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Spec mapping: OCR H556 Module 5.1 — Thermal physics (specific latent heat of fusion and vaporisation; the equation Q=mL; the constant-temperature nature of a phase change; experimental determination of L). PAG 7 anchors the practical work alongside specific heat capacity. Refer to the official OCR H556 specification document for exact wording.
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 of the OCR A-Level Physics A specification (H556) requires you to understand specific latent heat of fusion and vaporisation, to use the equation Q=mL, and to explain what happens to the particles during a change of state. The practical work is anchored in PAG 7 (alongside the specific-heat-capacity electrical method of lesson 3).
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 L of 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:
If a mass m of a substance undergoes a change of state completely, the energy required is
Q=mL,
where L is the specific latent heat of fusion or vaporisation, whichever is appropriate.
Notice the contrast with the specific heat capacity formula Q=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 |
|---|---|
| Q=mcΔθ | Single-phase heating (no change of state) |
| Q=mLf | Melting / freezing at the melting point |
| Q=mLv | Boiling / condensing at the boiling point |
| Substance | Lf / kJ kg−1 | Melting point / ∘C | Lv / kJ kg−1 | Boiling point / ∘C |
|---|---|---|---|---|
| Water | 334 | 0 | 2260 | 100 |
| Ethanol | 108 | −114 | 855 | 78 |
| Lead | 23 | 327 | 858 | 1750 |
| Copper | 205 | 1085 | 4720 | 2560 |
| Nitrogen | 25.7 | −210 | 199 | −196 |
Two points stand out:
How much energy is needed to melt 0.50 kg of ice at 0∘C to water at 0∘C? (Lf=334 kJ kg−1.)
Q=mLf=(0.50)(334×103)=167000 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 s 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? (Lv=2260 kJ kg−1.)
Q=mLv=(0.40)(2260×103)=904000 J=904 kJ.
t=PQ=3000904000≈301 s≈5 minutes.
Contrast this with the time required to heat the water from 20∘C to 100∘C in the first place:
Q=mcΔθ=(0.40)(4200)(80)=134400 J≈134 kJ,t=134000/3000≈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 Lv 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? (cice=2100, cwater=4200, csteam=2000 J kg−1 K−1; Lf=334 kJ kg−1, Lv=2260 kJ kg−1.)
Break the process into five stages:
| Stage | Process | Formula | Energy / kJ |
|---|---|---|---|
| 1 | Warm ice from −10 to 0∘C | mcΔθ=(0.20)(2100)(10) | 4.2 |
| 2 | Melt ice at 0∘C | mLf=(0.20)(334) | 66.8 |
| 3 | Warm water from 0 to 100∘C | mcΔθ=(0.20)(4200)(100) | 84.0 |
| 4 | Boil water at 100∘C | mLv=(0.20)(2260) | 452.0 |
| 5 | Warm steam from 100 to 110∘C | mcΔθ=(0.20)(2000)(10) | 4.0 |
| Total | 611.0 |
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 diagrammatic summary:
flowchart LR
S1["-10 deg ice"] -- "warm ice (4.2 kJ)" --> S2["0 deg ice"]
S2 -- "melt (66.8 kJ)" --> S3["0 deg water"]
S3 -- "warm water (84.0 kJ)" --> S4["100 deg water"]
S4 -- "boil (452.0 kJ)" --> S5["100 deg steam"]
S5 -- "warm steam (4.0 kJ)" --> S6["110 deg steam"]
The stages where the temperature is constant (melting, boiling) correspond to flat sections (plateaus) on the temperature-vs-energy graph above. The steepest sloped sections are stages 1 and 5 (warming ice and warming steam), because ice and steam have lower specific heat capacities than liquid water.
flowchart TD
Q["Energy supplied: which equation?"] --> P{"Is the substance changing phase?"}
P -- "No (single phase)" --> SHC["Q = m c Delta-theta"]
P -- "Yes (melting / boiling)" --> LH["Q = m L"]
SHC --> ST["Temperature changes"]
LH --> SC["Temperature constant"]
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.
In a solid, particles are held at fixed positions by intermolecular forces. To melt the solid, these forces must be partially overcome. The particles do not need to be separated all the way — they remain in close contact in the liquid — but they must be released from their fixed lattice sites. Energy goes into raising their potential energy.
The kinetic energy (and hence the temperature) does not change. This is the key: a molecule of ice at 0∘C and a molecule of water at 0∘C have the same average kinetic energy; they differ in their potential energy.
In a liquid, particles are still close to their neighbours. To evaporate the liquid, the molecules must escape entirely from the attractive forces of their neighbours and move off into the vapour. This requires much more energy than melting, because the molecules must move from close contact to wide separation.
Again, the kinetic energy does not change, so the temperature stays at the boiling point throughout the process. The added energy is stored as potential energy of the gas molecules, which are now far from their original neighbours.
The ratio Lv/Lf is roughly 7 for water, around 8 for ethanol, around 20 for copper. A rough rule of thumb: the specific latent heat of vaporisation is always several times the specific latent heat of fusion.
The reason is that melting requires only that the rigid lattice be broken, while boiling requires that all the intermolecular forces be overcome. In a liquid, each molecule still has many neighbours in contact; in the vapour (at normal pressures), each molecule is essentially alone. You can think of it this way:
The far greater rearrangement of boiling requires far more energy.
In an experiment to measure the specific latent heat of fusion of ice, a 50 W heater is used to melt ice. Over 300 s, the mass of meltwater collected is 0.0480 kg. Find Lf.
Q=Pt=(50)(300)=15000 J. Lf=mQ=0.048015000≈3.13×105 J kg−1=313 kJ kg−1.
The accepted value is around 334 kJ kg−1. Our answer is about 6% lower. Why? Because some of the energy supplied by the heater was lost to the surroundings (conduction and convection from the heater and the container) rather than going into the ice. To reduce this, a control experiment is often used: the same apparatus without the heater on is run for the same duration, and the mass of meltwater due to heat from the surroundings alone is measured and subtracted. This is a standard technique for reducing systematic error.
A 0.25 kg block of ice at 0∘C is added to 0.50 kg of water at 20∘C in an insulated flask. Does all the ice melt? If so, what is the final temperature of the mixture? (Lf=334 kJ kg−1, cw=4200 J kg−1 K−1.)
First, check whether all the ice melts. The energy required to melt all 0.25 kg of ice is
Qmelt=mLf=(0.25)(334×103)=83500 J.
The energy available from cooling the water from 20∘C to 0∘C is
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