Lesson 1 · 25 min

Heating curves and the heat of phase change

The burner keeps burning, yet ice water stays at 0 °C. Heat a block of ice all the way to steam and see where the energy goes: warming stretches, the melting plateau, and the long boiling plateau.

In this lesson

Read heating and cooling curves: where the temperature rises, where it plateaus, and what a plateau means
Tell sensible heat ( $q = mc\Delta T$ ) from latent heat ( $q = n\Delta H$ ) and compute the heat for each stage
Convert heat into time at constant power, and explain why the boiling plateau is the longest
Explain at the molecular level where the heat goes during a phase change

Put a pan of ice water on the stove with a thermometer in it. The burner roars on, yet the thermometer sits at 0 °C until the last piece of ice has melted; only then does it climb. Near 100 °C it stops again, and stays there for the whole time the water boils. Heat is pouring in the entire time — so where is it going?

Predict first

A 500 W heater takes 100 g of ice at −20 °C all the way to steam at 120 °C (no heat losses). Which stage takes the longest?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Warming the ice from −20 °C to 0 °C Melting the ice at 0 °C Warming the water from 0 °C to 100 °C Boiling the water at 100 °C

Two kinds of heat: sensible and latent

A heating curve alternates between two kinds of stretch:

Warming stretches: one phase only, and the added heat raises the temperature. The heat is proportional to the mass, the specific heat capacity and the temperature change.
Plateaus: two phases coexist, and every joule goes into changing state while the temperature stays put. This is the latent heat of the transition, proportional to the amount of substance and its molar enthalpy of transition.

As long as two phases coexist, the temperature is pinned at the transition temperature — which is exactly why an ice–water mixture, fixed at 0 °C, became a thermometer calibration point.

q_{\text{warm}} = m\,c\,\Delta T \qquad q_{\text{plateau}} = n\,\Delta H_{\text{trs}} \qquad t = \frac{q}{P}

One formula for warming stretches and one for plateaus; at constant power, heat divided by power is time.

$m$: sample mass, g
$c$: specific heat capacity: ice 2.09, liquid water 4.18, steam 2.01 J g⁻¹ K⁻¹
$n$: amount of substance, mol
$\Delta H_{\text{trs}}$: molar enthalpy of transition: for water $\Delta_\text{fus}H = 6.01$ kJ/mol and $\Delta_\text{vap}H = 40.7$ kJ/mol
$P$: heater power, W (= J/s)

Why is the enthalpy of vaporization so much larger than that of fusion? Melting only loosens ice’s hydrogen-bonded network — liquid water keeps most of its hydrogen bonds and its molecules stay in contact. Vaporization must pull the molecules completely apart, overcoming almost all of their attractions. Heat added on a plateau raises the molecules’ potential energy, not their kinetic energy, so the temperature does not change. Every substance in the table below shows the same pattern: $\Delta_\text{vap}H$ is several times $\Delta_\text{fus}H$ .

Transition temperatures and molar enthalpies at 1 atm
Substance	Melting point / °C	ΔfusH / (kJ/mol)	Boiling point / °C	ΔvapH / (kJ/mol)
Water	0.0	6.01	100.0	40.65
Ethanol	−114.1	4.93	78.3	38.56
Benzene	5.5	9.87	80.1	30.72
Ammonia	−77.7	5.66	−33.3	23.33

Source: CRC Handbook of Chemistry and Physics, 97th ed. (2016): enthalpies of fusion and vaporization.

Worked example

From ice to superheated steam

How much heat turns 50.0 g of ice at −10.0 °C into steam at 110.0 °C (1 atm)? What fraction goes into boiling?

Constant power · water at 1 atm

Heating-curve simulator

Sample mass · 100 g Cooling power · 500 W Starting temperature of the ice · -20 °C Final temperature of the steam · 120 °C

Steam cools

Time elapsed · 0.0 s

Temperature 120.0°C

Elapsed time 0s

Heat removed 0.00kJ

Ice 0 %

Liquid 0 %

Steam 100 %

Molecular arrangement (schematic)

Stage	Temperature	Heat	Time	Share
Steam cools	120 → 100 °C	4.02 kJ	8.0 s	1.3 %
Condensing	100 °C	225.92 kJ	451.8 s	73.0 %
Water cools	100 → 0 °C	41.80 kJ	83.6 s	13.5 %
Freezing	0 °C	33.36 kJ	66.7 s	10.8 %
Ice cools	0 → −20 °C	4.18 kJ	8.4 s	1.4 %
Total	−20 ⇄ 120 °C	309.28 kJ	618.6 s	100 %

Switch to Cool: the same plateaus appear at the same temperatures, run backwards — condensing steam releases 40.7 kJ/mol and freezing water 6.01 kJ/mol. With heat on the horizontal axis, each plateau’s length is the molar enthalpy times the amount, whatever the power.

Concept check

Ice is melting at 0 °C while it absorbs heat. What happens to the molecules’ average kinetic and potential energy?

Kinetic energy rises; potential energy unchanged Kinetic energy unchanged; potential energy rises Both rise Neither changes; the heat disappears

Try the calculation

A 1000 W kettle (assume all its heat goes into the water) must boil away 0.500 kg of water already at 100 °C. How many seconds does it take? ( $\Delta_\text{vap}H = 40.7$ kJ/mol)

Key takeaways

A heating curve alternates warming stretches ( $q = mc\Delta T$ ) with plateaus ( $q = n\Delta H$ ).
While two phases coexist the temperature holds; the heat raises intermolecular potential energy and is released again on cooling.
Water’s enthalpy of vaporization (40.7 kJ/mol) is about seven times its enthalpy of fusion (6.01 kJ/mol), so the boiling plateau is the longest.
At constant power a plateau lasts $t = n\Delta H/P$ : more power shortens it but never moves its temperature.

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Data and references

W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016): enthalpies of fusion and vaporization, heat capacity of water.
N. J. Tro, Chemistry: A Molecular Approach, 5th ed. (2020), Section 11.7 (heating curve of water; cp of ice 2.09, water 4.18, steam 2.01 J g⁻¹ °C⁻¹).
R. Feistel & W. Wagner, “A new equation of state for H2O ice Ih”, J. Phys. Chem. Ref. Data 35, 1021 (2006) — heat capacity of ice (IAPWS R10-06).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 3B and 4A (phase transitions).