Visual walkthrough — Latent heat — phase transitions

This page assumes you have read Latent heat — phase transitions (the parent). Here we build its central formula from the ground up, in pictures.

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Step 0 — The two questions heat can answer

Before any formula, understand this: when you pour heat energy into a substance, that energy always does exactly one of two jobs:

• Job A — speed the molecules up. Faster molecules = higher temperature. This is the kinetic energy story.
• Job B — pull the molecules apart. No speed change, just breaking the grip between molecules. This is stored as potential energy.

So heat can either make things hotter (Job A) or rearrange things (Job B). Every step below is just asking: which job is happening right now?

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Step 1 — The picture that starts it all: the heating curve

WHAT. We plot temperature (vertical axis, in $°C$) against heat added (horizontal axis, in joules) as we steadily heat $1$ block of water-stuff from cold ice to hot steam.

WHY start with a graph? Because the shape of this curve — where it climbs and where it goes flat — is the derivation. Every climb is Job A; every flat stretch is Job B. Read the shape, and the formula writes itself.

PICTURE. Look at the staircase below. It has five pieces: climb, flat, climb, flat, climb. Trace it left to right with your finger.

Notice: heat is always flowing in (we never stop heating), yet on the flat parts the temperature refuses to move. That refusal is the whole mystery. We will now explain each of the five pieces separately, because — and this is the key insight — the total heat is just the five pieces added up:

$Q_{\text{total}}=Q_1+Q_2+Q_3+Q_4+Q_5$

Each $Q$ is the heat for one piece of the staircase. We build them one at a time.

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Step 2 — A climbing piece: warming the ice ($Q_1$)

WHAT. The ice starts below freezing, say at $-10°C$. We heat it up to $0°C$. This is the first climb on the graph.

WHY this formula and not another? The temperature is changing, so Job A is happening — heat is becoming kinetic energy. The tool that connects heat to a temperature change is specific heat capacity $c$: it answers the question "how many joules raise unit mass by one degree?"

PICTURE. The red arrow climbs the slope; each degree of rise costs a fixed chunk of heat.

$Q_1=\underset{\begin{array}{c}\text{mass of}\\ \text{the block (kg)}\end{array}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to warm 1 kg}\\ \text{of ice by 1}°\text{C}\end{array}}{\underbrace{c_{\text{ice}}}}\underset{\begin{array}{c}\text{temperature rise}\\ =0-(-10)=10°\text{C}\end{array}}{\underbrace{\Delta T}}$

• $m$ — more stuff, more heat needed. Doubling the block doubles $Q_1$.
• $c_{\text{ice}}$ — the "price per degree per kilogram" for solid ice.
• $\Delta T$ — how far up the slope we climb. Here $10°C$.

If the ice already started at $0°C$, then $\Delta T=0$ and $Q_1=0$ — this step simply vanishes. (That is exactly the situation in the parent's Worked Example 1.)

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Step 3 — The first flat piece: melting ($Q_2$)

WHAT. Now the ice sits at $0°C$. Keep heating — but the temperature stops rising. We have hit the first flat plateau. Here the ice turns to water.

WHY does the climb stop? Job A has been suspended; Job B has taken over. Every incoming joule is now spent prying molecules out of their rigid lattice, not speeding them up. Since temperature = average kinetic energy, and kinetic energy is not changing, the thermometer holds dead still.

WHY a new formula? Because $\Delta T=0$ here, the formula $mc\Delta T$ would give $Q=0$ — clearly wrong, since heat is pouring in! We need a formula that measures phase change, not temperature change. That is latent heat of fusion $L_f$.

PICTURE. The amber flat line: heat flows in (arrow), temperature flat (dashed level line), bonds loosening (lattice → liquid).

$Q_2=\underset{\begin{array}{c}\text{mass being}\\ \text{melted (kg)}\end{array}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to melt 1 kg}\\ \text{at 0}°\text{C, no }\Delta T\end{array}}{\underbrace{L_f}}$

• No $\Delta T$ appears — temperature is not a variable here. Only how much mass changes phase matters.
• $L_f\approx 3.34\times 10^5$ J/kg for water. This is the length of the amber plateau.

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Step 4 — A climbing piece: warming the water ($Q_3$)

WHAT. All the ice is now water at $0°C$. Keep heating and the temperature climbs again — this time all the way to $100°C$. Second climb.

WHY back to $mc\Delta T$? Temperature is moving again, so Job A resumed. The lattice is already broken; extra heat now just speeds up the liquid molecules. We use $c_{\text{water}}$ (a different price-per-degree than ice — liquid water is harder to heat).

PICTURE. Same climbing arrow as Step 2, but longer ($100°$ of rise) and steeper price ($c_{\text{water}}>c_{\text{ice}}$).

$Q_3=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J per kg per}°\text{C}\\ \text{for }liquid\text{ water}\end{array}}{\underbrace{c_{\text{water}}}}\underset{100-0=100°\text{C}}{\underbrace{\Delta T}}$

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Step 5 — The second flat piece: boiling ($Q_4$)

WHAT. Water reaches $100°C$. Temperature flattens again — the second plateau. Here liquid becomes gas (steam).

WHY is this plateau so much longer? Melting only loosened a lattice; the molecules still touch. Boiling must rip molecules completely apart and shove back the surrounding atmosphere (see First law of thermodynamics). Full separation costs far more energy, so this flat stretch is huge.

WHY latent again? Same reasoning as Step 3: $\Delta T=0$, phase is changing, so $Q=mL$ — now with the latent heat of vaporisation $L_v$.

PICTURE. A very long amber plateau — about $7\times$ the melting one — molecules flying fully apart.

$Q_4=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J to boil 1 kg}\\ \approx 22.6\times 10^5\text{ J/kg}\end{array}}{\underbrace{L_v}}$

Compare the numbers: $L_v\approx 22.6\times 10^5$ versus $L_f\approx 3.34\times 10^5$ — vaporisation is roughly seven times larger. That ratio is drawn into the width of the two plateaus on the master curve.

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Step 6 — The last climbing piece: superheating the steam ($Q_5$)

WHAT. Now it is all steam at $100°C$. Heat it further, say to $110°C$. Final climb.

WHY $mc\Delta T$ one last time? Temperature rising ⇒ Job A ⇒ specific heat. Steam has its own price-per-degree $c_{\text{steam}}$.

PICTURE. The short final climb at the top-right of the staircase.

$Q_5=\underset{\text{mass (kg)}}{\underbrace{m}}\underset{\begin{array}{c}\text{J per kg per}°\text{C}\\ \text{for steam}\end{array}}{\underbrace{c_{\text{steam}}}}\underset{110-100=10°\text{C}}{\underbrace{\Delta T}}$

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Step 7 — Adding the pieces: the master formula

WHAT. We now stack all five heats. Because the pieces happen one after another and energy simply accumulates, we add them.

WHY addition works. Heat is conserved and additive — the heat to do the whole journey equals the heat for each leg summed. No overlaps, no double counting: each leg either climbs (Job A) or flattens (Job B), never both.

$Q_{\text{total}}=\underset{Q_1:\text{warm ice}}{\underbrace{m{c}_{\text{ice}}\Delta T}}+\underset{Q_2:\text{melt}}{\underbrace{m{L}_f}}+\underset{Q_3:\text{warm water}}{\underbrace{m{c}_{\text{water}}\Delta T}}+\underset{Q_4:\text{boil}}{\underbrace{m{L}_v}}+\underset{Q_5:\text{warm steam}}{\underbrace{m{c}_{\text{steam}}\Delta T}}$

Numeric check (parent Worked Example 2, $m=0.05$ kg, $-10°C\to 110°C$):

Leg	Term	Value (J)
$Q_1$ warm ice	$0.05\cdot 2100\cdot 10$	$1050$
$Q_2$ melt	$0.05\cdot 3.34\times 10^5$	$16700$
$Q_3$ warm water	$0.05\cdot 4186\cdot 100$	$20930$
$Q_4$ boil	$0.05\cdot 22.6\times 10^5$	$113000$
$Q_5$ warm steam	$0.05\cdot 2010\cdot 10$	$1005$
Total		$\approx 1.527\times 10^5$

The boiling leg ($Q_4=113$ kJ) dwarfs everything — exactly what the long amber plateau predicted.

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Step 8 — Degenerate and edge cases (never get surprised)

Real problems rarely run the full staircase. Handle each stub:

• Ice already at $0°C$ → $Q_1=0$ (drop the first climb). Start at Step 3. This is Worked Example 1.
• Stop at water at $25°C$ → keep $Q_1,Q_2$ and part of $Q_3$; never reach $Q_4,Q_5$. Truncate the staircase where you stop.
• Only partial melting (mixing problems, Calorimetry — method of mixtures) → in $Q_2=m_{\text{ice}}{L}_f$, solve for the mass $m_{\text{ice}}$ that the available heat can melt: $m_{\text{ice}}=Q/L_f$. In Worked Example 3, $31395/(3.34\times 10^5)\approx 94$ g melts.
• Cooling / freezing → run the staircase backwards: heat is released, every $Q$ carries a minus sign. Freezing gives back exactly the $m{L}_f$ that melting absorbed.
• Exactly at a plateau temperature but no heat added → nothing changes; you sit still on the flat until either heat arrives or leaves.

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The one-picture summary

Everything above, compressed into a single annotated staircase — the two tools ($mc\Delta T$ on climbs, $mL$ on flats), the five legs, and the reason each flat exists.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a crowd of kids frozen in a huddle, all holding hands. First you give them energy and they shiver faster — that is the ice warming up, temperature climbing (Step 2). Then something odd: you keep giving energy, but they stop shivering faster — instead they use it to let go of each other's hands. The "temperature" holds still while they break free (melting, Step 3). Once everyone's hands are free (now they're a milling liquid), fresh energy makes them run faster again — the water warming (Step 4). Then the biggest pause of all: they use a huge amount of energy to fly completely apart and scatter into a room (boiling, Step 5) — that's why the flat stretch is so long. Finally the scattered kids (steam) speed up once more (Step 6). Add up the energy for every "run faster" leg ($mc\Delta T$) and every "let go / fly apart" leg ($mL$), and you get the total — the master formula (Step 7). Missing a leg is like forgetting the kids still had to warm up between letting go and flying apart.

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Connections

• Specific heat capacity — the $Q=mc\Delta T$ tool used on every climbing leg.
• Calorimetry — method of mixtures — partial-melting and cooling edge cases (Step 8).
• Kinetic theory of gases — why temperature = average kinetic energy (Step 0).
• Evaporation vs boiling — surface vs bulk vaporisation, both use $L_v$.
• First law of thermodynamics — why boiling also does work on the atmosphere (Step 5).
• Phase diagrams — where these five legs live in $P$–$T$ space.