Specific Latent Heat

Latent heat is the energy associated with changes of state. The word "latent" means "hidden" — the energy is hidden in the sense that it does not cause a temperature change. This topic is closely linked to internal energy and is frequently tested in AQA exams.

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Definition and Equation

Key Definition: The specific latent heat (L) of a substance is the energy required to change the state of 1 kg of the substance at constant temperature.

Q = mL

where:

• Q is the energy transferred (J)
• m is the mass of the substance (kg)
• L is the specific latent heat (J kg⁻¹)

There are two types of specific latent heat:

• Specific latent heat of fusion (L_f) — the energy per kg to change between solid and liquid (melting or freezing).
• Specific latent heat of vaporisation (L_v) — the energy per kg to change between liquid and gas (boiling/evaporating or condensing).

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Why Is L_v Always Greater Than L_f?

During melting, the molecules are loosened from the rigid lattice structure but remain close together — intermolecular forces are only partially overcome. During vaporisation, the molecules must be completely separated from one another, overcoming all remaining intermolecular forces. Additionally, during vaporisation the substance expands greatly against atmospheric pressure, doing work on the surroundings (W = pΔV). This work also requires energy.

For water:

• L_f = 3.34 × 10⁵ J kg⁻¹ (334 kJ kg⁻¹)
• L_v = 2.26 × 10⁶ J kg⁻¹ (2260 kJ kg⁻¹)

The latent heat of vaporisation is about 6.8 times larger than the latent heat of fusion for water.

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Latent Heat Values for Common Substances

Substance	Melting Point (°C)	L_f (kJ kg⁻¹)	Boiling Point (°C)	L_v (kJ kg⁻¹)
Water	0	334	100	2260
Ethanol	−114	109	78	855
Lead	327	23	1750	858
Aluminium	660	397	2470	10 900
Copper	1083	205	2567	5070
Nitrogen	−210	25.7	−196	199
Oxygen	−219	13.9	−183	213

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Heating and Cooling Curves

A heating curve (temperature against time) for a substance heated at a constant rate shows a characteristic stepped shape:

1. Rising section (solid phase): Temperature increases linearly. The gradient depends on the specific heat capacity of the solid and the mass.
2. First flat section (melting): Temperature remains constant at the melting point. Energy is used to break bonds (latent heat of fusion). Duration depends on mL_f and the heating power.
3. Rising section (liquid phase): Temperature rises again, usually with a different gradient (different c value).
4. Second flat section (boiling): Temperature remains constant at the boiling point. This section is longer than the first because L_v > L_f.
5. Rising section (gas phase): Temperature increases once more.

Interpreting the Gradient

During a rising section, Q = mcΔθ and Q = Pt (where P is the constant heating power). Therefore:

Pt = mcΔθ → Δθ/t = P/(mc)

The gradient of the temperature-time graph equals P/(mc). A steeper gradient means either a smaller mass, a lower specific heat capacity, or a higher heating power.

Interpreting Flat Sections

During a flat section, Q = mL and Q = Pt. The duration of the flat section is:

t = mL/P

A longer flat section means a larger latent heat (for the same mass and power).

Exam Tip: Examiners often ask you to identify states, explain why sections are flat, or compare gradients. Always connect gradients to c and flat section lengths to L. If two substances are heated with the same power, the one with the longer flat section has the larger latent heat.

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Worked Examples

Worked Example 1 — Energy to Melt Ice

Question: Calculate the energy required to melt 0.50 kg of ice at 0 °C. (L_f = 3.34 × 10⁵ J kg⁻¹)

Solution:

Q = mL_f = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J = 167 kJ

Worked Example 2 — Complete Heating Process

Question: A 0.20 kg block of ice at −15 °C is heated until it becomes steam at 100 °C. Calculate the total energy required. (c_ice = 2100 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹, L_f = 3.34 × 10⁵ J kg⁻¹, L_v = 2.26 × 10⁶ J kg⁻¹)

Solution:

Step 1: Heat ice from −15 °C to 0 °C. Q₁ = mcΔθ = 0.20 × 2100 × 15 = 6300 J