Chemistry: Atoms First 2e · Liquids and Solids
Phase Transitions
On this page 9 sections
In 30 seconds
A phase transition Change between solid, liquid, and gas states Full entry → is a change of state — melting, freezing, vaporization, condensation, sublimation, or deposition — each carrying an enthalpy change: heat is absorbed for the "upward" transitions (melting, vaporization, sublimation) and released for the reverse ones. Since enthalpy of sublimation Heat to convert one mole of solid directly to gas Full entry → equals fusion plus vaporization, the six transitions form a closed energy loop (Hess's law).
The central idea is energy accounting: during a phase change at constant pressure the temperature stays constant — added heat breaks intermolecular attractions rather than raising kinetic energy. Only within a single phase does temperature rise. This topic develops the heating curve Temperature vs. added heat, with plateaus at transitions Full entry →, the equations q = mcΔT and q = nΔH, and multi-step calculations — the classic "heat ice to steam" problem.
Why this matters
Phase-change energy is everywhere: sweating cools you because vaporization absorbs heat from skin (about 2.26 kJ per gram of water); steam burns worse than boiling water because condensation releases the full enthalpy of vaporization Heat to vaporize one mole of liquid at its boiling point Full entry →; refrigerators and heat pumps are vaporization/condensation engines; freeze-drying preserves food by subliming ice. These heats let engineers size cooling systems — and explain why phase changes dominate cooking and weather energy budgets.
The college version
Core Concepts
The six transitions and their enthalpy changes
| Transition | Direction | Enthalpy change |
|---|---|---|
| Melting (fusion) | solid → liquid | endothermic, ΔHfus > 0 |
| Freezing | liquid → solid | exothermic, -ΔHfus |
| Vaporization | liquid → gas | endothermic, ΔHvap > 0 |
| Condensation | gas → liquid | exothermic, -ΔHvap |
| Sublimation | solid → gas | endothermic, ΔHsub > 0 |
| Deposition | gas → solid | exothermic, -ΔHsub |
For a given substance, ΔHsub = ΔHfus + ΔHvap (Hess's law) — for water, 46.66 kJ/mol. Vaporization dominates: nearly seven times the heat of fusion.
Why temperature stays constant during a transition
During melting, added heat breaks the hydrogen bonds of the ice lattice, so temperature does not rise until the last crystal dissolves; the same plateau occurs at the boiling point. This is why an ice–water mixture stays at 0 °C while melting. At the transition, the two phases coexist in dynamic equilibrium Two phases interconverting at equal rates Full entry →.
Heating curves: two kinds of heat
A heating curve plots temperature versus added heat. Sloped segments are single-phase regions where
q = m c ΔT
with m mass, c specific heat capacity Heat to raise 1 g of a substance by 1 °C Full entry →, ΔT temperature change. Flat segments are phase changes where:
q = n ΔH
with n moles and ΔH molar enthalpy. Multi-step problems sum the heat per segment: heating absorbs (+), cooling releases (−).
Supercooling and superheating (model limits)
Liquids can be cooled below their freezing point without freezing (supercooling) or heated above their boiling point without boiling (superheating) — the phase change needs a nucleation site to start. These metastable states don't violate equilibrium; boiling chips in the lab provide nucleation sites that prevent violent "bumping."
Vapor pressure and temperature: the Clausius–Clapeyron link
Vapor pressure's temperature dependence follows:
ln(P2P1) = -ΔHvapR(1T2 - 1T1)
At the boiling point, vapor pressure equals external pressure — which is why the boiling point moves with altitude, and why the vapor-pressure curve on a phase diagram doubles as the boiling-point curve.
How It Works / Step-by-Step Process
Heating 18.0 g of ice from −20 °C to steam at 120 °C:
- Divide the process into segments: ice warming (−20 → 0 °C), melting at 0 °C, water warming (0 → 100 °C), vaporizing at 100 °C, steam warming (100 → 120 °C).
- For warming segments use q = mcΔT with the right specific heat (ice 2.09, water 4.18, steam 2.01 J/(g·°C)).
- For phase changes use q = nΔH.
- Sum all five heats; heating = +, cooling = −.
- Check: vaporization should dominate.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| Temperature rises while ice melts. | No — during melting temperature stays at 0 °C (constant pressure); heat breaks the lattice, not kinetic energy. |
| Boiling water gets hotter the longer it boils. | At constant pressure, no — temperature is pinned at the boiling point. |
| Evaporation requires reaching the boiling point. | Evaporation is surface-only, at any temperature; boiling is bulk vaporization at the boiling point. |
| Sublimation is special to dry ice. | Any solid with significant vapor pressure sublimes — iodine, snow in dry air, freezer frost. |
| Condensation and freezing absorb heat. | They release heat (exothermic) — the reverse of vaporization and melting. |
| ΔHsub = ΔHfus + ΔHvap only sometimes. | It always holds for the same substance — a Hess's-law path. |
| All heating curves have plateaus at the same temperatures. | Same shape, but plateau temperatures and lengths depend on the substance's ΔH and specific heats. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Changing phase is like breaking a chain of paper clips: melting snaps the chains so pieces slide (that takes energy), boiling snaps almost all links so pieces fly free (even more). While you're snapping clips, temperature doesn't rise — the heat is busy breaking links. Freezing and condensing re-link the pieces and give the energy back.
Worked example
Example 1: Heating 18.0 g of ice from −20 °C to steam at 120 °C
Given: m = 18.0 g, n ≈ 1.00 mol; specific heats: ice 2.09, water 4.18, steam 2.01 J/(g·°C); ΔHfus = 6.01, ΔHvap = 40.65 kJ/mol.
Step 1 — warm ice: q1 = (18.0)(2.09)(20) = 752 J
Step 2 — melt ice: q2 = (1.00)(6.01) = 6010 J
Step 3 — warm water: q3 = (18.0)(4.18)(100) = 7524 J
Step 4 — vaporize water: q4 = (1.00)(40.65) = 40,650 J
Step 5 — warm steam: q5 = (18.0)(2.01)(20) = 724 J
Total:
qtotal = 752 + 6010 + 7524 + 40,650 + 724 = 55,660 J ≈ 55.7 kJ
Vaporization alone contributes 73% — the key lesson of the heating curve. Run in reverse, the same calculation releases exactly 55.7 kJ.
Example 2: Why do steam burns hurt so much?
A 150 g jet of steam at 100 °C condenses to water at 100 °C: n = 150 g / 18.02 g/mol = 8.32 mol.
Write the condensation heat:
q = n(-ΔHvap) = (8.32 mol)(-40.65 kJ/mol) = -338 kJ
The negative sign means 338 kJ is released onto skin. For comparison, heating 1.0 kg of water from 20 °C to 100 °C requires:
q = mcΔT = (1000 g)(4.18 J/(g·°C))(80 °C) = 334 kJ
So condensing 150 g of steam delivers roughly the energy of boiling an entire liter of water — in an instant. That is why steam scolds are so much worse than boiling-water splashes of the same size.
Key takeaways
- Six transitions: melting/freezing, vaporization/condensation, sublimation/deposition; upward = endothermic, downward = exothermic.
- ΔHsub = ΔHfus + ΔHvap for the same substance.
- Water: ΔHfus = 6.01 kJ/mol, ΔHvap = 40.65 kJ/mol.
- Temperature is constant during a phase change at constant pressure — the heating-curve plateaus.
- q = mcΔT for single-phase heating; q = nΔH for phase changes.
- Vaporization dominates the energy budget: ~7× the heat of fusion for water.
- Condensation and freezing release heat — the basis of steam burns and frost warming.
- Supercooling/superheating are metastable states needing nucleation sites; boiling chips prevent bumps.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Name the six phase transitions and state whether each is endothermic or exothermic.
Show answer
Melting (+), freezing (−), vaporization (+), condensation (−), sublimation (+), deposition (−).
Why does temperature remain constant while a pure substance melts, even as heat is added?
Show answer
At constant pressure, added heat breaks intermolecular attractions (potential energy), not average kinetic energy — so temperature holds steady until the transition completes.
Which single step dominates the total heat for ice at −20 °C → steam at 120 °C?
Show answer
Vaporization (40.65 kJ) — by far the largest step (73% of the total).
How much heat is released when 1.00 mol of steam at 100 °C condenses to water at 100 °C?
Show answer
−40.65 kJ — condensation releases -ΔHvap.
Why are boiling chips added to liquids being heated in the lab?
Show answer
Boiling chips provide nucleation sites for bubbles, preventing superheating and violent bumping.
Write ΔHsub for water in terms of ΔHfus and ΔHvap, and give the value.
Show answer
ΔHsub = ΔHfus + ΔHvap = 46.66 kJ/mol.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- phase transition
- Change between solid, liquid, and gas states
- enthalpy of fusion
- Heat to melt one mole of solid at its melting point
- enthalpy of vaporization
- Heat to vaporize one mole of liquid at its boiling point
- enthalpy of sublimation
- Heat to convert one mole of solid directly to gas
- heating curve
- Temperature vs. added heat, with plateaus at transitions
- specific heat capacity
- Heat to raise 1 g of a substance by 1 °C
- dynamic equilibrium
- Two phases interconverting at equal rates
- supercooling / superheating
- Metastable states where a phase persists past its transition point
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.