6.5 Phase Changes

Heating & Cooling Curve

Scrub through a heating or cooling curve to read live temperature, phase, and formula per segment. Slopes and plateaus show why T stalls during phase changes.

q = mcΔTPhase PlateausHeating & Cooling4 Substances
Topic 6.5

Energy of Phase Changes

Explain changes in the heat q absorbed or released by a system undergoing a phase transition based on the amount of the substance in moles and the molar enthalpy of the phase transition.

Energy must be transferred to a system to melt or boil a substance, so the energy of the system increases during a solid-to-liquid or liquid-to-gas transition. Likewise the system releases energy when it freezes or condenses.

The crucial experimental fact in the same EK: the temperature of a pure substance remains constant during a phase change. All the transferred energy goes into changing the potential energy of the particles — separating them against their intermolecular attractions — rather than into their kinetic energy, which is what a thermometer reads.

EK 6.5.A.2 gives the reversibility rule: the energy absorbed during a phase change equals the energy released during the complementary change in the opposite direction. The molar enthalpy of condensation is the negative of the molar enthalpy of vaporization; the molar enthalpy of fusion works for both melting and freezing.

The calculation:

q = n × ΔHtransition

Reading a heating curve is the standard AP task. It has two kinds of segment:

  • Sloped segments — one phase is being heated. Temperature rises, kinetic energy rises. Use q = mcΔT. A shallower slope means a larger specific heat capacity.
  • Flat plateaus — a phase change is occurring. Temperature is constant, potential energy rises. Use q = nΔH.

Why the boiling plateau is longer than the melting plateau: ΔHvap is always much larger than ΔHfus for the same substance. Melting only loosens the particles enough to flow; boiling must separate them completely against all intermolecular attractions. For water, ΔHfus = 6.01 kJ/mol while ΔHvap = 40.7 kJ/mol.

Key points

  • Temperature is constant during a phase change — energy goes into potential, not kinetic, energy.
  • Sloped segment → q = mcΔT. Flat segment → q = nΔHtransition.
  • ΔHvap ≫ ΔHfus for any given substance.
  • Forward and reverse transitions have equal magnitudes and opposite signs.

Equations

  • not on the sheet
    • moles undergoing the transition
    • molar enthalpy of fusion or vaporization (kJ/mol)
  • not on the sheet

Common mistakes

  • Using q = mcΔT during a phase change. ΔT is zero there.
  • Using mass with a molar enthalpy. Convert to moles first.
  • Assuming temperature rises during melting. It does not, for a pure substance.
  • Forgetting a segment in a multi-stage problem. Ice at −10 °C to steam at 110 °C has five separate calculations.

Worked example

How much energy is required to convert 36.0 g of ice at −15.0 °C to steam at 120.0 °C? Data: cice = 2.09, cwater = 4.18, csteam = 2.01 J·g⁻¹·°C⁻¹; ΔHfus = 6.01 kJ/mol; ΔHvap = 40.7 kJ/mol.

n = 36.0 g ÷ 18.02 g/mol = 2.00 mol. Five segments:

1. Heat ice, −15.0 → 0.0 °C:
q = (36.0)(2.09)(15.0) = 1129 J = 1.13 kJ

2. Melt ice at 0.0 °C:
q = (2.00)(6.01) = 12.02 kJ

3. Heat water, 0.0 → 100.0 °C:
q = (36.0)(4.18)(100.0) = 15 048 J = 15.05 kJ

4. Vaporize water at 100.0 °C:
q = (2.00)(40.7) = 81.4 kJ

5. Heat steam, 100.0 → 120.0 °C:
q = (36.0)(2.01)(20.0) = 1447 J = 1.45 kJ

Total = 1.13 + 12.02 + 15.05 + 81.4 + 1.45 = 111 kJ

Notice that vaporization alone accounts for about 73% of the total — the boiling plateau dominates the curve.

Full notes for topic 6.5 →