Chemistry 2e · Liquids and Solids

Phase Transitions

8 min read

Constants cross-checked against NIST-referenced data: ΔH_fus = 333.55 J/g, ΔH_vap = 2257 J/g, c_p(liquid) = 4.184 J/(g · K) (2026-08).

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
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In 30 seconds

A is a change of state — melting, freezing, vaporizing, condensing, subliming, or depositing. What makes these changes special is that they happen at constant temperature: ice and liquid water coexist at 0 °C while heat pours in and melts the ice, and the temperature does not budge until the last crystal is gone. The energy absorbed or released during a transition is called ("hidden" heat), because it changes potential energy — the energy of molecular arrangement — rather than kinetic energy, which is what a thermometer measures.

This topic explains the six transitions, the enthalpy changes that accompany them, and how to calculate the heat required to move a substance through temperature changes and phase changes using heat capacities and enthalpies of fusion and vaporization.

Why this matters

Phase-transition energy is enormous compared to ordinary warming. Boiling away a pot of water takes about five times the energy needed to heat it from room temperature to boiling — which is why steam burns are so dangerous: steam at 100 °C carries far more energy than water at 100 °C. Refrigeration, air conditioning, and heat pumps all exploit vaporization (cooling) and condensation (heating) cycles. In medicine, steam sterilization, cryotherapy, and fever-reducing cooling pads rely on the same physics. And the energy of freezing and thawing governs everything from ice-cream making to frost heave in roads.

The college version

Core Concepts

The six transitions and their enthalpy changes

Every transition pairs a "heating" direction with a "cooling" direction:

Transition	Direction	Enthalpy change	Sign
melting / fusion	solid → liquid	enthalpy of fusion, ΔH_fus	endothermic (+)
freezing	liquid → solid	-ΔH_fus	exothermic (−)
vaporization	liquid → gas	enthalpy of vaporization, ΔH_vap	endothermic (+)
condensation	gas → liquid	-ΔH_vap	exothermic (−)
sublimation	solid → gas	enthalpy of sublimation, ΔH_sub	endothermic (+)
deposition	gas → solid	-ΔH_sub	exothermic (−)

Melting and vaporization are endothermic — they require heat input because molecules must be pulled apart against intermolecular forces. The reverse transitions release exactly the same amount of energy: ΔH_fus and ΔH_vap are equal in magnitude for the forward and reverse processes. Because vaporization separates molecules completely (breaking essentially all intermolecular contacts) while melting only loosens them, ΔH_vap is much larger than ΔH_fus. For water: ΔH_fus = 6.01 kJ/mol (333.55 J/g) versus ΔH_vap = 40.65 kJ/mol (2257 J/g) — about seven times larger.

Why temperature stays constant during a transition

While a pure substance undergoes a transition, added heat goes into breaking intermolecular attractions (increasing potential energy), not into raising molecular kinetic energy. The temperature therefore stays pinned at the transition temperature until one phase is entirely converted. Only after the transition completes does further heating raise the temperature. This is why a pot of boiling water stays at 100 °C no matter how high the burner is set — the extra heat only vaporizes water faster.

The reverse is equally important: freezing releases heat. Water freezing in a pipe releases ΔH_fus per gram, and that released heat is what keeps a body of water from freezing solid overnight.

Heating curves and the energy budget

A plots temperature versus heat added and shows the signature flat regions at each transition. Calculating the total energy to take a substance from one temperature to another through transitions requires breaking the process into segments:

Temperature-change segments (no transition): q = m c ΔT, where m is mass, c is specific heat capacity, and ΔT is the temperature change.
Transition segments (constant temperature): q = m ΔH, using the appropriate enthalpy per gram (or per mole with n instead of m).

Heat capacities differ by phase — for water, c_ice = 2.09 J/(g · K), c_liquid = 4.184 J/(g · K), and c_steam = 2.01 J/(g · K) — so each temperature segment must use the capacity for the phase present.

Sublimation and deposition

Some substances skip the liquid phase entirely at certain pressures. Sublimation is solid → gas directly; deposition is gas → solid directly. Dry ice (solid CO₂) sublimes at −78.5 °C at 1 atm because liquid CO₂ cannot exist below 5.11 atm. Freeze-drying (lyophilization) uses sublimation to remove water from food and pharmaceuticals while preserving structure. Frost forming on a cold windshield is deposition — water vapor going straight to ice.

How It Works / Step-by-Step Process

To calculate the heat required for a multi-step warming (or cooling) process:

Identify every segment: temperature changes and transitions on the path.
List the phase present in each segment and its heat capacity or enthalpy.
Compute temperature segments with q = mcΔT, transitions with q = mΔH.
Verify units: g × J/(g · K) × K = J; g × J/g = J.
Sum all segments. The largest term is usually the vaporization term.

Common Confusions

Common Confusion	Correct Understanding
"Temperature rises while a substance is melting/boiling."	No — during a pure transition, temperature stays constant until the phase change completes.
"Evaporation and boiling are the same."	Evaporation occurs at the surface at any temperature; boiling occurs throughout the liquid when vapor pressure equals external pressure.
"Melting and freezing are different processes with different energies."	They are the same transition in opposite directions with equal-magnitude energy changes.
"Ice at 0 °C and water at 0 °C have the same energy."	They have the same temperature (kinetic energy) but different potential energy — liquid water holds the extra ΔH_fus.
"Sublimation happens only for dry ice."	Many solids sublime under the right conditions (iodine, solid CO₂, snow in dry air); frost formation is deposition, the reverse.
"Using q = mcΔT for the whole heating curve."	It works only for temperature-change segments; transitions need q = mΔH.

Eli explains

The same idea, in plain words

Explain it like I’m 10

Think of molecules as kids holding hands in a chain. Melting is loosening the grip so they can shuffle around but stay touching; vaporizing is letting go completely so they run off. Both take energy — that's why ice in your drink melts and why boiling water needs constant heat. Freezing and condensing are the same steps backward, and they give off energy, which is why steam and boiling water burn so badly.

Worked example

Example 1: Energy to convert 50.0 g of ice at −20 °C to steam at 120 °C

Segment 1 — warm ice from −20 °C to 0 °C:

$q 1 = mc ice Δ T = (50.0 g)(2.09 J /(g \cdot K))(20 K) = 2090 J$

Segment 2 — melt ice at 0 °C:

$q 2 = m Δ H fus = (50.0 g)(333.55 J / g) = 16, 700 J$

Segment 3 — warm water from 0 °C to 100 °C:

$q 3 = mc water Δ T = (50.0 g)(4.184 J /(g \cdot K))(100 K) = 20, 900 J$

Segment 4 — vaporize water at 100 °C:

$q 4 = m Δ H vap = (50.0 g)(2257 J / g) = 112, 850 J$

Segment 5 — warm steam from 100 °C to 120 °C:

$q 5 = mc steam Δ T = (50.0 g)(2.01 J /(g \cdot K))(20 K) = 2010 J$

Total:

$q total = 2090 + 16, 700 + 20, 900 + 112, 850 + 2010 = 154, 550 J \approx 155 kJ$

The vaporization step alone (113 kJ) supplies about 73% of the total — a striking demonstration of why ΔH_vap dominates every heating budget.

Example 2: How much ice can condensing steam melt?

What mass of ice at 0 °C can be melted by the energy released when 100.0 g of steam at 100 °C condenses to water at 100 °C?

First, the condensation energy:

$q = m Δ H vap = (100.0 g)(2257 J / g) = 225, 700 J$

Then the mass of ice this energy can melt:

$m ice = q Δ H fus = 225, 700 J 333.55 J / g = 676.7 g$

Condensing just 100 g of steam releases enough energy to melt nearly 677 g of ice — roughly seven times its own mass. This ratio (2257/333.55 ≈ 6.77) is exactly the ratio of the two enthalpies, and it explains why steam scalds are so much worse than hot-water burns: the condensation step dumps the large vaporization enthalpy directly onto skin.

Key takeaways

Transitions happen at constant temperature; heat input during a transition changes potential energy, not temperature.
Melting, vaporization, and sublimation are endothermic; freezing, condensation, and deposition are exothermic.
ΔH_vap > ΔH_fus because vaporization breaks all intermolecular contacts. For water: 40.65 vs 6.01 kJ/mol.
Temperature segments: q = mcΔT. Transition segments: q = mΔH (or nΔH).
Steam carries far more energy than liquid water at the same temperature — the condensation enthalpy adds on top.
Sublimation bypasses the liquid phase; dry ice and freeze-drying rely on it.
Heat capacities differ by phase: ice 2.09, liquid water 4.184, steam 2.01 J/(g·K).

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Why does a pot of boiling water stay at 100 °C even on a high burner?
Show answer
Boiling is a transition: extra heat goes into vaporizing water (potential energy), not raising kinetic energy, so the temperature stays pinned at 100 °C until the water is gone.
Which is larger for a given substance, ΔH_fus or ΔH_vap, and why?
Show answer
ΔH_vap is much larger: vaporization separates molecules completely, breaking essentially all intermolecular contacts, while melting only loosens them. For water the ratio is about 6.8:1.
In Example 1, which segment contributes the most energy, and what fraction of the total?
Show answer
The vaporization segment (112,850 J out of 154,550 J total) — about 73%.
Is freezing endothermic or exothermic? Give a real-world consequence.
Show answer
Freezing is exothermic: it releases ΔH_fus. This released heat slows the freezing of lakes, and water pipes can burst when ice forms because the freezing water expands.
How much heat is needed to warm 25.0 g of liquid water from 25 °C to 75 °C? (c = 4.184 J/(g · K))
Show answer
q = mcΔT = (25.0)(4.184)(50) = 5230 J ≈ 5.23 kJ.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

phase transition: A change between solid, liquid, and gas states
enthalpy of fusion, Δ Hfus: Heat absorbed to melt one mole (or gram) of solid
enthalpy of vaporization, Δ Hvap: Heat absorbed to vaporize one mole (or gram) of liquid
enthalpy of sublimation, Δ Hsub: Heat absorbed for solid → gas directly
latent heat: "Hidden" heat absorbed or released during a transition at constant temperature
heating curve: Plot of temperature vs heat added showing transition plateaus
specific heat capacity, c: Heat needed to raise 1 g of a substance by 1 K (no transition)

Sources & references

openstax.org — Chemistry 2e

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.

In 30 seconds

Why this matters

The college version

Core Concepts

The six transitions and their enthalpy changes

Why temperature stays constant during a transition

Heating curves and the energy budget

Sublimation and deposition

How It Works / Step-by-Step Process

Common Confusions

The same idea, in plain words

Worked example

Example 1: Energy to convert 50.0 g of ice at −20 °C to steam at 120 °C

Example 2: How much ice can condensing steam melt?

Key takeaways

Check yourself

Keep learning

Key vocabulary

Related lessons

Sources & references