Specific Latent Heat

Heating a substance either makes it warmer or makes it change state. These two effects are handled by two separate equations, and knowing which to use is the key skill in this topic.

• Raising the temperature without changing state uses specific heat capacity, $\Delta E=mc\Delta \theta$.
• Changing the state without changing temperature uses specific latent heat, $E=mL$.

$\Delta E=mc\Delta \theta$ is given on the Physics equation sheet. It is the same equation covered in the Specific Heat Capacity lesson, where the required practical for measuring $c$ is set out in full.

$E=mL$ is also given on the Physics equation sheet. You do not have to memorise either equation, but you must know when to apply each one.

The rest of this lesson focuses on latent heat and on choosing between the two equations when a question involves both warming and a change of state.

Latent heat is the energy needed to change the state of a substance. The word "latent" means hidden, because this energy does not raise the temperature: a thermometer shows no change while the state is changing.

Latent heat changes the internal energy of a substance by changing the particles' potential energy, but it does not change their kinetic energy, so the temperature stays constant.

When a solid melts, the energy supplied breaks the fixed arrangement holding the particles in place. When a liquid boils, energy is used to pull the particles fully apart into a gas. In both cases the energy goes into potential energy, not kinetic energy, so no temperature rise occurs.

The specific latent heat of a substance is the energy per kilogram:

Specific latent heat is the amount of energy required to change the state of one kilogram of a substance with no change in temperature.

The energy to change the state of a mass $m$ is found from the specific latent heat equation.

$E=mL$

• $E$ is the energy transferred, in joules (J)
• $m$ is the mass, in kilograms (kg)
• $L$ is the specific latent heat, in joules per kilogram (J/kg)

This equation is given on the Physics equation sheet.

Worked example — the specific latent heat of fusion of water is 334 000 J/kg. How much energy is needed to melt 0.5 kg of ice at 0 °C?

$E=mL=0.5\times 334000=167000\text{J}$

So 167 000 J (167 kJ) melts the ice, all at 0 °C. Notice the temperature does not appear in this equation, because it does not change during melting.

A substance has two specific latent heats, one for each kind of change of state.

Specific latent heat of fusion is the energy to change 1 kg between solid and liquid (melting or freezing).

Specific latent heat of vaporisation is the energy to change 1 kg between liquid and gas (boiling or condensing).

For a given substance, vaporisation usually needs much more energy than fusion, because boiling has to pull the particles completely apart, while melting only loosens their fixed arrangement.

Change of state	Latent heat used	Water value
Melting / freezing	Fusion	334 000 J/kg
Boiling / condensing	Vaporisation	2 260 000 J/kg

Worked example — how much energy is needed to boil 2 kg of water already at 100 °C? Use the specific latent heat of vaporisation, 2 260 000 J/kg.

$E=mL=2\times 2260000=4520000\text{J}$

That is 4 520 000 J (4.52 MJ), far more than would be needed to melt the same mass, exactly as expected.

A graph of temperature against time (or against energy supplied) for a substance being heated steadily shows a distinctive stepped shape.

Temp
 |            ____ (gas heating)
 |           /
 |      ____/  <- boiling (flat)
 |     /
 |    /  <- liquid heating (sloped)
 |___/
 |  <- melting (flat)
 |_/  <- solid heating (sloped)
 |________________________ Time

• Sloped sections are where the substance is warming up within one state. The temperature rises, so use $\Delta E=mc\Delta \theta$.
• Flat (horizontal) sections are where the substance is changing state. The temperature is constant, so use $E=mL$.

The first flat section is melting, at the melting point; the second, longer flat section is boiling, at the boiling point. A cooling graph is the mirror image: flat sections appear as the substance freezes and condenses, releasing latent heat while the temperature holds steady.

The boiling plateau is longer than the melting plateau because vaporisation needs far more energy, so it takes longer at the same rate of heating.

Harder questions combine warming and a change of state, so you must split the process into stages and pick the right equation for each.

Use $\Delta E=mc\Delta \theta$ for a sloped part of the graph (temperature changing). Use $E=mL$ for a flat part (state changing).

Worked example — how much energy is needed to heat 0.5 kg of water from 20 °C to 100 °C, then boil it all away? Use $c=4200$ J/kg °C and $L_{\text{vap}}=2260000$ J/kg.

Stage 1 — warm the water from 20 °C to 100 °C ($\Delta \theta =80$ °C):

$\Delta E=mc\Delta \theta =0.5\times 4200\times 80=168000\text{J}$

Stage 2 — boil all the water at 100 °C:

$E=mL=0.5\times 2260000=1130000\text{J}$

Total energy = $168000+1130000=1298000\text{J}$ (about 1.3 MJ).

Splitting the process into a warming stage and a boiling stage keeps each equation and its units under control.

1. Using the wrong equation for the stage

If the temperature is changing, use $\Delta E=mc\Delta \theta$. If the state is changing at constant temperature, use $E=mL$. Check the graph: sloped means heat capacity, flat means latent heat.

2. Putting a temperature change into E = mL

The latent heat equation has no $\Delta \theta$ term because the temperature does not change during a change of state. Do not multiply by a temperature.

3. Confusing fusion with vaporisation

Fusion is melting and freezing (solid to liquid); vaporisation is boiling and condensing (liquid to gas). Vaporisation needs much more energy per kilogram.

4. Forgetting to add up all the stages

When a question involves warming and then a change of state, work out each stage separately and add the energies together. Missing a stage gives an answer that is far too small.

5. Mixing up mass units

Both equations use mass in kilograms. If a question gives the mass in grams, convert it first, or the answer will be a thousand times out.