Latent heat and phase changes

Temperature tracks the mean translational kinetic energy of the molecules. A phase change spends its energy somewhere else: on prising molecules apart against the forces holding them together, which is potential energy, not speed. Melting loosens a rigid lattice into a liquid whose molecules still touch; boiling removes each molecule from its neighbours entirely and also pushes the surrounding atmosphere back to make room for vapour. Neither raises the mean kinetic energy, so neither moves the thermometer, and the reading stays clamped at the transition temperature until the last gram has converted. The energy flow is real and measurable. A 2.0 kW element raising 1.00 kg of water from 20 °C to 100 °C delivers 3.35 × 10⁵ J and takes 167 s, with the thermometer climbing the whole time. Boiling that same kilogram away takes 2.26 × 10⁶ J, 19 minutes at the same power, with the thermometer motionless at 100 °C throughout.

Because the transition happens at one temperature, its energy cost cannot depend on a ΔT. What it does depend on is how much matter converts, so Q = mL defines a specific latent heat L in J kg⁻¹: the latent heat of fusion L_f for solid to liquid, the latent heat of vaporisation Lᵥ for liquid to vapour. For water at 1 atm, L_f = 3.34 × 10⁵ J kg⁻¹ and Lᵥ = 2.26 × 10⁶ J kg⁻¹, a ratio of 6.77. The gap is structural, not accidental. Melting only has to break the lattice's rigidity; the molecules stay in contact and keep most of their mutual binding. Vaporisation has to remove a molecule from all of its neighbours, and the specific volume jumps from 1.04 × 10⁻³ to 1.67 m³ kg⁻¹, a factor of 1600. Making that room costs p Δv = 1.69 × 10⁵ J kg⁻¹ of work against the atmosphere, which is only 7.5% of Lᵥ; the other 92.5% is the internal energy of separation.

Plot temperature against energy supplied for 1.00 kg of water taken from −20 °C ice to 120 °C steam and you get five legs. Warming the ice takes 42 kJ, melting takes 334 kJ, warming the liquid takes 419 kJ, boiling takes 2260 kJ, superheating the steam takes 40 kJ: 3.09 MJ in all, and boiling alone is 73% of it. On the sloping legs the graph obeys dT/dQ = 1/(mc), so the ice leg, at c = 2100 J kg⁻¹ K⁻¹, rises very nearly twice as steeply as the liquid leg at 4186 J kg⁻¹ K⁻¹. On the flat legs dT/dQ = 0 and the length of the plateau, not its height, carries the physics. That makes the curve a measuring instrument: heat at a steady power P and the x-axis becomes time, so timing a plateau of duration t gives L = P t / m directly.

In a calorimeter problem with a phase change, the first question is not what the final temperature is but whether the change runs to completion. Compare the energy on offer with the energy the change demands. Drop 0.100 kg of ice at 0 °C into 0.300 kg of water at 20 °C: cooling that water all the way to 0 °C releases 0.300 × 4186 × 20 = 2.51 × 10⁴ J, while melting all the ice would need 0.100 × 3.34 × 10⁵ = 3.34 × 10⁴ J. The demand exceeds the supply, so the melt stalls partway. The melted fraction is x = 25116/33400 = 0.752, meaning 75.2 g becomes water and 24.8 g stays ice, and the final state is that mixture at exactly 0 °C. Had the inequality gone the other way, every gram would melt and the final temperature would rise above 0 °C, which is then the unknown to solve for.

For an insulated system, energy leaving one part enters another: ΣQ = 0. Give every term the sign of the energy gained by its own substance, so warming and melting count positive while cooling, freezing and condensing count negative, and never smuggle a second sign in by writing a temperature difference backwards. The discipline earns its keep when one substance crosses several segments. A gram of steam at 100 °C landing on skin at 45 °C first condenses, releasing 2260 J, and only then cools as liquid, releasing a further 4.186 × 55 = 230 J, for 2490 J in all. A gram of liquid water at 100 °C cooling over that same interval delivers only the 230 J. The steam carries 10.8 times as much energy, and the whole difference is the condensation term that a ΔT calculation never sees.

L is not a property of the substance alone. Because it includes the work p Δv, it depends on the pressure, and because the molecular binding is temperature-dependent, it depends on where the transition happens: water's Lᵥ is 2.50 × 10⁶ J kg⁻¹ at 0 °C but only 2.26 × 10⁶ J kg⁻¹ at 100 °C, and it falls to zero at the critical point, 374 °C and 22.1 MPa, where liquid and vapour cease to be distinguishable. Below the triple point a solid converts straight to vapour, with Lₛ ≈ L_f + Lᵥ, about 2.83 × 10⁶ J kg⁻¹ for ice. The plateau itself assumes equilibrium: carefully cooled water can supercool several degrees below 0 °C and then freeze in a rush, and evaporation removes fast molecules from a free surface at any temperature at all, which is why a puddle dries without ever reaching 100 °C.