Phase Changes and Latent Heat

Beginner

Step 1. Ice melts, and water boils. Matter comes in three everyday states. In a solid such as ice, the molecules hold fixed places in a regular pattern and can only jiggle on the spot. In a liquid they stay close but slide past one another. In a gas such as steam they are far apart and fly about freely.

Step 2. Warm some ice and the thermometer climbs, but only to 0 °C. Surprisingly, the flame keeps supplying heat, yet the thermometer stays at 0 °C until the last bit of ice has melted. The heat is busy breaking the links that hold the molecules in their pattern, not making them move faster. Because it seems to vanish, it is called latent (hidden) heat.

Step 3. The same happens at 100 °C: the water boils, bubbles of steam form inside it, and the temperature stays put until all the water has turned to steam. Boiling needs about seven times more heat than melting, because each molecule must tear itself completely away from its neighbours.

Step 4. Run the film backwards and the heat comes back: freezing water and condensing steam give it out. Water also evaporates below boiling: the fastest molecules escape from the surface and carry energy away, which is how sweat cools your skin.

Butter is a mixture of fats and glass has no regular pattern, so they soften gradually instead of melting at one temperature.

In the animation, watch the thermometer stop at 0 °C while the links of the crystal break, then at 100 °C while molecules escape one by one. The boiling stage is shown four times faster.

Check yourself: why does a drink with ice cubes stay near 0 °C until the ice is gone? Answer: the heat coming in from the warm room melts the ice instead of warming the drink.

Intermediate

Two kinds of heat

It is tempting to think that heating a substance must make it hotter. In fact the energy can do two things. It can make the molecules move faster, which a thermometer shows as a rising temperature: this is sensible heat, $Q=mc\Delta T$, met in Temperature and Heat. Or it can pull the molecules apart without speeding them up, and the substance changes state at constant temperature: this is latent, or hidden, heat.

The specific heat capacity $c$ belongs to a phase, not the substance. For water it is about 2100 J/(kg·K) in ice, 4180 J/(kg·K) in the liquid and 2010 J/(kg·K) in steam, so the same heat warms ice about twice as much as liquid water.

Latent heat changes the state

While a substance changes state, the heat is proportional to the mass that changes:
$Q=mL$
Here $L$ is the specific latent heat in J/kg: $L_f$ (fusion) for melting, $L_v$ (vaporisation) for boiling, with $\Delta T=0$.

For water at atmospheric pressure, $L_f=334kJ/kg$ at 0 °C and $L_v=2260kJ/kg$ at 100 °C. Boiling a kilogram takes 6.8 times the heat that melts it and 5.4 times the heat that warms it from 0 to 100 °C, because melting only loosens the bonds, while boiling tears each molecule away from its neighbours.

The change runs both ways at the same temperature: freezing a kilogram of water releases 334 kJ, condensing a kilogram of steam 2260 kJ. Heat given out counts as negative: $Q=-m{L}_f$ for freezing.

Heat balance with a change of state

Three relations, applied stage by stage, solve nearly every problem here:

• $Q=mc\Delta T$
• $Q=mL$
• $L_f=334kJ/kg,L_v=2260kJ/kg\text{(water)}$

Split the process into stages, each changing either the temperature of one phase, with its $c$, or the state at a fixed temperature, with $mL$; no single $mc\Delta T$ may cross 0 °C or 100 °C. In an insulated container all heats add up to zero, latent ones included. Before solving for a final temperature, check that there is enough heat for the change of state to finish; if not, the mixture ends at the transition temperature with both phases present.

Worked example: the ice in the animation

A kilogram of ice at −20 °C is heated at 1 atm into steam at 120 °C. In kilojoules, with $c$ in kJ/(kg·K) as on the screen:

• Ice from −20 to 0 °C: $Q_1=m{c}_{\mathrm{ice}}\Delta T=1\cdot 2.1\cdot 20=42\mathrm{kJ}$
• Melting at 0 °C: $Q_2=m{L}_f=1\cdot 334=334\mathrm{kJ}$
• Water from 0 to 100 °C: $Q_3=m{c}_{\mathrm{water}}\Delta T=1\cdot 4.18\cdot 100=418\mathrm{kJ}$
• Boiling at 100 °C: $Q_4=m{L}_v=1\cdot 2260=2260\mathrm{kJ}$
• Steam from 100 to 120 °C: $Q_5=m{c}_{\mathrm{steam}}\Delta T=1\cdot 2.01\cdot 20=40.2\mathrm{kJ}$

The total, $42+334+418+2260+40.2\approx 3094\mathrm{kJ}$, is the final value on the $Q$ readout, and the $T$ readout ends at 120 °C. Boiling alone takes 73 % of the heat.

The ideal burner supplies $P=2\mathrm{kW}$, so each stage lasts $t=Q/P$: melting $334000/2000=167\mathrm{s}$, boiling $2260000/2000=1130\mathrm{s}$, almost 19 minutes, and the whole journey 1547 s, the final 25.8 min on the clock readout. The film is not to scale: boiling runs four times faster and the short warmings four times slower, as the purple labels say.

On the sloped stages the formula line reads $Q=mc\Delta T$ with the current phase's $c$; on the flat ones it switches to $Q=mL$ with $\Delta T=0$, and the $T$ readout stops while $Q$ keeps climbing. The gold arrows beside the flame never change: heat flows in steadily. In the box the grey bonds of the crystal break one by one at 0 °C, and at 100 °C the molecules escape one at a time, each of the 24 taking $2260/24\approx 94\mathrm{kJ}$.

Worked example: ice in a jug of lemonade

You add 100 g of ice at −10 °C to 300 g of lemonade at 24 °C in an insulated jug. Treating the lemonade as water, what is the final temperature?

First compare what the ice needs with what the lemonade can give. Warming the ice to 0 °C takes $0.10\cdot 2100\cdot 10=2100\mathrm{J}$ and melting it all $0.10\cdot 334000=33400\mathrm{J}$, 35 500 J in total, but the lemonade, cooling to 0 °C, can give at most $0.30\cdot 4180\cdot 24\approx 30100\mathrm{J}$. So the drink ends at exactly 0 °C: the remaining 28 000 J melts $28000/334000\approx 0.084\mathrm{kg}$, and 16 g of ice still floats.

With only 50 g of ice, warming and melting need $1050+16700=17750\mathrm{J}$, so all of it melts, and the energy balance reads
$0.30\cdot 4180\cdot (24-T)=17750+0.05\cdot 4180\cdot T$
This gives about 8.4 °C, whereas 50 g of water at 0 °C would leave the drink at 20.6 °C: most of the cooling comes from the melting.

Reading the heating curve

Plot the $T$ readout against the $Q$ readout for the heating curve, made of straight pieces. A rising piece has slope $1/(mc)$, so the water piece is half as steep as the ice and steam pieces. A flat piece is $mL$ long: 334 kJ for melting, 2260 kJ for boiling. With a steady burner the graph against time looks the same, since $t=Q/P$.

Read backwards, it is a cooling curve: a freezing liquid levels off because the latent heat it releases makes up for the heat it loses, which is how melting points are measured. Butter and wax, being mixtures, and glass, which has no regular crystal, show no flat piece; they soften gradually.

Common mistakes

• Turning up the flame does not heat boiling water above 100 °C at 1 atm; it only boils it away faster: a 4 kW burner boils away the animation's kilogram in 9.4 minutes, not 19.
• Heat is not temperature: while the ice melts, 334 kJ flow in and the thermometer stays at 0 °C. The energy rearranges the molecules instead of speeding them up.
• Watch the units of $L$: 334 kJ/kg equals 334 J/g, so melting 20 g of ice takes $20\cdot 334=6680\mathrm{J}$. Mixing grams with joules per kilogram gives errors by a factor of 1000.
• The values 0 °C and 100 °C hold only at 1 atm. On Everest, at about a third of an atmosphere, water boils near 71 °C; in a pressure cooker at about 2 atm, near 120 °C.
• Evaporation is not boiling: it happens at the surface at any temperature, while boiling needs bubbles of vapour inside the liquid.

Where you meet it

Steam scalds because each gram condensing on the skin releases 2260 J, while a gram of water at 100 °C gives about 260 J as it cools to body temperature. Sweat cools you by evaporating: with $L_v\approx 2420kJ/kg$ at skin temperature, a litre an hour carries away about 670 W. On frosty nights growers spray fruit trees, because freezing water releases 334 kJ per kilogram and holds the buds near 0 °C. A refrigerator evaporates a fluid inside, taking latent heat from the food, and condenses it at the back, releasing the heat into the kitchen.

In clouds, condensing vapour releases the latent heat that powers storms. The ideas return in later lessons: Conduction, Convection and Radiation; Kinetic Theory of Gases; and First Law of Thermodynamics and Heat Engines.

Advanced

The molecular picture of a plateau

A plateau on a heating curve is a stretch where heat keeps flowing in but the temperature does not rise. To see where that energy goes, look at the molecules.

In ice each water molecule is held by hydrogen bonds to four neighbours, in an open hexagonal lattice, and vibrates about its fixed site. Temperature measures the mean kinetic energy of this jiggling. Below 0 °C, added heat makes the vibrations larger, and the thermometer rises. At the melting point the added energy goes into breaking bonds instead: it becomes potential energy as the lattice comes apart, the mean kinetic energy stays the same, and so the temperature stays put until the crystal is gone.

The animation draws this as a cartoon. The crystal's molecules vibrate with an amplitude that grows as $\sqrt{T}$, because the mean square displacement of a vibration is proportional to the temperature. On the plateau the bonds break one by one, and the liquid packs more closely than the open crystal, because water is denser than ice.

Two latent heats measure the two jobs. The specific latent heat of fusion, $L_f=334kJ/kg$, melts ice at 0 °C; the specific latent heat of vaporisation, $L_v=2260kJ/kg$, boils water at 100 °C. Melting only loosens the hydrogen-bond network, so $L_f$ is just 12 % of the 2834 kJ/kg that takes ice straight to vapour. Boiling must free every molecule from all of its neighbours, so $L_v\approx 6.8L_f$.

The energy per molecule is $L_{v}M/N_A$, where $M=0.018kg/mol$ is the molar mass and $N_A$ is Avogadro's number: $L_{v}M/N_A\approx 6.8\times 10^{-20}\mathrm{J}\approx 0.42\mathrm{eV}$. That is about 13 times $k_{B}T$ at 100 °C, the typical thermal energy, with $k_B$ Boltzmann's constant. A molecule escapes only when it happens to gather far more than its average share of energy. Escape is therefore a rare event whose rate depends steeply on temperature, and this returns below as the steep vapour-pressure curve.

Latent heat as a jump in enthalpy

Legend

$p$	Constant pressure of the surroundings, pushing down on the piston.	$\mathrm{Pa}$
$L$	Heat that turns one kilogram of liquid into vapour: the latent heat.	$J/kg$
$\Delta v$	Volume gained per kilogram: vapour minus liquid.	${\mathrm{m}}^3/kg$
$\Delta u$	Rise in internal energy per kilogram: the energy that pulls the molecules apart.	$J/kg$
$p\Delta v$	Work per kilogram done on the surroundings as the vapour pushes the piston up.	$J/kg$

Where exactly does the latent heat go? Part of it separates the molecules, and part of it pays for the room the vapour takes up. Splitting the two shows that latent heat is a change in enthalpy, the quantity used for every phase change in the rest of this lesson.

Figure 1 shows one kilogram of liquid boiling in a cylinder closed by a piston that slides freely. The arrows on the piston stand for the pressure $p$ of the surroundings, which stays constant. The arrows at the bottom are the heat $L$ flowing in. As liquid turns into vapour the volume grows by $\Delta v$, the bracket beside the cylinder, so the piston rises against $p$. Lifting it takes the work $p\Delta v$, energy that leaves the substance and goes to the surroundings. The rest, $\Delta u$, stays inside as a rise in internal energy: the energy of pulling the molecules apart. The bar on the right stacks the two parts, and together they make up $L$.

To write this as an equation, use lower-case letters for quantities per kilogram: $u$ for internal energy, $v$ for volume and $h=u+pv$ for enthalpy. The first law says that the heat taken in equals the rise in internal energy plus the work done, $L=\Delta u+p\Delta v$. Because $p$ is constant, $p\Delta v$ is the same as $\Delta (pv)$, so the heat is exactly the jump in enthalpy:

$L=\Delta h=\Delta u+p\Delta v$

For water boiling at 1 atm (101 325 Pa), each kilogram grows from 0.00104 m³ of liquid to 1.673 m³ of vapour, so $p\Delta v=101325\cdot 1.672\approx 169kJ/kg$. Only 7.5 % of $L_v$ goes into pushing back the atmosphere; the other 2091 kJ/kg separates the molecules. The bar in Figure 1 is drawn in these proportions.

Melting is different. Ice shrinks by about 8 % as it melts, so $\Delta v$ is small and negative and $p\Delta v\approx -9J/kg$: the atmosphere does a tiny amount of work on the ice, and $L_f$ is almost entirely internal energy.

Why the temperature stalls

Legend

$g$	Specific Gibbs energy, $h-Ts$; at fixed $T$ and $p$ the phase with the lower $g$ is stable.	$J/kg$
$T$	Temperature.	$\mathrm{K}$
$g_s$	Specific Gibbs energy of the solid.	$J/kg$
$g_l$	Specific Gibbs energy of the liquid.	$J/kg$
$T_m$	Temperature where the curves cross: the melting point at this pressure.	$\mathrm{K}$
$dT$	A small step in temperature.	$\mathrm{K}$
$s_{s}dT$	How far $g_s$ falls over the step $dT$: the solid's entropy times the step.	$J/kg$
$s_{l}dT$	How far $g_l$ falls over the same step: more, because the liquid's entropy $s_l$ is larger.	$J/kg$

The molecular picture says what the heat does on a plateau. Thermodynamics says why the substance must stay at one temperature while it changes phase. The tool is the specific Gibbs energy $g=h-Ts$, where $s$ is the specific entropy. At fixed temperature and pressure a system settles in the state of lowest Gibbs energy, just as a mechanical system settles at its lowest potential energy.

Figure 2 plots $g$ against $T$ at one fixed pressure, for the solid, $g_s$, and for the liquid, $g_l$. At each temperature the phase whose curve lies lower is the stable one, and that part of its curve is drawn thick: the solid's to the left of $T_m$, the liquid's to the right.

Why do the curves cross? At fixed pressure a small change of $g$ is $dg=-sdT$, so each curve slopes down with gradient $-s$. Look at the two small triangles: over the same step $dT$ the solid's curve drops by $s_{s}dT$ and the liquid's by $s_{l}dT$. The liquid is the more disordered phase, so $s_l>s_s$ and its curve falls faster. At low temperature the liquid's curve lies higher, so the steeper curve must cut the shallower one, and it does so once, at the melting point $T_m$: below it the solid has the lower $g$ and is stable, above it the liquid. Both curves also bend downwards, because each phase's entropy grows as it warms.

Now put ice and water in contact. They share the same $T$ and $p$ and exchange molecules freely. Moving a kilogram from solid to liquid changes the Gibbs energy by $g_l-g_s$, so molecules drift towards the phase with the lower curve until the other phase is used up. The two can stay together only where neither direction gains anything, $g_s=g_l$: at the crossing, at $T_m$ and nowhere else.

That is why the temperature stalls. While both phases are present it is pinned at $T_m$, and added heat can only change the proportions: after a heat $Q$ on the melting plateau, a fraction $Q/(m{L}_f)$ of the mass $m$ has melted.

The crossing also fixes the jump in entropy. Writing $g_s=g_l$ as $h_s-T{s}_s=h_l-T{s}_l$ gives $\Delta h=T\Delta s$, and since $\Delta h=L$,

$\Delta s=\frac{L}{T}$

For water this is 1.22 kJ/(kg·K) on melting and 6.06 kJ/(kg·K) on boiling. Per mole the boiling value is 109 J/(mol·K). Trouton's rule, an observation rather than a law, says that many simple liquids gain roughly 87 J/(mol·K) on boiling. Water gains more because its hydrogen bonds make the liquid unusually ordered, so turning it into vapour adds more disorder.

The heating curve

The expert graph of the animation plots the temperature $T$ against the heat supplied $Q$ while 1 kg of ice at −20 °C becomes steam at 120 °C. It has two kinds of stretch. Within one phase, heat raises the temperature and the line is straight, with slope

$\frac{dT}{dQ}=\frac{1}{mc}$

where $c$ is the specific heat capacity. With $m=1\mathrm{kg}$ and $c$ equal to 2.1, 4.18 and 2.01 kJ/(kg·K), the slopes are 0.48 K/kJ for ice, 0.24 K/kJ for water and 0.50 K/kJ for steam: liquid water is the hardest to warm. On a plateau the temperature is fixed and only the width matters: $m{L}_f=334\mathrm{kJ}$ for melting and $m{L}_v=2260\mathrm{kJ}$ for boiling. The whole axis spans 3094 kJ, and the boiling plateau alone fills 73 % of it.

The phase diagram

Legend

$p$	Pressure, on a logarithmic axis.	$\mathrm{Pa}$
$T$	Temperature.	$\mathrm{K}$
$t$	The triple point, the only point where solid, liquid and vapour coexist.
$p_1$	Sea-level pressure, 1 atm.	$\mathrm{Pa}$
$p_2$	Pressure on a high mountain, a third of $p_1$.	$\mathrm{Pa}$
$T_m$	Melting point of ice, almost the same at every pressure shown.	$\mathrm{K}$
$T_1$	Boiling point at $p_1$.	$\mathrm{K}$
$T_2$	Boiling point at $p_2$, lower than $T_1$.	$\mathrm{K}$

A heating curve is one experiment at one pressure. The phase diagram shows, for every pressure and temperature at once, which phase is stable and where two phases can coexist.

Figure 3 is the $p$–$T$ diagram of water, with pressure on a logarithmic axis so that very low pressures and atmospheric pressure both fit. Each phase owns a region: the cube marks the solid, the drop the liquid and the bubbles the vapour. Along the lines between the regions two phases coexist, and the three lines meet at the triple point $t$, 0.01 °C and 612 Pa, the only point where all three phases coexist.

A heating experiment at constant pressure is a horizontal path from left to right. The animation's dot runs along the dashed line $p_1$, 1 atm. It stops where the path meets the nearly vertical melting line, at $T_m$, 0 °C, and again where it meets the vaporisation curve, at $T_1$, 100 °C: these two crossings are the two plateaus.

On a high mountain the pressure is lower, the dashed line $p_2$, a third of an atmosphere. That path meets the rising vaporisation curve sooner, so the boiling point $T_2$ is lower. The melting line is drawn leaning back far more than it really does; its true slope is worked out later in the lesson.

Below the triple-point pressure a horizontal path never enters the liquid region. It crosses the sublimation curve instead, where ice turns straight into vapour. Snow does this on a cold, dry, sunny day, when the water vapour in the air has a pressure well below 612 Pa. At the triple point the sublimation curve is steeper than the vaporisation curve, because its latent heat is larger, and the next section shows that the slope grows with the latent heat. That latent heat is about 2834 kJ/kg: $L_f$ plus $L_v$ measured at 0 °C, which is 2501 kJ/kg, not the 2260 kJ/kg of 100 °C.

Derivation: the Clausius–Clapeyron equation

Legend

$p$	Pressure.	$\mathrm{Pa}$
$v$	Volume per kilogram.	${\mathrm{m}}^3/kg$
$v_l$	Volume of one kilogram of liquid.	${\mathrm{m}}^3/kg$
$v_g$	Volume of one kilogram of vapour, in reality far larger than $v_l$.	${\mathrm{m}}^3/kg$
$\Delta p$	Drop in vapour pressure from $T$ to $T-\Delta T$.	$\mathrm{Pa}$
$T$	Temperature of the evaporation: the upper isotherm.	$\mathrm{K}$
$T-\Delta T$	Slightly lower temperature of the condensation: the lower isotherm.	$\mathrm{K}$
$L$	Heat absorbed while one kilogram evaporates.	$J/kg$
$w$	Net work of one cycle per kilogram, the area of the strip: $\Delta p(v_g-v_l)$.	$J/kg$

The lines of Figure 3 are not arbitrary. Their slope $dp/dT$ follows from the latent heat and the change in volume alone, and the argument needs nothing beyond the second law. The idea is to run a tiny Carnot engine that uses boiling as its working stroke.

Figure 4 is a $p$–$v$ diagram. Inside the dome, liquid and vapour coexist; there the pressure at a given temperature is fixed at the vapour pressure, so each isotherm is a horizontal line. Follow one kilogram round the cycle. Along the top edge, at temperature $T$ and vapour pressure $p$, the liquid evaporates completely: its volume grows from $v_l$ to $v_g$ while it absorbs the heat $L$. Down the slanted right side the wet vapour expands adiabatically, with no heat exchanged, and cools to $T-\Delta T$, so its pressure falls by $\Delta p$. Along the bottom edge, at $T-\Delta T$ and the lower pressure $p-\Delta p$, it condenses and gives out heat. Up the slanted left side an adiabatic compression brings it back to the start.

Now find the work of one cycle in two ways. Geometrically, the net work is the area enclosed: a thin strip of base $v_g-v_l$ and height $\Delta p$, the shaded strip, so $w=\Delta p(v_g-v_l)$. Thermodynamically, any reversible engine working between $T$ and $T-\Delta T$ has the Carnot efficiency $\Delta T/T$, so it turns that fraction of the absorbed heat into work: $w=L\Delta T/T$. Setting the two equal and letting $\Delta T$ shrink to zero gives the Clausius–Clapeyron equation, with $\Delta v=v_g-v_l$:

$\frac{dp}{dT}=\frac{L}{T\Delta v}$

The Gibbs energy gives the same result. Along a coexistence line the two phases keep $g_l=g_g$, so their changes match, $d{g}_l=d{g}_g$. In general $dg=vdp-sdT$, so $v_{l}dp-s_{l}dT=v_{g}dp-s_{g}dT$, which gives $dp/dT=\Delta s/\Delta v$; with $\Delta s=L/T$ this is the same equation.

Boiling points and vapour pressures

The equation becomes easy to use with two approximations, both good far from the critical point. The vapour takes up far more room than the liquid, $v_g\gg v_l$, so $\Delta v\approx v_g$. And the vapour is nearly an ideal gas, $v_g\approx RT/(Mp)$, with $R$ the gas constant. Substituting both gives

$\frac{d(\ln p)}{dT}=\frac{LM}{R{T}^2}$

and, if $L$ is treated as constant, integrating between two points $(T_1,p_1)$ and $(T_2,p_2)$ of the curve gives

$\ln \left(\frac{p_2}{p_1}\right)=-\left(\frac{LM}{R}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

For water $LM/R\approx 4900\mathrm{K}$. Start from sea level, where $p_1$ is 1 atm and the boiling point is $T_1=373\mathrm{K}$ (Figure 3). On the mountain $p_2=p_1/3$, so $\ln (p_2/p_1)=-\ln 3$ and the formula gives $1/T_2=1/373+\ln 3/4900$ (in ${\mathrm{K}}^{-1}$): $T_2=344\mathrm{K}$, about 71 °C. At 2 atm the same formula gives about 121 °C, roughly the temperature inside a pressure cooker.

The curve can be read in two ways: as the vapour pressure of the liquid at each temperature, or as its boiling point at each outside pressure. These are one curve because a liquid boils when bubbles of its own vapour can grow inside it, and that happens once its vapour pressure reaches the pressure around it.

For melting, the same equation explains the near-vertical line. With $\Delta v=-9.07\times 10^{-5}{\mathrm{m}}^3/kg$ at 273 K it gives $dp/dT\approx -13.5MPa/K$. The slope is huge and negative: it takes 133 atmospheres to lower the melting point by 1 K, and the line leans slightly back, towards lower temperature at higher pressure. Water is unusual here; for most substances the solid is denser than the liquid, $\Delta v>0$, and the melting line leans forwards.

Kinetic theory explains why the boiling curve rises so steeply. A molecule needs an energy $W\approx LM/N_A$ to escape from the liquid, and the share of molecules with that much energy goes roughly as the Boltzmann factor $e^{-W/k_{B}T}$. Since $W/k_{B}T=LM/(RT)$, this is the same exponential that the integrated formula gives, $p\propto e^{-LM/(RT)}$. As a result the vapour pressure of water rises by 3.5 % per kelvin at 100 °C and by about 6 % per kelvin at 20 °C. And because the molecules that escape are the fastest ones, those left behind are slower on average: an evaporating liquid cools.

Worked problem: the same ice on a mountain

Take the animation's kilogram of ice at −20 °C and its ideal 2 kW burner to the summit of Everest, where the pressure is about a third of an atmosphere: the dashed line $p_2$ in Figure 3. How much heat, and how much time, does it take to turn all the ice into vapour? At sea level it takes 3054 kJ, the $Q$ readout when the last water boils away, and 1527 s: 42 kJ to warm the ice, 334 kJ to melt it, 418 kJ to warm the water and 2260 kJ to boil it.

Warming and melting the ice. The melting point barely moves, as the nearly vertical melting line of Figure 3 suggests. The pressure drops by 68 kPa, which raises the melting point by only $68000/(13.5\times 10^6)\approx 0.005\mathrm{K}$, so the first two stages still need 42 kJ and 334 kJ.

Warming the water. The water now boils at about 71 °C, the point $T_2$ of Figure 3, so warming it from 0 °C takes $4.18\cdot 71\approx 297\mathrm{kJ}$ instead of 418 kJ.

Boiling it. The latent heat at 71 °C is not the 2260 kJ/kg of 100 °C. To find it, use the fact that enthalpy is a function of state: any route between the same two end states gives the same change. So instead of boiling at 71 °C, take a detour: warm the liquid to 100 °C, boil it there, and cool the vapour back to 71 °C. The vapour behaves almost as an ideal gas, whose enthalpy hardly depends on pressure, so the different pressures along the detour do not matter:

$L_v(71{}^{\circ }\mathrm{C})\approx 4.18\cdot 29+2260-2.01\cdot 29\approx 2323kJ/kg$

This is close to the measured 2331 kJ/kg. In general $L_v$ grows at the rate $c_{\mathrm{water}}-c_{\mathrm{steam}}\approx 2.2kJ/(\mathrm{kg}\cdot \mathrm{K})$ as the boiling point is lowered, that is by about 2.2 kJ/kg for every kelvin: at a lower temperature both the liquid and the vapour hold less enthalpy, but the liquid loses more, so the gap between them, $L_v$, widens.

In total $42+334+297+2323\approx 2996\mathrm{kJ}$, which the burner supplies in 1498 s. On the heating curve the water-warming stretch shrinks by 121 kJ, while the boiling plateau, now at 71 °C, grows by 63 kJ, so the whole job is only 2 % cheaper. That is why cooking at altitude is slow: the energy hardly changes, but the pot never gets hotter than 71 °C while it holds water, and the speed of the chemistry of cooking is set by temperature, not by energy.

Limits of validity

Everything above treats the latent heat as a constant and the vapour as an ideal gas. Both are approximations, and a phase change need not even happen where the diagram says.

The latent heat depends on temperature: $L_v$ is 2501 kJ/kg at 0 °C and 2260 kJ/kg at 100 °C. Integrated down from 100 °C with a constant 2260 kJ/kg, the vapour-pressure formula therefore gives 2.8 kPa at 20 °C instead of the measured 2.34 kPa. The animation's diagram uses a mean value of 2.40 MJ/kg, chosen so that its curve passes both through the triple point and through 100 °C at 1 atm. At the critical point, 374 °C and 22.06 MPa, liquid and vapour become identical: $\Delta v$ and $L_v$ vanish, the dome of Figure 4 closes at its top, and the boiling plateau disappears.

A new phase needs a seed, such as a dust grain or a crystal already present. Without one, tiny droplets of water stay liquid far below 0 °C, as in clouds at −20 °C; such water is called supercooled. When supercooled water at −10 °C finally freezes, the latent heat it releases warms it back to 0 °C. The heat from freezing a fraction of the water must warm all of it by 10 K, so only $c\Delta T/L_f=4.18\cdot 10/334\approx 1/8$ freezes at once, and the rest stays liquid at 0 °C. In the same way, water heated in a smooth cup in a microwave can rise above 100 °C without boiling, becoming superheated, and then boil violently.

Pressure melting is weak. A 70 kg skater on a blade contact of 3 mm by 10 cm presses on the ice with about 2.3 MPa, which lowers the melting point by only 0.17 K. Friction and a thin, disordered surface layer on the ice explain skating far better.

Not every transition has a sharp plateau. Sea water starts to freeze at about −1.9 °C and goes on freezing over a range of temperatures, because the brine left behind grows saltier. Glass softens gradually, with no latent heat at all. And iron, which loses its magnetism at 770 °C, shows a peak in its specific heat there rather than a plateau.

Where this leads

The first law and the Carnot argument are the subject of First Law of Thermodynamics and Heat Engines, and every steam power station works by boiling and condensing water. The entropy jump $\Delta s=L/T$ belongs to The Second Law of Thermodynamics and Entropy, and the Boltzmann factor and the Maxwell speeds of the escaping molecules to Kinetic Theory of Gases. Water's 9 % expansion on freezing, which tilts the melting line back, was met earlier in Thermal Expansion. Clausius–Clapeyron also shapes the weather: air can hold about 6–7 % more vapour for each kelvin of warming, so a warmer atmosphere carries more latent heat upwards by convection, treated in Conduction, Convection and Radiation, and releases it in heavier downpours.