Entropy change in irreversible processes — always - 0: Worked examples

This page is the drill hall for the parent topic. We take the single law $Δ S_{univ} \geq 0$ and hit it from every angle: system entropy up, system entropy down, zero (reversible), degenerate, real-world, and an exam twist. Before we start, a few symbols in plain words.

Definition Every symbol we use, defined once

$Q$ = heat: energy that crosses a boundary because of a temperature difference. Sign convention (used everywhere on this page): $Q > 0$ means heat enters the object we are looking at; $Q < 0$ means heat leaves it. So if a hot body gives up heat $Q$ to a cold body, the hot body sees $- Q$ and the cold body sees $+ Q$ .
$W$ = work done by the system on its surroundings (e.g. a gas pushing a piston outward). First law: $Δ U = Q - W$ — internal energy $U$ rises with heat added and falls by the work the system gives away.
$n$ = number of moles of gas (one mole $= 6.02 \times 1 0^{23}$ molecules); it just counts how much gas there is.
$R = 8.314$ J·mol⁻¹·K⁻¹ = the ideal-gas constant, the fixed conversion factor in $p V = n R T$ . It is what turns "moles and kelvin" into "joules."
$c$ = specific heat capacity: joules needed to warm one kilogram by one kelvin (units J·kg⁻¹·K⁻¹). A mass $m$ warmed by $d T$ soaks up $δ Q = m c d T$ .
$C \equiv m c$ = heat capacity of a whole object (units J·K⁻¹): the joules to warm that object by one kelvin. When a problem hands us $C$ directly we don't need $m$ and $c$ separately.

Two tools we will reuse. Both are just the definition $d S = δ Q_{rev} / T$ evaluated on a chosen reversible path (because $S$ is a state function — path doesn't matter, only endpoints).

The scenario matrix

Cell	What makes it special	Example
A $Δ S_{sys} > 0$ , no surroundings	isolated, irreversible	Ex 1 (free expansion)
B two reservoirs, finite $Δ T$	classic "hot→cold"	Ex 2 (heat conduction)
C $Δ S_{sys} < 0$ but $Δ S_{univ} > 0$	system entropy falls	Ex 3 (water freezing)
D temperature changes → need $ln$	thermal equilibration	Ex 4 (two blocks)
E limiting case $\to$ reversible	$Δ S_{univ} \to 0$	Ex 5 (shrink the gap)
F degenerate / zero input	$Δ T = 0$ , or $V$ unchanged	Ex 6 (equal-temp mixing)
G real-world word problem	numbers, units, phase change	Ex 7 (coffee cooling)
H exam twist — mixing counting	Boltzmann view cross-check	Ex 8 (gas mixing)

We now cover all eight.

Ex 1 — Cell A: free expansion (isolated, entropy up)

Worked example Statement

$n = 2$ mol of ideal gas at $T = 300$ K expands from $V$ to $3 V$ into vacuum. Rigid insulated box. Find $Δ S_{univ}$ .

Forecast: $Q = 0$ and $W = 0$ . Guess whether $Δ S_{univ}$ is $0$ or $> 0$ before reading on.

Steps.

$Δ U = Q - W = 0 - 0 = 0$ , so for an ideal gas $T$ stays $300$ K. Why this step? First law fixes the endpoint state; we need both endpoints before we can use a state function.
Pick a reversible isothermal path between the same endpoints ( $V \to 3 V$ at $300$ K). Along it $δ Q_{rev} = p d V$ and $p = n R T / V$ (here $n = 2$ mol and $R = 8.314$ J·mol⁻¹·K⁻¹, both defined above): $Δ S_{sys} = \int_{V}^{3 V} \frac{p d V}{T} = n R ln \frac{3 V}{V} = n R ln 3.$ Why this step? The real path is irreversible so $δ Q / T$ can't be used directly; but $S$ only cares about endpoints, so we invent a reversible route.
Numbers: $2 \times 8.314 \times ln 3 = 18.27$ J/K.
Surroundings: nothing crossed the boundary, $Δ S_{surr} = 0$ . So $Δ S_{univ} = 18.27$ J/K $> 0$ .

Verify: positive, as an irreversible process must be. Units: $mol \cdot J mol^{- 1} K^{- 1} = J/K$ ✔. See Free expansion of an ideal gas.

Read the figure below: the left box is the gas packed into $V$ (red dots crowded left); the right box is the same molecules after the partition is pulled, now roaming the full $3 V$ . The picture is the entropy rise — more room to be in means more ways to be arranged, and the black "expand" arrow marks the one-way street: the dots never spontaneously crowd back left.

Figure — Entropy change in irreversible processes — always - 0

Ex 2 — Cell B: heat flow across a finite temperature gap

Worked example Statement

$Q = 1000$ J of heat conducts from a hot reservoir at $T_{H} = 400$ K to a cold one at $T_{C} = 300$ K. Find $Δ S_{univ}$ .

Forecast: the hot body loses entropy, the cold body gains. Which wins? (Hint: same $Q$ , smaller $T$ on the cold side.)

Steps.

Hot reservoir loses $Q$ at its own temperature (heat leaves, so its entropy change uses $- Q$ ): $Δ S_{H} = - \frac{Q}{T _{H}} = - \frac{1000}{400} = - 2.500 J/K .$ Why this step? A reservoir is huge; its temperature doesn't budge, so the fixed- $T$ recipe applies exactly.
Cold reservoir gains the same $Q$ at its temperature: $Δ S_{C} = + \frac{Q}{T _{C}} = + \frac{1000}{300} = + 3.333 J/K .$ Why this step? Conservation of energy: every joule that left the hot side arrives at the cold side, so the same $Q$ enters here — but divided by the cold temperature, which is smaller.
Add: $Δ S_{univ} = - 2.500 + 3.333 = + 0.833 J/K$ . Why this step? The universe is just system + surroundings, so its entropy change is the plain sum of the two reservoirs' changes.

Verify: $Δ S_{univ} = Q (\frac{1}{T _{C}} - \frac{1}{T _{H}})$ , and since $T_{C} < T_{H}$ the bracket is positive ✔.

Read the figure below: the red arrow is the $1000$ J heat-packet crossing from the hot box to the cold box. The same packet "costs" $Q / T_{H}$ where it leaves but "buys" $Q / T_{C}$ where it lands; because the landing temperature is lower, the purchase beats the cost — that surplus is the entropy the universe manufactures.

Ex 3 — Cell C: system entropy goes DOWN (water freezing)

Worked example Statement

$m = 0.100$ kg of water at $0^{\circ}$ C ( $T_{m} = 273$ K) freezes in a freezer whose air is at $T_{env} = 263$ K. Latent heat of fusion $L = 3.34 \times 1 0^{5}$ J/kg. Show $Δ S_{sys} < 0$ yet $Δ S_{univ} > 0$ .

Forecast: freezing = molecules lining up = more ordered = system entropy drops. Does the law break? Guess before reading.

Steps.

Heat released by the water as it freezes: $Q = m L = 0.100 \times 3.34 \times 1 0^{5} = 3.34 \times 1 0^{4}$ J. This leaves the water (so for the water $Q$ is negative). Why this step? Freezing dumps latent heat; that heat is what the surroundings receive.
Water (system) loses $Q$ at constant $273$ K: $Δ S_{sys} = - \frac{Q}{273} = - 122.3 J/K (< 0) .$ Why this step? A phase change happens at one fixed temperature ( $273$ K), so the fixed- $T$ recipe applies; the minus sign is because heat leaves the water.
Freezer air gains $Q$ at $263$ K: $Δ S_{surr} = + \frac{Q}{263} = + 127.0 J/K .$ Why this step? The very same $Q$ now enters the freezer's cold air (a reservoir at fixed $263$ K); plus sign because heat is received.
Total: $Δ S_{univ} = - 122.3 + 127.0 = + 4.66 J/K > 0$ . Why this step? Summing system and surroundings gives the universe; the cold air's gain (heat banked at the lower $263$ K) outweighs the water's loss.

Verify: system entropy is negative (order increased) but the universe still rose ✔. This is the textbook proof that "entropy always rises" refers only to the universe, not the system. Compare Reversible vs irreversible processes.

Ex 4 — Cell D: temperatures CHANGE → the $ln$ tool

Worked example Statement

Two identical copper blocks, each with heat capacity $C = m c = 500$ J/K (e.g. $m = 0.176$ kg of copper with $c = 2840$ J·kg⁻¹·K⁻¹ — but only the product $C$ matters here). One starts at $T_{1} = 400$ K, one at $T_{2} = 200$ K. They touch and reach a common final temperature $T_{f}$ . Find $T_{f}$ and $Δ S_{univ}$ .

Forecast: guess $T_{f}$ (is it the plain average?) and guess the sign of $Δ S_{univ}$ .

Steps.

Equal heat capacities and no heat lost outside ⇒ energy balance $C (T_{f} - T_{1}) + C (T_{f} - T_{2}) = 0$ , giving $T_{f} = \frac{T _{1} + T _{2}}{2} = 300$ K. Why this step? We need the endpoint temperature before any entropy formula; equal $C$ makes it the arithmetic mean.
Each block's temperature moves, so we must use $Δ S = C ln (T_{f} / T_{i})$ , not $Q / T$ . Why this step? Heat enters/leaves over a range of temperatures; only the $ln$ recipe sums the slices correctly.
Hot block cools $400 \to 300$ : $Δ S_{1} = 500 ln \frac{300}{400} = - 143.8$ J/K. Cold block warms $200 \to 300$ : $Δ S_{2} = 500 ln \frac{300}{200} = + 202.7$ J/K. Why this step? We now put actual numbers into the $ln$ tool for each block; the ratio $< 1$ (cooling) gives a negative log, the ratio $> 1$ (warming) gives a positive log — the signs must come out this way.
$Δ S_{univ} = - 143.8 + 202.7 = + 58.9 J/K > 0$ . Why this step? Adding the two block changes gives the universe; the cold block warmed while it was cold (dividing by small $T$ ), so it gained more entropy than the hot block lost — that is why the total is positive.

Verify: general form $C ln \frac{T _{f}^{2}}{T _{1} T _{2}} = 500 ln \frac{30 0 ^{2}}{400 \cdot 200} > 0$ since $T_{f}^{2} = 90000 > T_{1} T_{2} = 80000$ (arithmetic mean beats geometric mean) ✔.

Read the figure below: the black curve is $Δ S = 500 ln (T /300)$ ; the two red dots mark the blocks' starting temperatures. The hot dot slides down to $300$ K along a shallow part of the curve (small entropy loss), while the cold dot climbs up to $300$ K along a steep part (large entropy gain). The lopsided steepness of the $ln$ curve is exactly why the net is positive.

Ex 5 — Cell E: limiting case, shrink the gap → reversible

Worked example Statement

Redo Ex 2 but let the cold reservoir sit at $T_{C} = 399$ K (gap only $1$ K below $T_{H} = 400$ K), still $Q = 1000$ J. Then imagine $T_{C} \to 400$ K.

Forecast: as the gap closes, does $Δ S_{univ}$ shrink toward $0$ or blow up?

Steps.

$Δ S_{univ} = Q (\frac{1}{399} - \frac{1}{400}) = 1000 \times (0.0025063 - 0.0025000) = + 0.00627 J/K$ . Why this step? Same recipe as Ex 2; we're just watching the number as the driving gradient shrinks.
Take the limit $T_{C} \to T_{H}$ : the bracket $\to 0$ , so $Δ S_{univ} \to 0$ . Why this step? A vanishing temperature difference is the definition of a reversible heat exchange — no wasted gradient, no entropy produced.

Verify: tiny positive ( $0.00627 ≪ 0.833$ of Ex 2), heading to $0$ ✔. This is the $Δ S_{univ} = 0$ boundary that reversible processes live on — see Carnot cycle and efficiency.

Ex 6 — Cell F: degenerate input, zero gradient

Worked example Statement

$Q = 1000$ J flows between two reservoirs both at $T = 300$ K. And separately: $n = 1$ mol of gas undergoes a reversible isothermal step in which its volume does not change ( $V \to V$ ). Find all three entropies in each case.

Forecast: with no gradient and no change, what must every entropy be?

Steps.

Reservoirs: hot side $- Q /300$ , cold side $+ Q /300$ , so $Δ S_{univ} = Q (\frac{1}{300} - \frac{1}{300}) = 0.$ Why this step? Equal temperatures = no direction for heat to prefer = zero entropy production (reversible limit).
Gas with $V \to V$ (with $n = 1$ mol): the system's entropy is $Δ S_{sys} = n R ln \frac{V}{V} = n R ln 1 = 0.$ Why this step? $ln 1 = 0$ : identical endpoints mean zero change for the state function $S$ .
Since the volume never changed, no heat crossed the boundary, so the surroundings also see nothing: $Δ S_{surr} = 0, Δ S_{univ} = Δ S_{sys} + Δ S_{surr} = 0.$ Why this step? We must state all three; the law's equality case $Δ S_{univ} = 0$ is precisely what a degenerate (no-change) input produces.

Verify: every entropy is exactly $0$ ✔. Degenerate inputs land precisely on the reversible boundary — no violation, just the equality case.

Ex 7 — Cell G: real-world word problem (coffee cooling)

Worked example Statement

A $0.250$ kg cup of coffee (treat as water, $c = 4186$ J·kg⁻¹·K⁻¹) cools from $T_{1} = 350$ K to room temperature $T_{room} = 293$ K. Find $Δ S_{sys}$ , $Δ S_{surr}$ , $Δ S_{univ}$ .

Forecast: the coffee's temperature changes (use $ln$ ), but the room is a giant reservoir at fixed $293$ K (use $Q / T$ ). Which side of the mismatch produces the entropy?

Steps.

Heat capacity of the coffee: $C = m c = 0.250 \times 4186 = 1046.5$ J/K. Heat leaving the coffee as it cools: $Q = C (T_{1} - T_{room}) = 1046.5 \times (350 - 293) = 5.965 \times 1 0^{4}$ J. Why this step? This total $Q$ is what the room receives; we need it for the reservoir side.
Coffee cools over a range ⇒ $ln$ tool: $Δ S_{sys} = C ln \frac{T _{room}}{T _{1}} = 1046.5 ln \frac{293}{350} = - 185.3 J/K .$ Why this step? The coffee's temperature slides continuously from $350$ to $293$ K, so each scoop of heat leaves at a different $T$ ; only the $ln$ tool adds those slices up correctly (the fixed- $T$ recipe would be wrong here).
Room (fixed $T$ ) absorbs all $Q$ : $Δ S_{surr} = + \frac{Q}{293} = + 203.6 J/K .$ Why this step? Reservoir temperature never moves, so the fixed- $T$ recipe is exact.
$Δ S_{univ} = - 185.3 + 203.6 = + 18.3 J/K > 0$ . Why this step? Summing gives the universe; the room banks every joule at the single low temperature $293$ K, while the coffee shed those joules at temperatures as high as $350$ K — that mismatch is the entropy produced.

Verify: system down (cooling = less spread), universe up ✔. The room "collects" heat that departed at up to $350$ K but banks it all at $293$ K — the temperature mismatch is exactly the irreversibility.

Ex 8 — Cell H: exam twist, mixing (Boltzmann cross-check)

Worked example Statement

A box is split in two equal halves. Left half: $n = 1$ mol of gas A. Right half: $n = 1$ mol of gas B, same $T$ , same $p$ . Remove the partition; the gases interdiffuse to fill the whole box. Find $Δ S_{univ}$ , and cross-check with counting.

Forecast: nothing gets hotter or colder, no heat flows — so is $Δ S_{univ}$ zero? (Trap!)

Steps.

Each gas free-expands from $V /2$ to $V$ (each species now occupies the whole box). Using Ex 1's tool per gas (each with $n = 1$ mol): $Δ S = n R ln \frac{V}{V /2} + n R ln \frac{V}{V /2} = 2 n R ln 2.$ Why this step? Mixing is two simultaneous free expansions; each doubles its available volume, so $ln (2)$ appears twice.
For $n = 1$ mol each: $Δ S_{univ} = 2 R ln 2 = 11.53$ J/K $> 0$ (surroundings unchanged, since no heat crossed any boundary). Why this step? We put a concrete number in ( $n = 1$ mol per gas, $R = 8.314$ ) to see the size of the effect and confirm it is strictly positive — mixing really did produce entropy despite zero heat flow.
Boltzmann cross-check (Statistical interpretation of entropy (Boltzmann S = k ln W)): entropy of mixing is $- R n_{tot} \sum x_{i} ln x_{i}$ . With mole fractions $x_{A} = x_{B} = \frac{1}{2}$ and $n_{tot} = 2$ mol: $Δ S = - 2 R (\frac{1}{2} ln \frac{1}{2} + \frac{1}{2} ln \frac{1}{2}) = 2 R ln 2 = 11.53 J/K .$ Why this step? The macroscopic "reversible path" answer must equal the microscopic counting answer — two roads, one number.

Verify: both routes give $2 R ln 2 = 11.53$ J/K ✔. Mixing raises entropy with zero heat flow — the classic exam trap the " $Q = 0 \Rightarrow Δ S = 0$ " mistake falls into.

Recall One-line self-test per cell

Ex 1 free expansion ::: $Δ S = n R ln 3 = 18.27$ J/K, surroundings $0$ . Ex 2 heat flow ::: $Q (1/ T_{C} - 1/ T_{H}) = + 0.833$ J/K. Ex 3 freezing ::: $Δ S_{sys} < 0$ ( $- 122.3$ ) but $Δ S_{univ} = + 4.66$ J/K. Ex 4 two blocks ::: $T_{f} = 300$ K, $Δ S_{univ} = + 58.9$ J/K. Ex 5 tiny gap ::: $+ 0.00627$ J/K $\to 0$ as gap closes. Ex 6 degenerate ::: all three entropies $= 0$ . Ex 7 coffee ::: $+ 18.3$ J/K. Ex 8 mixing ::: $2 R ln 2 = 11.53$ J/K, matches counting.