Phase Changes and Heating Curves

Core Concept

When a substance is heated, its temperature rises only while it stays within a single phase; during an actual phase change (melting, vaporizing), all added heat goes into breaking intermolecular forces rather than raising temperature, so the temperature plateaus. A heating curve plots temperature versus heat added, showing sloped segments (q = mcΔT, sensible heat) and flat segments (q = nΔHfus or nΔHvap, latent heat).

Key Ideas

The six transitions

• Melting/freezing (solid ↔ liquid): ΔHfus ≈ 6.01 kJ/mol for water.
• Vaporization/condensation (liquid ↔ gas): ΔHvap ≈ 40.7 kJ/mol for water.
• Sublimation/deposition (solid ↔ gas): ΔHsub = ΔHfus + ΔHvap.

Sign convention

• Endothermic (energy in, positive q): melting, vaporization, sublimation.
• Exothermic (energy out, negative q): freezing, condensation, deposition.

Two kinds of heat

• Sensible heat (temperature change within a phase): q = m·c·ΔT.
• Latent heat (phase change at constant T): q = n·ΔH (or q = m·ΔH per gram).

Why plateaus occur

• During a phase change, added energy breaks IMFs instead of increasing molecular kinetic energy, so temperature holds constant until the transition completes.

Equations and Variables

• Sensible heat (slopes): q = m·c·ΔT, where q = heat (J), m = mass (g), c = specific heat capacity (J/g·°C), ΔT = temperature change.
• Latent heat of fusion (melting plateau): q = n·ΔHfus (or q = m·ΔHfus,per gram).
• Latent heat of vaporization (boiling plateau): q = n·ΔHvap.
• Sublimation: ΔHsub = ΔHfus + ΔHvap (Hess's law).
• Water constants: c_ice ≈ 2.09 J/g·°C, c_water = 4.184 J/g·°C, c_steam ≈ 2.01 J/g·°C; ΔHfus = 6.01 kJ/mol (= 334 J/g); ΔHvap = 40.7 kJ/mol (= 2260 J/g).

How It Works

1. Heat a solid: temperature rises along a slope, q = m·c_ice·ΔT.
2. Reach the melting point: added heat now breaks the lattice; temperature plateaus at 0 °C while ice melts (q = n·ΔHfus).
3. Heat the liquid: temperature rises again, q = m·c_water·ΔT.
4. Reach the boiling point: temperature plateaus at 100 °C while water vaporizes (q = n·ΔHvap).
5. Heat the gas: temperature rises once more, q = m·c_steam·ΔT.
6. Sum all five contributions to get the total heat for the entire process.

Worked Example

How much heat is required to convert 25.0 g of ice at −10.0 °C to steam at 110.0 °C?

1. Warm ice −10 → 0 °C: q₁ = m·c_ice·ΔT = 25.0 × 2.09 × 10.0 = 522.5 J.
2. Melt ice: q₂ = m·ΔHfus = 25.0 × 334 = 8,350 J.
3. Warm water 0 → 100 °C: q₃ = 25.0 × 4.184 × 100 = 10,460 J.
4. Vaporize water: q₄ = m·ΔHvap = 25.0 × 2260 = 56,500 J.
5. Warm steam 100 → 110 °C: q₅ = 25.0 × 2.01 × 10.0 = 502.5 J.
6. Total: q = 522.5 + 8,350 + 10,460 + 56,500 + 502.5 = 76,335 J ≈ 76.3 kJ.

Notice that vaporization (56.5 kJ) is by far the largest term — breaking the liquid's IMFs dominates the energy budget.

How it works

1. Heat a solid: temperature rises along a slope, q = m·c_ice·ΔT.
2. Reach the melting point: added heat now breaks the lattice; temperature plateaus at 0 °C while ice melts (q = n·ΔHfus).
3. Heat the liquid: temperature rises again, q = m·c_water·ΔT.
4. Reach the boiling point: temperature plateaus at 100 °C while water vaporizes (q = n·ΔHvap).
5. Heat the gas: temperature rises once more, q = m·c_steam·ΔT.
6. Sum all five contributions to get the total heat for the entire process.

Common confusions

• "Temperature rises during melting/boiling." — Wrong. Temperature is constant throughout a phase change; added heat breaks IMFs.
• "The plateau means heat isn't being absorbed." — Wrong. Heat is absorbed the whole time (q = nΔH); it just doesn't raise temperature.
• "Melting and boiling need about the same energy." — Wrong. ΔHvap (40.7 kJ/mol) is about 6.8× ΔHfus (6.01 kJ/mol) for water.
• "ΔHsub = ΔHfus × ΔHvap." — Wrong. They add: ΔHsub = ΔHfus + ΔHvap.
• "Specific heat is the same for ice, water, and steam." — Wrong. They differ (2.09, 4.184, 2.01 J/g·°C).

Quick review

• Six transitions; melting/vaporization/sublimation are endothermic.
• Slopes q = mcΔT; plateaus q = nΔHfus / nΔHvap.
• T constant during phase change; ΔHsub = ΔHfus + ΔHvap.
• Water: 6.01 and 40.7 kJ/mol.
• Worked: 25.0 g ice (−10 °C) → steam (110 °C) ≈ 76.3 kJ.