Worked examples — Latent heat — phase transitions

This is a problem clinic for the parent topic. We will not re-teach the theory — instead we hunt down every kind of situation a latent-heat problem can be, put each in a box, and solve one example per box. When you finish, you should never meet a phase-change problem that surprises you.

Before we start, keep these two tools side by side — they are the only two formulas in play:

Water constants we reuse everywhere (memorise these once):

Symbol	Meaning	Value
$c_{\text{ice}}$	specific heat of ice	$2100$ J/kg·K
$c_w$	specific heat of liquid water	$4186$ J/kg·K
$c_{\text{steam}}$	specific heat of steam	$2010$ J/kg·K
$L_f$	latent heat of fusion (melt/freeze)	$3.34\times 10^5$ J/kg
$L_v$	latent heat of vaporisation (boil/condense)	$22.6\times 10^5$ J/kg

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The scenario matrix

Every latent-heat problem is one of these case classes. The examples below are labelled with the cell they cover, and together they hit all of them.

#	Case class	What makes it different	Example
A	Pure plateau, heat in	mass already at transition temp, only $mL$	Ex 1
B	Pure plateau, heat out (freezing/condensing)	sign flips: heat released, $L$ same magnitude	Ex 2
C	Multi-region climb (slope→flat→slope→…)	sum of several terms across kinks	Ex 3
D	Mixing / calorimetry, no leftover	heat lost = heat gained, everything used up	Ex 4
E	Mixing where one substance runs out first	must check how far the reaction gets	Ex 5
F	Degenerate / limiting input (zero mass, exact-boundary energy)	edge cases that trip formulas	Ex 6
G	Real-world word problem	sweating / cooling by evaporation	Ex 7
H	Exam-style twist: final temperature unknown	you must find where the mixture settles	Ex 8

The two big ideas hiding under all of it:

1. Direction of heat flow decides the sign, but $L$ is always the same size (melting absorbs $L$, freezing releases the identical amount).
2. You never guess the final state — you compute how much energy is available and see how far it gets you along the melt–warm–boil ladder.

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Example 1 — Cell A: pure plateau, heat in

Forecast: Only one term or several? Guess the number of joules before reading on.

1. Identify the state. The ice sits exactly at its melting point $0°C$. Why this step? If the substance is already at the transition temperature, there is no slope to climb — every joule goes straight into the plateau ($mL$).
2. Pick the tool. Phase change ⇒ use $Q=m{L}_f$. Why this step? Temperature does not change while melting, so $\Delta T=0$ and $Q=mc\Delta T$ would falsely give zero. We must switch to the plateau formula.
3. Plug in. $Q=m{L}_f=0.400\times 3.34\times 10^5=1.336\times 10^5\text{ J}=133.6\text{ kJ}$

Verify: Units: $\text{kg}\times \text{J/kg}=\text{J}$ ✓. Ex 1 of the parent melted $0.200$ kg for $66.8$ kJ; doubling the mass should double the heat — and $2\times 66.8=133.6$ kJ ✓.

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Example 2 — Cell B: pure plateau, heat OUT (freezing)

Forecast: Bigger, smaller, or equal to the melting energy for the same mass?

1. Recognise the reverse trip. Freezing is melting run backwards. Why this step? Latent heat is symmetric: the energy released on freezing equals the energy absorbed on melting for the same mass. Only the direction flips.
2. Compute the magnitude. $|Q|=m{L}_f=0.150\times 3.34\times 10^5=5.01\times 10^4\text{ J}=50.1\text{ kJ}$
3. Attach the sign. Heat is released by the water, so from the water's viewpoint $Q=-5.01\times 10^4$ J. Why this step? Sign bookkeeping (see First law of thermodynamics) matters when you later add this to other flows; the magnitude alone is what the freezer must pump away.

Verify: Same $m$ and $L_f$ as a melting problem ⇒ same magnitude, opposite sign — exactly what symmetry demands ✓. Units J ✓.

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Example 3 — Cell C: full multi-region climb

Forecast: Which single term will dominate the bill?

Look at the heating-curve figure — we cross five regions, splitting the journey at each kink.

1. Warm ice $-20°C\to 0°C$ (slope, $\Delta T=20$): $Q_1=m{c}_{\text{ice}}\Delta T=0.030\times 2100\times 20=1260\text{ J}$ Why this step? Temperature is rising ⇒ slope tool $mc\Delta T$.
2. Melt at $0°C$ (plateau): $Q_2=m{L}_f=0.030\times 3.34\times 10^5=10020\text{ J}$ Why this step? At the first kink the curve goes flat — phase change, so switch to $mL$.
3. Warm water $0°C\to 100°C$ (slope, $\Delta T=100$): $Q_3=m{c}_w\Delta T=0.030\times 4186\times 100=12558\text{ J}$ Why this step? The classic mistake is skipping this: liquid water must still climb 100 degrees before it can boil.
4. Boil at $100°C$ (plateau): $Q_4=m{L}_v=0.030\times 22.6\times 10^5=67800\text{ J}$
5. Warm steam $100°C\to 120°C$ (slope, $\Delta T=20$): $Q_5=m{c}_{\text{steam}}\Delta T=0.030\times 2010\times 20=1206\text{ J}$
6. Sum. $Q_{\text{total}}=1260+10020+12558+67800+1206=92844\text{ J}\approx 9.28\times 10^4\text{ J}$

Verify: The boiling term ($67800$ J) is the biggest — matches "Vaporise = Vast" ✓. All slopes use $mc\Delta T$, both plateaus use $mL$ ✓. Units J throughout ✓.

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Example 4 — Cell D: mixing with nothing left over

Forecast: More or less than the water's own mass?

1. Write energy conservation (see Calorimetry — method of mixtures): $\underset{m{c}_w\Delta T}{\underbrace{\text{heat lost by warm water}}}=\underset{m_{\text{ice}}{L}_f}{\underbrace{\text{heat gained by melting ice}}}$ Why this step? In an insulated container, no energy escapes — the warm side's loss equals the cold side's gain.
2. Heat released by the water cooling $25°C\to 0°C$: $Q=0.300\times 4186\times 25=31395\text{ J}$
3. Set equal and solve for the ice mass: $m_{\text{ice}}=\frac{Q}{L_f}=\frac{31395}{3.34\times 10^5}=0.0940\text{ kg}=94.0\text{ g}$ Why this step? The ice only melts (stays at $0°C$), so every joule it receives is latent — divide by $L_f$.

Verify: $94$ g of ice melted by $300$ g of water — the water carries plenty of energy but $L_f$ is large, so it melts less than a third of its mass. Reasonable ✓. Units: $\text{J}÷(\text{J/kg})=\text{kg}$ ✓.

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Example 5 — Cell E: one substance runs out first

Forecast: Fully melted pond, or an ice-water slush left over?

1. Budget the available heat. The warm water can at most fall to $0°C$, releasing: $Q_{\text{avail}}=m_w{c}_w\Delta T=0.050\times 4186\times 30=6279\text{ J}$ Why this step? Water cannot go below $0°C$ without itself freezing, so this is the maximum energy the ice can draw.
2. Heat needed to melt ALL the ice: $Q_{\text{needed}}=m_{\text{ice}}{L}_f=0.100\times 3.34\times 10^5=33400\text{ J}$ Why this step? We compare demand vs supply. This is the crucial check that catches the runs-out case.
3. Compare. $Q_{\text{avail}}=6279$ J $<$ $Q_{\text{needed}}=33400$ J. Not enough heat — only part of the ice melts. Why this step? When supply < demand, the phase change stalls; the mixture settles at the plateau temperature $0°C$ with leftover ice.
4. How much ice actually melts? $m_{\text{melted}}=\frac{Q_{\text{avail}}}{L_f}=\frac{6279}{3.34\times 10^5}=0.01880\text{ kg}=18.8\text{ g}$
5. Final state: temperature $=0°C$; ice remaining $=100-18.8=81.2$ g floating in $50+18.8=68.8$ g of water.

Verify: Melted mass ($18.8$ g) is far below the total ice ($100$ g), consistent with "not enough heat" ✓. Energy check: $18.8\text{ g}\times L_f=0.0188\times 3.34\times 10^5=6279$ J $=Q_{\text{avail}}$ ✓.

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Example 6 — Cell F: degenerate & boundary inputs

Forecast: Do the formulas break, or behave gracefully?

Part (a):

1. $Q=m{L}_f=0\times 3.34\times 10^5=0$ J. Why this step? Zero mass means zero substance to change phase — the formula returns $0$, which is physically correct, not a bug.

Part (b): 2. Energy to melt all this ice: $Q_{\text{melt}}=0.030\times 3.34\times 10^5=10020$ J. Why this step? We compare the supplied energy to the plateau's total demand. 3. Compare. Supplied $=10020$ J exactly equals the melt demand. Why this step? This is the razor's-edge boundary: just enough to finish melting, with zero left over to warm the resulting water. 4. Final state: $0.030$ kg of liquid water at exactly $0°C$ — every joule spent on the plateau, temperature not yet risen.

Verify: (a) any formula times zero mass is zero ✓. (b) $10020=0.030\times 3.34\times 10^5$ ✓; since nothing remains after melting, $\Delta T=0$ and the water stays at $0°C$ ✓.

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Example 7 — Cell G: real-world word problem (sweat cooling)

Forecast: A few joules, or tens of thousands?

1. Spot the phase change. Liquid sweat → water vapour: this is vaporisation (see Evaporation vs boiling). Why this step? Evaporation is a phase change, so the plateau tool $Q=m{L}_v$ applies — even though it happens below $100°C$, at the surface.
2. Compute the heat carried away: $Q=m{L}_v=0.020\times 2.43\times 10^6=4.86\times 10^4\text{ J}=48.6\text{ kJ}$ Why this step? The evaporating molecules steal this latent energy from your skin, which is why you feel cooler.

Verify: $48.6$ kJ from just $20$ g — huge, because $L_v$ is huge. This is exactly why sweating is such an effective coolant ✓. Units: $\text{kg}\times \text{J/kg}=\text{J}$ ✓.

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Example 8 — Cell H: exam twist, final temperature unknown

Forecast: Will the mixture end below, at, or above $0°C$? Will any ice survive?

1. Melt-first check. Heat needed to melt all the ice: $Q_{\text{melt}}=0.040\times 3.34\times 10^5=13360\text{ J}$ Heat the warm water can give while cooling to $0°C$: $Q_{\text{water}\to 0}=0.200\times 4186\times 50=41860\text{ J}$ Why this step? Before hunting for a temperature, confirm the ice fully melts. Since $41860>13360$, all ice melts and there is leftover energy — so the final temperature is above $0°C$.
2. Set up the balance for the mixed final temperature $T_f$. After melting, we have $0.040$ kg of melt-water (starting at $0°C$) warming up, and $0.200$ kg of original water cooling down: $\underset{\text{spent melting}}{\underbrace{Q_{\text{melt}}}}+\underset{\text{warm the melt-water}}{\underbrace{m_{\text{ice}}{c}_w(T_f-0)}}=\underset{\text{heat from original water}}{\underbrace{m_w{c}_w(50-T_f)}}$ Why this step? Energy conservation with every term: the ice's energy budget is "latent to melt" plus "sensible to warm up to $T_f$," paid for by the original water cooling from $50°C$ to $T_f$.
3. Substitute numbers ($c_w=4186$): $13360+0.040\times 4186\times T_f=0.200\times 4186\times (50-T_f)$ $13360+167.44T_f=41860-837.2T_f$
4. Solve for $T_f$: $167.44T_f+837.2T_f=41860-13360$ $1004.64T_f=28500⟹T_f=28.37°C$

Verify: $T_f$ lies between $0$ and $50°C$ — physically sensible ✓. Left side: $13360+167.44\times 28.37=13360+4750=18110$ J; right side: $837.2\times (50-28.37)=837.2\times 21.63=18110$ J — balanced ✓.

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Recall Which matrix cell am I in? (decision checklist)

Reveal the guiding question for each cell. Substance already at transition temp, only phase changing ::: Cell A/B — pure plateau, $Q=mL$ (sign = direction). Heating across several regions from one solid state to a hot gas ::: Cell C — sum slopes ($mc\Delta T$) and plateaus ($mL$). Two bodies mixed, told they reach a boundary temperature exactly ::: Cell D — heat lost = heat gained, solve for the unknown mass. Two bodies mixed, unsure if the phase change even finishes ::: Cell E — compare available vs needed heat first. Zero mass, or energy exactly equal to a plateau demand ::: Cell F — formulas still hold; watch the "nothing left over" boundary. Real cooling by evaporation ::: Cell G — surface vaporisation, $Q=m{L}_v$. Final temperature not given ::: Cell H — melt-check first, then energy balance for $T_f$.

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Connections

• Latent heat — phase transitions — the parent topic these examples drill.
• Specific heat capacity — the $Q=mc\Delta T$ tool used on every slope.
• Calorimetry — method of mixtures — the "heat lost = heat gained" backbone of Cells D, E, H.
• Kinetic theory of gases — why plateaus sit at fixed temperatures.
• Evaporation vs boiling — the physics behind the sweat-cooling problem (Cell G).
• First law of thermodynamics — sign conventions for heat in vs out (Cell B).
• Phase diagrams — where each transition temperature comes from.