Phase Changes and Latent Heat

Heat a solid steadily and its temperature rises — until it starts to melt, at which point the temperature stops rising even though heat is still pouring in. It stays fixed until the last of the solid has become liquid, then begins to climb again. The same thing happens, more dramatically, at the boiling point. This chapter explains where the "hidden" energy goes during those flat stretches, quantifies it with the specific latent heat, and connects the macroscopic picture to the microscopic one built in Chapter 1: the supplied energy is doing work against the intermolecular forces, raising the potential-energy part of the internal energy while leaving the kinetic part — and therefore the temperature — untouched.

The chapter builds directly on the last one. There, $Q = mc\,\Delta T$ handled every temperature change; here, $Q = mL$ handles every change of state. Real problems — a block of ice heated to steam, a hailstone forming in a cloud — chain the two formulas together, and knowing which one applies on each segment is the whole skill.

Reading note. The word "latent" means hidden: the energy is hidden in the sense that it produces no temperature change while it is being absorbed. It reappears in full when the phase change is reversed.

1. The Phases and Their Transitions

Matter commonly exists in three phases — solid, liquid, gas — distinguished by how tightly the molecules are bound:

• Solid: molecules locked in a fixed lattice, vibrating about mean positions; strong intermolecular forces, deep potential wells.
• Liquid: molecules still close-packed but free to move past one another; forces still significant.
• Gas: molecules far apart, moving freely; intermolecular forces negligible except in brief collisions.

The transitions are melting (solid → liquid) and its reverse freezing; boiling or vaporisation (liquid → gas) and its reverse condensation; and sublimation (solid → gas directly, as with dry ice). At a given pressure each transition happens at a definite temperature — the melting point and boiling point — which stays constant throughout the change.

2. Specific Latent Heat

Definition. The specific latent heat $L$ of a substance is the energy required to change the phase of one kilogram of it, with no change in temperature:

$$L = \frac{Q}{m} \qquad \left[\dfrac{\text{J}}{\text{kg}}\right] \qquad\Longleftrightarrow\qquad Q = mL$$

Two values are distinguished:

• the specific latent heat of fusion $L_f$, for melting or freezing;
• the specific latent heat of vaporisation $L_v$, for boiling or condensing.

For water, $L_f = 3.34\times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$ and $L_v = 2.26\times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$. The second is nearly seven times the first, and Section 4 explains why. On freezing or condensing, the same energy is released rather than absorbed — which is why steam at $100\ ^{\circ}\text{C}$ scalds far more severely than water at $100\ ^{\circ}\text{C}$, and why frost forming on a field releases enough energy to slow further cooling.

3. The Microscopic Explanation: Energy Goes Into Potential Energy

During a temperature rise, the supplied heat increases the molecules' random kinetic energy: they vibrate or move faster, and the thermometer registers it.

During a phase change, the story is different. The molecules must be pulled apart against the intermolecular forces that bind them — bonds in the lattice are broken on melting, and essentially all intermolecular bonds are broken on boiling. That work goes into the potential-energy part of the internal energy (Chapter 1), increasing the mean separation of the molecules. Their mean kinetic energy does not change, so — since temperature is a measure of mean molecular kinetic energy — the temperature holds constant.

Key distinction. A temperature change alters the kinetic part of the internal energy ($Q = mc\,\Delta T$). A change of state alters the potential part ($Q = mL$), at constant kinetic energy and therefore constant temperature. The two processes never overlap: a substance is either changing temperature or changing state, not both at once.

4. Why the Latent Heat of Vaporisation Exceeds That of Fusion

Two effects make $L_v \gg L_f$ for essentially every substance.

• Bonds broken. Melting only loosens the lattice — the molecules stay in contact and many bonds survive. Boiling separates the molecules to distances where the forces are negligible, so all the intermolecular bonds must be broken.
• Work done on the atmosphere. When a liquid boils, its volume increases enormously — for water, roughly by a factor of $1600$ at atmospheric pressure. The expanding vapour pushes back the surrounding air, doing external work $p\,\Delta V$. Melting involves only a tiny volume change, so this term is negligible for fusion.

The energy supplied during boiling therefore splits into a large part that breaks bonds (increasing internal energy) and a smaller part that does expansion work on the surroundings.

5. Heating Curves

Plot temperature against time while supplying heat at a constant rate to a substance taken from solid, through liquid, to gas. The graph has five segments:

1. Solid warming — a sloped line; the slope is $\dfrac{\text{rate of heating}}{mc_{\text{solid}}}$.
2. Melting — a horizontal plateau at the melting point; its length is proportional to $mL_f$.
3. Liquid warming — a sloped line, generally less steep than segment 1 if $c_{\text{liquid}} > c_{\text{solid}}$ (true for water).
4. Boiling — a horizontal plateau at the boiling point, usually the longest, since $L_v$ is large.
5. Gas warming — a sloped line again.

Reading a heating curve is a standard exam task: the slopes give the specific heat capacities of each phase (steeper slope ⇒ smaller $mc$), and the plateau lengths give the latent heats (longer plateau ⇒ larger $mL$).

6. Evaporation Versus Boiling

Boiling occurs throughout the bulk of a liquid, at a single fixed temperature (the boiling point) for a given external pressure, with bubbles of vapour forming inside the liquid.

Evaporation occurs only at the surface, at any temperature. Within any liquid the molecules have a spread of speeds; the fastest surface molecules occasionally have enough kinetic energy to escape the intermolecular attraction and leave. Because it is the fastest molecules that go, the mean kinetic energy of those left behind falls, and so does the temperature: evaporation cools the remaining liquid. This is the mechanism of sweating, of a wet towel feeling cold, and of the refrigerant in a fridge.

The rate of evaporation increases with:

• higher liquid temperature (more molecules near escape energy);
• larger surface area;
• moving air above the surface (which carries escaped molecules away before they can return);
• lower humidity and lower external pressure.

Worked example. A runner loses heat at $650\ \text{W}$, entirely by the evaporation of sweat. Taking the latent heat of vaporisation of sweat as $2.4\times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$, find the rate at which sweat must evaporate.

about a litre per hour — which is why endurance athletes must drink continuously.

7. Worked Examples Chaining $mc\,\Delta T$ and $mL$

Example 1 — ice to warm water. Find the energy needed to convert $0.30\ \text{kg}$ of ice at $-12\ ^{\circ}\text{C}$ into water at $25\ ^{\circ}\text{C}$. Take $c_{\text{ice}} = 2100$, $c_{\text{water}} = 4180\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$, $L_f = 3.34\times 10^{5}\ \dfrac{\text{J}}{\text{kg}}$.

Three stages, in order:

The melting stage alone accounts for about $70\%$ of the total, even though it involves no temperature change at all.

Example 2 — condensing steam. A jet of steam at $100\ ^{\circ}\text{C}$ is passed into $0.400\ \text{kg}$ of water at $20\ ^{\circ}\text{C}$ until the water reaches $50\ ^{\circ}\text{C}$. Ignoring the container, find the mass of steam that condensed. Take $L_v = 2.26\times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$, $c_{\text{water}} = 4180\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}$.

The steam first condenses at $100\ ^{\circ}\text{C}$, then the condensed water cools from $100\ ^{\circ}\text{C}$ to $50\ ^{\circ}\text{C}$:

The cool water absorbs

Setting released equal to absorbed:

Of the $\approx 4.9\times 10^{4}\ \text{J}$ this $20\ \text{g}$ of steam delivers, roughly $92\%$ is the latent term $m_sL_v$ and only $8\%$ is the subsequent cooling. The same energy taken from liquid water cooling by $50\ \text{K}$ would require about $0.24\ \text{kg}$ — more than ten times the mass. That dominance of the latent term is why steam burns are so much worse than hot-water burns.

Common Exam Pitfalls

• Expecting the temperature to rise during melting or boiling. It does not; the supplied energy is going into potential energy, not kinetic. The heating-curve plateaus are horizontal.
• Using $Q = mc\,\Delta T$ across a phase change, or $Q = mL$ for a temperature change. Each formula applies to exactly one kind of process.
• Omitting a stage in a multi-step problem. Warming ice to $0\ ^{\circ}\text{C}$, melting it, and warming the meltwater are three separate terms; forgetting the sensible-heat stages either side of a phase change is the most common slip.
• Confusing $L_f$ and $L_v$. Fusion is melting/freezing; vaporisation is boiling/condensing. For water $L_v$ is about seven times $L_f$.
• Forgetting that freezing and condensing release energy. The sign flips: $Q = mL$ is energy given out, not taken in.
• Believing evaporation requires the boiling point. Evaporation happens at any temperature, only at the surface, and it cools the liquid left behind.
• Leaving mass in grams, or forgetting that $L$ is quoted per kilogram.
• Assuming the boiling point is fixed. It depends on the external pressure — water boils below $100\ ^{\circ}\text{C}$ on a mountain.

Summary

• A change of state occurs at a fixed temperature (for a given pressure); the temperature is constant throughout melting or boiling.
• Specific latent heat: $L = \dfrac{Q}{m}$, so $Q = mL$, with $L_f$ for fusion and $L_v$ for vaporisation.
• The energy supplied during a phase change increases the potential-energy part of the internal energy — pulling molecules apart against their mutual forces — while the kinetic part, and hence the temperature, stays constant.
• $L_v \gg L_f$ because boiling breaks essentially all intermolecular bonds and does large expansion work $p\,\Delta V$ on the atmosphere, whereas melting only loosens the lattice.
• On a heating curve, sloped segments (slope set by $mc$) are temperature changes and horizontal plateaus (length set by $mL$) are phase changes.
• Evaporation occurs only at the surface, at any temperature, and cools the remaining liquid because the fastest molecules escape; boiling occurs throughout the liquid at a fixed temperature.
• Multi-stage problems chain $Q = mc\,\Delta T$ (each phase) with $Q = mL$ (each transition); the latent terms usually dominate the total.