The big idea: Entropy (symbol S) measures the disorder of a system — or more precisely, 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. A tidy, ordered state has low entropy; a spread-out, jumbled state has high entropy.
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.
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)
Worked example — entropy gained by a reservoir
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.
Solution
- Write the given formula first:
- Heat flows in, so ΔQ = +600 J; substitute the values:
- Work it out — keep the unit:
Final answer
ΔS = +2.0 J K⁻¹. The reservoir's entropy increases because heat entered it.
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The second law: The second law of thermodynamics states:
The entropy of an isolated system never decreases.
It stays the same for an ideal (reversible) process, and 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 that exchanges no heat with its surroundings).
Worked example — net entropy change of the universe
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.
Solution
- Apply the given formula to each body and add (hot body loses heat, cold body gains it):
- Substitute ΔQ = 1200 J, Thot = 400 K, Tcold = 300 K:
- Add — keep the unit:
Final answer
ΔStotal = +1.0 J K⁻¹. It is positive, so this process is allowed and irreversible — heat naturally flows from hot to cold.
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 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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Where it shows up: Entropy and the second law are HL only (B.4):
- Paper 1A — a one-step 'what is ΔS?' or '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).
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.
IB-style question — is the process possible?
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.
Solution
- Apply the given formula to each part and add (block loses heat, water gains it):
- Substitute ΔQ = 2000 J, Tblock = 500 K, Twater = 250 K:
- Add — keep the unit:
Final answer
ΔStotal = +4.0 J K⁻¹. The total entropy increases, so the process can occur — heat flows from the hot block to the cold water.