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If you place a mug of water in a freezer, you can watch it slowly cool, droplet by droplet, on an hour-by-hour timescale. When the temperature reaches 0 °C, however, something curious happens: the cooling slows dramatically, and the temperature sticks at 0 °C while ice begins to form. Only when the water is completely frozen does the temperature of the ice resume falling. The same thing happens, in reverse, if you heat a pan of water on a hob: the temperature rises rapidly up to 100 °C and then sticks there, while bubbles of vapour form and escape, until the pan is nearly dry. The process of freezing or boiling happens at constant temperature, yet it clearly involves a transfer of energy. The energy involved is the latent heat, the topic of this lesson.
Module 5.1.2 of the OCR A-Level Physics A specification (H556) requires you to understand specific latent heat of fusion and vaporisation, to use the equation E = mL, and to explain what happens to the particles during a change of state.
The word latent means "hidden". When you supply energy to a body, you expect the temperature to rise. Sometimes, however, the temperature does not rise at all — the energy appears to vanish. What has happened to it? It has not really vanished; it has gone into changing the state of the substance (melting, boiling, sublimation), and we call it latent because it is hidden from the thermometer.
More precisely:
Latent heat is the energy transferred to or from a substance during a change of state at constant temperature, without any change in the kinetic energy of the particles.
At the melting point, the energy supplied goes into breaking down the lattice structure of the solid: increasing the potential energy of the particles, not their kinetic energy. At the boiling point, the energy supplied goes into pulling molecules apart against the attractive forces that hold the liquid together: again, an increase in potential energy.
Specific latent heat
Lof a substance is the energy required to change the state of 1 kg of the substance without any change in temperature.
The symbol is L, the units are J kg⁻¹, and there are two distinct values, one for each type of phase transition:
L_f: the energy per kilogram required to change a substance from solid to liquid at its melting point.L_v: the energy per kilogram required to change a substance from liquid to gas at its boiling point.If a mass m of a substance undergoes a change of state completely, the energy required is
E = m L
where L is the specific latent heat of fusion or vaporisation, whichever is appropriate.
Notice the contrast with the specific heat capacity formula E = mcΔθ: here there is no Δθ, because the temperature does not change during the phase transition. Latent heat is a price you pay per kilogram to change state, regardless of temperature.
| Formula | When it applies |
|---|---|
E = m c Δθ | Heating (no change of state) |
E = m L_f | Melting / freezing at melting point |
E = m L_v | Boiling / condensing at boiling point |
| Substance | L_f (kJ kg⁻¹) | Melting point | L_v (kJ kg⁻¹) | Boiling point |
|---|---|---|---|---|
| Water | 334 | 0 °C | 2260 | 100 °C |
| Ethanol | 108 | -114 °C | 855 | 78 °C |
| Lead | 23 | 327 °C | 858 | 1750 °C |
| Copper | 205 | 1085 °C | 4720 | 2560 °C |
| Nitrogen | 25.7 | -210 °C | 199 | -196 °C |
Two points stand out:
L_v is always much larger than L_f. For water, L_v = 2260 kJ kg⁻¹ is about seven times L_f = 334 kJ kg⁻¹. The reason is physical: melting requires only that neighbours slide past one another while remaining in close contact, so only a fraction of the intermolecular forces must be overcome. Boiling requires that molecules escape entirely from the liquid into the vapour phase, separating by many molecular diameters — a much larger job.L_v. Water's latent heat of vaporisation is second only to ammonia among common substances. This has huge consequences: sweating cools us very efficiently (each kilogram of sweat evaporated absorbs 2.26 MJ), steam burns are far more dangerous than boiling-water burns (the condensing steam deposits a huge additional latent-heat energy on the skin), and storms transport huge amounts of energy from the tropics via latent heat of water vapour.How much energy is needed to melt 0.50 kg of ice at 0 °C to water at 0 °C? (L_f = 334 kJ kg⁻¹.)
E = m L_f
= (0.50)(334 × 10³)
= 167 000 J
= 167 kJ
Notice: the temperature does not change, yet 167 kJ is required. If you have only a 2 kW heater, it takes 167/2 ≈ 84 seconds simply to melt the ice — time during which the thermometer reads 0 °C the whole way.
A 3.0 kW electric kettle contains 0.40 kg of water at its boiling point of 100 °C. Ignoring heat loss, how long does it take to boil all the water off as steam? (L_v = 2260 kJ kg⁻¹.)
E = m L_v
= (0.40)(2260 × 10³)
= 904 000 J
= 904 kJ
t = E/P = 904000 / 3000 ≈ 301 s ≈ 5 minutes
Contrast this with the time required to heat the water from 20 °C to 100 °C in the first place:
E = mcΔθ = (0.40)(4200)(80) = 134 400 J ≈ 134 kJ
t = 134 000 / 3000 ≈ 45 s
It takes about 45 s to heat the water up but nearly 5 minutes to boil it away — about seven times longer. The reason is the factor of roughly 7 between the energies cΔθ and L_v for water over this temperature change.
How much energy is needed to convert 0.20 kg of ice at -10 °C to steam at 110 °C? (c_ice = 2100, c_water = 4200, c_steam = 2000 J kg⁻¹ K⁻¹; L_f = 334 kJ kg⁻¹, L_v = 2260 kJ kg⁻¹.)
Break the process into five stages:
| Stage | Process | Formula | Energy (kJ) |
|---|---|---|---|
| 1 | Warm ice from -10 °C to 0 °C | mcΔθ = (0.20)(2100)(10) | 4.2 |
| 2 | Melt ice at 0 °C | mL_f = (0.20)(334) | 66.8 |
| 3 | Warm water from 0 °C to 100 °C | mcΔθ = (0.20)(4200)(100) | 84.0 |
| 4 | Boil water at 100 °C | mL_v = (0.20)(2260) | 452.0 |
| 5 | Warm steam from 100 °C to 110 °C | mcΔθ = (0.20)(2000)(10) | 4.0 |
| Total | 611.0 kJ |
The dominant contribution is stage 4: boiling the water. This is very typical. Whenever a phase change is involved, the latent heat term usually dwarfs the sensible-heat terms.
A plot of temperature against energy input looks like this:
graph LR
S1[-10 °C<br/>ice] -- warm ice --> S2[0 °C<br/>ice]
S2 -- melt --> S3[0 °C<br/>water]
S3 -- warm water --> S4[100 °C<br/>water]
S4 -- boil --> S5[100 °C<br/>steam]
S5 -- warm steam --> S6[110 °C<br/>steam]
The stages where the temperature is constant (melting, boiling) correspond to flat sections on the temperature vs energy graph. The steepest sections are stages 1 and 5 (warming ice and warming steam), because ice and steam have lower specific heat capacities than liquid water.
During a change of state, the added energy goes into changing the arrangement of the particles — specifically, increasing their potential energy without changing their kinetic energy.
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