Why does temperature stay constant during a phase change?
When ice is heated steadily, its temperature climbs to 0 °C and then stops, rising again only after all the ice has melted. The heat absorbed during that time doesn’t increase the molecules’ average kinetic energy (so the temperature doesn’t change). Instead it overcomes attractions between molecules (U3.1), letting them leave their fixed positions in the crystal. That energy is stored as potential energy between the molecules.
- Melting, vaporization, and sublimation overcome intermolecular forces: endothermic, ΔH > 0.
- Freezing, condensation, and deposition form intermolecular attractions: exothermic, ΔH < 0. Each reverse process has a ΔH of the same size and opposite sign.
A phase change only affects intermolecular forces; it does not break the covalent bonds inside molecules. When water boils, H₂O molecules are still H₂O molecules.
Heat of a phase change
The heat absorbed when 1 mol of a substance melts is its molar enthalpy of fusion, ΔH(fus)摩尔熔化焓, and the heat absorbed when 1 mol of liquid vaporizes is its molar enthalpy of vaporization, ΔH(vap)摩尔汽化焓. For a phase change:
For water, ΔH(fus) = 6.01 kJ/mol and ΔH(vap) = 40.7 kJ/mol (CRC Handbook). The enthalpy of vaporization is about 7 times the enthalpy of fusion. On melting, molecules only become free to move while staying close together, so only some of the hydrogen bonds must be overcome. On vaporizing, molecules must be separated completely, which overcomes almost all of the attractions between them.
Heating curves
Heating at a constant rate and plotting temperature against time (or heat added) gives a heating curve加热曲线:
- Sloped segments: a single state whose temperature rises, with q = m·c·ΔT. The larger the specific heat, the more slowly the temperature rises and the gentler the slope. Liquid water’s specific heat (4.18) is about twice that of ice (2.09) and steam (1.86), so the liquid segment is the gentlest.
- Plateaus: two states coexist and the temperature stays constant, with q = n·ΔH. The longer the plateau, the larger the enthalpy of the phase change, so the boiling plateau is much longer than the melting plateau.
A cooling curve is the reverse: the sloped segments go down, and the plateaus are condensation and freezing, which release heat.
Move the heat slider to see the temperature and how much of each state is present; use “Zoom in on the start” to compare the slopes for ice and liquid water:
Heat needed for each segment
Specific heats: ice 2.09, water 4.18, steam 1.86 J/(g·°C) (OpenStax Chemistry 2e, Table 5.1); ΔH(fus) = 6.01 kJ/mol and ΔH(vap) = 40.7 kJ/mol (CRC Handbook).
Example 1: Calculating the total heat
How much heat is needed to turn 36.0 g of ice at −10.0 °C into water at 25.0 °C? (c(ice) = 2.09 and c(water) = 4.18 J/(g·°C); ΔH(fus) = 6.01 kJ/mol)
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Split the process into three stages. Warming stages use q = mcΔT; the phase change (constant temperature) uses q = nΔH:
- ① ice warms from −10.0 °C to 0 °C
- ② ice melts at 0 °C
- ③ water warms from 0 °C to 25.0 °C
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① Warm the ice. ΔT = 10.0 °C (heat in J):
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② Melt the ice. Find moles, then multiply by the enthalpy of fusion (in kJ):
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③ Warm the water. ΔT = 25.0 °C:
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Total. Put everything in kJ before adding:
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Check. Melting takes the most heat (12.0 kJ, about 73%); a common mistake is adding 752 J to 12.0 kJ without converting units.
Example 2: Reading a heating curve
On a pure substance’s heating curve (constant heating rate), the melting plateau lasts 2 min and the boiling plateau lasts 13 min. The liquid segment is gentler than the solid segment. What can you conclude?
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Identify: what a plateau means. On a plateau the temperature is constant while the substance changes phase; at a constant heating rate, the heat absorbed is proportional to time.
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Compare the two plateaus.
The enthalpy of vaporization is about 6.5 times the enthalpy of fusion.
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Connect to intermolecular forces. Melting only frees molecules to move past one another while they stay in contact; vaporization separates them completely, overcoming far more intermolecular attraction, so ΔH(vap) is much larger.
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Compare the slopes. At the same heating rate, the liquid segment is gentler, meaning more heat is needed for the same temperature rise: the liquid has a larger specific heat than the solid.
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Why the temperature is constant on a plateau. The heat absorbed overcomes intermolecular forces (raising potential energy), so the molecules’ average kinetic energy, and hence the temperature, doesn’t change.
- Ice at −20 °C being heated
Only solid is present, so the temperature rises.
- At 0 °C, with ice and water together
The melting plateau.
- Water at 50 °C being heated
Only liquid is present.
- At 100 °C, with water boiling
The vaporization plateau.
- Steam at 110 °C being heated
Only gas is present.
- At −78 °C, with dry ice subliming
Solid turns directly to gas at constant temperature.
- Wax still melting (solid and liquid together)
A phase change in progress, at constant temperature.
- Ethanol heated from 20 °C to 60 °C
A liquid warming up.
Common mistakes
Mistake 1: “If you keep heating, the temperature must rise.” During a phase change the temperature stays constant; the heat absorbed overcomes intermolecular forces instead of increasing average kinetic energy.
Mistake 2: “When water boils, bonds inside the H₂O molecules break.” Boiling overcomes only the forces between molecules (hydrogen bonds and so on); the O–H bonds inside the molecules stay intact, and the gas formed is still H₂O.
Mistake 3: “Freezing and condensing also absorb heat.” They are the reverses of melting and vaporizing, so they release heat. That’s why steam condensing on skin releases a lot of heat and causes severe burns.
Quick check
How much heat is needed to melt 27.0 g of ice at 0 °C? (ΔH(fus) = 6.01 kJ/mol; M = 18.0 g/mol)
n = 27.0 ÷ 18.0 = 1.50 mol; q = 1.50 × 6.01 = 9.02 kJ. B multiplies ΔH by the mass.
On water’s heating curve, the boiling plateau is much longer than the melting plateau. Which is the best explanation?
Plateau length reflects the enthalpy of the phase change. A explains the slopes of sloped segments; B is a common mistake.
With constant heating, the liquid-water segment is gentler than the ice segment. Why?
For the same heat, a substance with a larger specific heat warms more slowly.
How much heat is released when 9.0 g of steam at 100 °C condenses to water at 100 °C? (ΔH(vap) = 40.7 kJ/mol)
n = 9.0 ÷ 18 = 0.50 mol, which releases 0.50 × 40.7 ≈ 20 kJ. No temperature change doesn’t mean no heat change.
A beaker of water is heated at a constant rate. Which change would make the boiling plateau on its heating curve longer?
The plateau lasts for the time it takes to supply n·ΔHvap at the heating rate. Doubling the water doubles the heat needed, so the plateau is longer; heating faster makes it shorter; the starting temperature affects only the warming segment before the plateau.
Which statement is correct as ice melts at 0 °C?
The temperature is constant, so the average kinetic energy is too; the heat absorbed overcomes intermolecular forces, raising the potential energy between molecules.
Which phase change has a negative ΔH?
Deposition is the reverse of sublimation and releases heat, so ΔH < 0.
AP-style free response
A heater with a constant power of 50.0 J/s heats 46.0 g of solid substance X (molar mass 46.0 g/mol). The heating curve shows a melting plateau lasting 100. s; while liquid, the temperature rises 60.0 °C in 132 s; and the boiling plateau lasts 790. s. (Hypothetical data; ignore heat loss.)
(a) During which two parts of the heating curve does the temperature stay constant? What is happening then?
(b) Calculate the molar enthalpy of fusion of X.
(c) Calculate the molar enthalpy of vaporization of X, and explain why it is much larger than the enthalpy of fusion.
(d) Calculate the specific heat of liquid X.
(e) On the melting plateau, how does the average kinetic energy of X’s molecules change? Where does the absorbed heat go?
Model answer and scoring
(a) The melting plateau and the boiling plateau. Two states coexist, and the heat absorbed overcomes intermolecular forces as the phase changes.
(b) q = 50.0 J/s × 100. s = 5.00 × 10³ J; n = 46.0 g ÷ 46.0 g/mol = 1.00 mol; ΔH(fus) = 5.00 kJ/mol.
(c) q = 50.0 J/s × 790. s = 3.95 × 10⁴ J, so ΔH(vap) = 39.5 kJ/mol. Vaporization separates the molecules completely, overcoming far more intermolecular attraction than melting (after melting, the molecules are still close together).
(d) q = 50.0 J/s × 132 s = 6.60 × 10³ J; c = 6600 J ÷ (46.0 g × 60.0 °C) = 2.39 J/(g·°C).
(e) The average kinetic energy doesn’t change (the temperature is constant). The absorbed heat overcomes intermolecular forces and becomes potential energy between the molecules.
Scoring: (a) both plateaus and the phase changes; (b) q = power × time, divided by n; (c) the value and a particle-level explanation; (d) the specific heat; (e) unchanged average kinetic energy and increased potential energy.
Summary
- During a phase change the temperature stays constant: the heat absorbed overcomes intermolecular forces (raising potential energy), while average kinetic energy is unchanged. Phase changes don’t break bonds inside molecules.
- Melting, vaporization, and sublimation are endothermic; freezing, condensation, and deposition are exothermic, with ΔH of the same size and opposite sign.
- q = n·ΔH(fus) or n·ΔH(vap). For water, ΔH(fus) = 6.01 kJ/mol and ΔH(vap) = 40.7 kJ/mol; vaporization separates molecules completely, so it is much larger.
- On a heating curve, plateaus are phase changes, with length ∝ the enthalpy of the phase change; sloped segments follow q = m·c·ΔT, and the larger the specific heat, the gentler the slope.
AP tip: For a multistep heat calculation, break it into “warm → change phase → warm,” use the right equation for each (m·c·ΔT or n·ΔH), and watch the conversion between J and kJ. To explain a plateau, say “average kinetic energy is unchanged; the heat overcomes intermolecular forces.”