it increases the total entropy of the universe.
Heat flow: Heat flowing hot → cold raises total entropy (the worked example gave +1.0 J K⁻¹), so it happens by itself. The reverse, heat flowing cold → hot on its own, would lower total entropy, so it never happens.
Happens by itself (ΔS_total > 0)
- Heat flows from a hot body to a cold body
- A gas expands to fill an empty space
- A drop of ink spreads through water
- These are irreversible — they never un-happen on their own
Never happens by itself (would give ΔS_total < 0)
- Heat flowing from a cold body to a hot body unaided
- A spread-out gas collecting back into one corner
- Ink un-mixing out of the water
- Each would decrease total entropy, so it is forbidden
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How this is tested — entropy and the second law are HL only (B.4) and appear at two levels:
Paper 1A
- A one-step 'what is ΔS?' using ΔS = ΔQ/T.
- 'What does the second law state?' multiple-choice.
Paper 2
- Calculate a net entropy change for two bodies.
- Then comment on whether the process is possible (look for ΔStotal > 0).
The classic trap: Leaving temperature in °C. ΔS = ΔQ/T needs T in kelvin (add 273). Forgetting to convert throws every entropy value out — and if your total ΔS comes out negative for a real process, a sign or a unit has slipped.
Three easy marks: (1) Always put T in kelvin (add 273 to °C). (2) Heat in is +ΔQ, heat out is −ΔQ. (3) For a real process the total ΔS must come out positive — if yours is negative, check a sign.
The words to know
| Word | What it means |
|---|---|
| Microstate | One particular arrangement of the particles (or their energies); all are equally likely |
| Entropy S | A measure of disorder: S = kB ln Ω, or ΔS = ΔQ/T for heat at temperature T |
| Isolated system | Neither mass nor energy crosses its boundary |
| Irreversible process | A real process that does not undo itself; it raises the total entropy |
A copper block at 500 K is placed in contact with a water bath at 250 K. In a short interval 2000 J of heat flows from the block to the water, and both temperatures stay essentially constant. Determine the net entropy change of the block–water system and state whether the process can occur.
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