Phase Changes and Heating Curves

What is it?

A phase change (change of state) turns a substance from one state into another without changing what it is: ice, liquid water and steam are all $\ce{H2O}$ .

Change	From → to	Energy
Melting (fusion)	solid → liquid	absorbed (endothermic)
Freezing	liquid → solid	released (exothermic)
Vaporization (boiling, evaporation)	liquid → gas	absorbed
Condensation	gas → liquid	released
Sublimation	solid → gas	absorbed
Deposition	gas → solid	released

Changes that pull particles apart (overcoming intermolecular forces) absorb energy; changes that bring them together release it.

Key idea

While a pure substance is melting or boiling, its temperature stays constant, even though heat is still being added. All the energy goes into overcoming the forces between particles, not into making them move faster. That’s why a heating curve has flat sections.

Why does it matter?

Cooling and heating. Sweat cools you because evaporation absorbs heat; ice keeps drinks at 0 °C while it melts; refrigerators work by evaporating and condensing a fluid.
Steam burns. Steam at 100 °C burns far worse than water at 100 °C, because it releases its large enthalpy of vaporization as it condenses on the skin.
Cooking and climate. Water boils at a lower temperature at altitude; huge amounts of energy are moved around the planet as water evaporates and condenses.

How does it work?

1. The heating curve

Heating ice at −15 °C steadily until it becomes steam at 125 °C gives five stages:

ice warms (sloped): $q = mc_\text{ice}\Delta T$
ice melts at 0 °C (flat): $q = n\Delta H_\text{fus}$
water warms (sloped): $q = mc_\text{water}\Delta T$
water boils at 100 °C (flat): $q = n\Delta H_\text{vap}$
steam warms (sloped): $q = mc_\text{steam}\Delta T$

The total heat is the sum of the five steps.

2. Enthalpies of fusion and vaporization

The enthalpy of fusion, $\Delta H_\text{fus}$ , is the energy to melt one mole of solid at its melting point; the enthalpy of vaporization, $\Delta H_\text{vap}$ , is the energy to vaporize one mole of liquid. For water, $\Delta H_\text{fus} = 6.01$ kJ/mol and $\Delta H_\text{vap} = 40.7$ kJ/mol. Vaporization needs far more energy because the molecules must be separated completely, while melting only loosens them.

3. Vapour pressure and boiling

Even below its boiling point, some molecules at the surface of a liquid have enough energy to escape: evaporation. In a closed container, the vapour builds up until evaporation and condensation balance. The pressure of the vapour then is the vapour pressure, which rises steeply with temperature.

A liquid boils when its vapour pressure equals the external pressure: bubbles of vapour can then form throughout the liquid. At 1 atm (760 mmHg) this is the normal boiling point: 100 °C for water. Where the air pressure is lower, such as on a mountain, water boils below 100 °C; in a pressure cooker it boils above 100 °C.

Think of it like this

Melting is like a crowd leaving a packed hall. While people are still squeezing out of the doors (the phase change), the crowd’s speed doesn’t increase: all the effort goes into getting through. Only once everyone is outside can they start to move faster (the temperature rises again).

More precisely

Evaporation happens at any temperature, from the surface only; boiling happens at one temperature (for a given pressure), throughout the liquid. The temperature stays constant during a phase change only while both phases are present and the pressure is constant. $\Delta H_\text{vap}$ itself changes slightly with temperature (water: 44.0 kJ/mol at 25 °C, 40.7 kJ/mol at 100 °C). The relationship between vapour pressure and temperature is described by the Clausius–Clapeyron equation.

Visualise it

Heating curve of water: temperature against heat added. A rising line for ice from minus 15 degrees to 0, a flat section at 0 degrees where ice melts, a rising line for liquid water from 0 to 100, a long flat section at 100 degrees where water boils, and a rising line for steam above 100. — Heating curve of water. The flat sections are phase changes; boiling takes by far the most energy.

Worked example

Worked example: Melting ice

Question: How much heat is needed to melt 100. g of ice at 0 °C?

Moles: $n = \dfrac{100.\ \text{g}}{18.02\ \text{g/mol}} = 5.549\ \text{mol}$
Heat:
$\small\begin{aligned} &q = n\Delta H_\text{fus} = 5.549\ \cancel{\text{mol}} \\[4pt] &\quad \times 6.01\ \frac{\text{kJ}}{\cancel{\text{mol}}} \\[4pt] &= 33.4\ \text{kJ} \end{aligned}$

Worked example: A full heating curve

Question: How much heat turns 25.0 g of ice at −15.0 °C into steam at 125.0 °C? ( $n = 25.0\ \text{g} \div 18.02\ \text{g/mol} = 1.387$ mol)

Warm ice:
$\small\begin{aligned} &q_1 = 25.0\ \text{g} \times 2.09\ \tfrac{\text{J}}{\text{g °C}} \\[4pt] &\quad \times 15.0\ \text{°C} \\[4pt] &= 784\ \text{J} = 0.784\ \text{kJ} \end{aligned}$
Melt: $q_2 = 1.387\ \text{mol} \times 6.01\ \text{kJ/mol} = 8.34\ \text{kJ}$
Warm water:
$\small\begin{aligned} &q_3 = 25.0\ \text{g} \times 4.184\ \tfrac{\text{J}}{\text{g °C}} \\[4pt] &\quad \times 100.0\ \text{°C} \\[4pt] &= 10\,460\ \text{J} = 10.46\ \text{kJ} \end{aligned}$
Boil: $q_4 = 1.387\ \text{mol} \times 40.7\ \text{kJ/mol} = 56.47\ \text{kJ}$
Warm steam:
$\small\begin{aligned} &q_5 = 25.0\ \text{g} \times 2.01\ \tfrac{\text{J}}{\text{g °C}} \\[4pt] &\quad \times 25.0\ \text{°C} \\[4pt] &= 1256\ \text{J} = 1.26\ \text{kJ} \end{aligned}$
Total:
$\small\begin{aligned} &q = (0.784 + 8.34 + 10.46 \\[4pt] &\qquad + 56.47 + 1.26)\ \text{kJ} \\[4pt] &= 77.3\ \text{kJ} \end{aligned}$
Boiling alone is about 73 % of the total.

Worked example: Why steam burns

Question: Compare the heat released to the skin (at 37 °C) by 5.00 g of steam at 100 °C and by 5.00 g of water at 100 °C.

Water cooling from 100 °C to 37 °C:
$\small\begin{aligned} &q = 5.00\ \text{g} \times 4.184\ \tfrac{\text{J}}{\text{g °C}} \\[4pt] &\quad \times 63\ \text{°C} \\[4pt] &= 1318\ \text{J} = 1.32\ \text{kJ} \end{aligned}$
Steam first condenses: $\dfrac{5.00\ \text{g}}{18.02\ \text{g/mol}} \times 40.7\ \text{kJ/mol} = 11.29\ \text{kJ}$ , then cools like the water (1.32 kJ): 12.6 kJ in total.
The steam delivers almost ten times as much heat.

Common mistake

Common mistake: Using q = mcΔT during a phase change

During melting or boiling, ΔT = 0, but heat is still absorbed. Use $q = n\Delta H$ for the flat parts and $q = mc\Delta T$ only for the sloped parts, with the right $c$ for each state.

Common mistake: Mixing J and kJ

$mc\Delta T$ usually gives joules, while $n\Delta H$ gives kilojoules. Convert to the same unit before adding.

Common mistake: Thinking boiling breaks bonds within molecules

When water boils, the bubbles contain $\ce{H2O}$ molecules. Only intermolecular forces are overcome.

Notation note

“Latent heat” is an older name for the enthalpy of a phase change.
ΔH(fus) is positive for melting; for freezing the same value is negative.

Remember this

Melting, vaporization, sublimation: endothermic. Freezing, condensation, deposition: exothermic.
Temperature is constant during a phase change: all the energy overcomes intermolecular forces.
Sloped parts: $q = mc\Delta T$ . Flat parts: $q = n\Delta H_\text{fus}$ or $n\Delta H_\text{vap}$ .
Water: $\Delta H_\text{fus} = 6.01$ kJ/mol; $\Delta H_\text{vap} = 40.7$ kJ/mol.
A liquid boils when its vapour pressure equals the external pressure.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/phase-changes-and-heating-curves/