Specific Latent Heat

Put a thermometer in a pan of melting ice and you will see something surprising: while the ice is melting, the temperature stays stubbornly at $0\,°\text{C}$ , even though the hob is pouring energy in. The temperature only starts to climb once all the ice has melted. Where is that energy going? The answer is latent heat — the "hidden" energy needed to change a substance from one state to another without changing its temperature. This lesson, part of Topic P1 (Matter) of OCR Gateway Science A, defines specific latent heat, works through the equation $E = mL$ , distinguishes the latent heat of fusion from that of vaporisation, explains why the temperature stays constant during a change of state, and reads the flat plateaus of full heating and cooling curves.

By the end of this lesson you should be able to define specific latent heat, use and rearrange $E = mL$ , distinguish the specific latent heats of fusion and vaporisation, explain why temperature stays constant during a change of state, and interpret heating and cooling curves.

What Is Latent Heat?

Latent heat is the energy needed to change the state of a substance while its temperature stays constant. The word latent means "hidden", because this energy does not show up as a temperature rise — the thermometer reading does not move while the change of state is happening.

The specific latent heat ( $L$ ) of a substance is the energy needed to change the state of $1 \text{ kg}$ of the substance with no change in temperature. Its unit is $\text{J/kg}$ (joules per kilogram).

There are two specific latent heats, one for each change of state:

the specific latent heat of fusion, $L_f$ — the energy to melt (or freeze) $1 \text{ kg}$ at the melting point;
the specific latent heat of vaporisation, $L_v$ — the energy to boil (or condense) $1 \text{ kg}$ at the boiling point.

("Fusion" is the scientific word for melting, and "vaporisation" for boiling.)

Exam Tip: Specific latent heat is the energy to change the state of $1 \text{ kg}$ with no temperature change. Compare this with specific heat capacity, which is the energy to raise the temperature of $1 \text{ kg}$ by $1\,°\text{C}$ — latent heat changes the state, capacity changes the temperature.

The Specific Latent Heat Equation

The energy needed to change the state of a substance is:

$E = mL$

where $E$ is the energy transferred (in $\text{J}$ ), $m$ is the mass (in $\text{kg}$ ) and $L$ is the specific latent heat (in $\text{J/kg}$ ). You use $L_f$ for melting or freezing and $L_v$ for boiling or condensing.

The equation rearranges to:

$E = mL \qquad m = \dfrac{E}{L} \qquad L = \dfrac{E}{m}$

Notice there is no $\Delta\theta$ in this equation — because the temperature does not change during a change of state, only the mass and the latent heat matter.

Substance / change	Specific latent heat / $\text{J/kg}$
Ice melting (fusion of water)	$3.34 \times 10^{5}$
Water boiling (vaporisation of water)	$2.26 \times 10^{6}$

Worked Example 1 — Energy to melt ice

How much energy is needed to melt $2 \text{ kg}$ of ice at $0\,°\text{C}$ ? (Specific latent heat of fusion of water $= 3.34 \times 10^{5} \text{ J/kg}$ .)

Step 1 — write the equation: $E = mL$ .

Step 2 — substitute: $E = 2 \times 3.34 \times 10^{5}$ .

Step 3 — calculate: $E = 6.68 \times 10^{5} \text{ J}$ (= $668\,000 \text{ J}$ ).

Answer: $6.68 \times 10^{5} \text{ J}$ .

Worked Example 2 — Energy to boil water

Calculate the energy needed to boil away $0.5 \text{ kg}$ of water already at $100\,°\text{C}$ . (Specific latent heat of vaporisation of water $= 2.26 \times 10^{6} \text{ J/kg}$ .)

Step 1 — write the equation: $E = mL$ .

Step 2 — substitute: $E = 0.5 \times 2.26 \times 10^{6}$ .

Step 3 — calculate: $E = 1.13 \times 10^{6} \text{ J}$ .

Answer: $1.13 \times 10^{6} \text{ J}$ (= $1\,130\,000 \text{ J}$ ). Notice how much more energy boiling needs than melting — vaporising completely separates the particles, which takes far more energy than just loosening them.

Worked Example 3 — Rearranging for mass

A freezer removes $1.002 \times 10^{6} \text{ J}$ of energy from water at $0\,°\text{C}$ to freeze it. What mass of ice is produced? (Specific latent heat of fusion $= 3.34 \times 10^{5} \text{ J/kg}$ .)

Step 1 — rearrange for $m$ : $m = \dfrac{E}{L}$ .

Step 2 — substitute: $m = \dfrac{1.002 \times 10^{6}}{3.34 \times 10^{5}}$ .

Step 3 — calculate: $m = 3 \text{ kg}$ .

Answer: $3 \text{ kg}$ of ice is produced. (Freezing releases the same latent heat that melting absorbs, so the same equation applies.)

Worked Example 4 — Rearranging for specific latent heat

It takes $84\,500 \text{ J}$ to completely melt $0.25 \text{ kg}$ of a substance at its melting point. Calculate its specific latent heat of fusion.

Step 1 — rearrange for $L$ : $L = \dfrac{E}{m}$ .

Step 2 — substitute: $L = \dfrac{84\,500}{0.25}$ .

Step 3 — calculate: $L = 338\,000 \text{ J/kg} = 3.38 \times 10^{5} \text{ J/kg}$ .

Answer: $3.38 \times 10^{5} \text{ J/kg}$ — very close to the value for water, suggesting the substance could be ice.

Exam Tip: There is no temperature term in $E = mL$ , so do not try to multiply by $\Delta\theta$ . Keep mass in kilograms, and watch the powers of ten in the latent-heat values — write them in standard form to avoid dropping a zero.

Why the Temperature Stays Constant During a Change of State

This is one of the most-asked "explain" questions in P1, so it is worth getting exactly right. When a substance is melting or boiling, energy is continually being supplied, yet the temperature does not rise. The reason lies in where that energy goes.

Recall that internal energy has two parts: the kinetic energy of the particles (which sets the temperature) and the potential energy stored in the forces between particles. During a change of state:

the energy supplied is used to overcome the forces of attraction between the particles — to pull them apart (melting) or separate them completely (boiling);
this increases the potential energy of the particles, not their kinetic energy;
because the kinetic energy of the particles is unchanged, their average speed is unchanged, so the temperature stays constant.

Only once the change of state is complete — all the bonds that need breaking have been broken — does any further energy go into making the particles move faster, raising the temperature again. In short: during a change of state the energy increases the particles' potential energy (breaking the forces between them) rather than their kinetic energy, so the temperature does not change.

Exam Tip: The mark-winning phrase is that during a change of state the energy goes into breaking the forces between particles (increasing potential energy), not into making the particles move faster (kinetic energy), so the temperature stays constant. Always mention both halves — what the energy does do and what it does not do.

Heating and Cooling Curves

A heating curve plots temperature against time as a substance is heated steadily from solid to gas; a cooling curve is the reverse, as a gas is cooled back to a solid. Both have flat plateaus at the changes of state, which is where latent heat is being absorbed (heating) or released (cooling).

Specific Latent Heat

Specific Latent Heat

What Is Latent Heat?

The Specific Latent Heat Equation

Worked Example 1 — Energy to melt ice

Worked Example 2 — Energy to boil water

Worked Example 3 — Rearranging for mass

Worked Example 4 — Rearranging for specific latent heat

Why the Temperature Stays Constant During a Change of State

Heating and Cooling Curves

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