Specific Latent Heat

This lesson covers specific latent heat as required by the Edexcel GCSE Physics specification (1PH0), Topic 3: Conservation of Energy. You need to understand what latent heat is, distinguish between latent heat of fusion and vaporisation, use the equation, and interpret heating curves.

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What Is Latent Heat?

When a substance changes state (e.g., solid to liquid, or liquid to gas), energy is needed — but the temperature does not change during the change of state.

Latent heat is the energy required to change the state of a substance without changing its temperature.

The word "latent" comes from Latin and means hidden — the energy is "hidden" because it does not cause a temperature rise.

Why Does the Temperature Stay Constant?

During a change of state:

• The energy supplied is used to break (or form) intermolecular bonds between particles.
• The energy goes into overcoming the attractive forces between particles, not into increasing their kinetic energy.
• Since temperature is a measure of the average kinetic energy of particles, and the kinetic energy does not change, the temperature remains constant.

Exam Tip: This is one of the most commonly tested explanations in the exam. You MUST state that the energy is used to "break intermolecular bonds" (or "overcome forces of attraction between particles") and that the kinetic energy of particles does not increase. Simply saying "energy is used to change state" is not enough for full marks.

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Specific Latent Heat of Fusion and Vaporisation

There are two types of specific latent heat:

Type	Change of State	Symbol	Definition
Specific latent heat of fusion ($L_f$Lf​)	Solid ↔ Liquid (melting/freezing)	$L_f$Lf​	Energy needed to change 1 kg of a substance from solid to liquid (or vice versa) at its melting point
Specific latent heat of vaporisation ($L_v$Lv​)	Liquid ↔ Gas (boiling/condensing)	$L_v$Lv​	Energy needed to change 1 kg of a substance from liquid to gas (or vice versa) at its boiling point

Key Point

The specific latent heat of vaporisation is always greater than the specific latent heat of fusion for the same substance. This is because:

• Changing from liquid to gas requires completely separating particles from each other.
• Changing from solid to liquid only requires partially overcoming the intermolecular forces — the particles are still close together.

Substance	$L_f$Lf​ (J/kg)	$L_v$Lv​ (J/kg)
Water	334,000	2,260,000
Ethanol	108,000	855,000
Lead	23,000	871,000

Exam Tip: Notice that for water, $L_v$Lv​ is about 6.8 times larger than $L_f$Lf​. This is why steam burns are so much worse than boiling water burns — steam releases a huge amount of latent heat when it condenses on your skin.

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The Equation

$E = m L$E=mL

Where:

• E = energy (joules, J)
• m = mass (kilograms, kg)
• L = specific latent heat (joules per kilogram, J/kg)

Units

• Specific latent heat (L) is measured in J/kg.
• Use $L_f$Lf​ for fusion (melting/freezing) and $L_v$Lv​ for vaporisation (boiling/condensing).

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Worked Examples

Example 1: Melting Ice

Question: How much energy is needed to melt 0.5 kg of ice at 0 °C? ($L_f$Lf​ for water = 334,000 J/kg)

Solution:

$E = m L_f = 0.5 \times 334{,}000 = 167{,}000 \text{ J} = 167 \text{ kJ}$E=mLf​=0.5×334,000=167,000 J=167 kJ

Example 2: Boiling Water

Question: How much energy is needed to boil 2 kg of water at 100 °C? ($L_v$Lv​ for water = 2,260,000 J/kg)

Solution:

$E = m L_v = 2 \times 2{,}260{,}000 = 4{,}520{,}000 \text{ J} = 4520 \text{ kJ}$E=mLv​=2×2,260,000=4,520,000 J=4520 kJ

Example 3: Finding Mass

Question: 668,000 J of energy is used to melt some ice. What mass of ice is melted? ($L_f$Lf​ = 334,000 J/kg)

Solution:

$m = \frac{E}{L_f} = \frac{668{,}000}{334{,}000} = 2 \text{ kg}$m=Lf​E​=334,000668,000​=2 kg

Example 4: Finding Latent Heat

Question: It takes 450,000 J to completely melt 3 kg of a substance at its melting point. What is its specific latent heat of fusion?

Solution:

$L_f = \frac{E}{m} = \frac{450{,}000}{3} = 150{,}000 \text{ J/kg}$Lf​=mE​=3450,000​=150,000 J/kg

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Heating Curves

A heating curve is a graph of temperature against time (or temperature against energy) when a substance is heated continuously.

Shape of a Heating Curve

flowchart LR
    A["Solid\n(temperature rising)"] --> B["Melting\n(temperature CONSTANT)"]
    B --> C["Liquid\n(temperature rising)"]
    C --> D["Boiling\n(temperature CONSTANT)"]
    D --> E["Gas\n(temperature rising)"]

Key Features

Section	What Happens	Temperature	Energy Goes Into
Solid heating	Particles vibrate faster	Rises	Increasing kinetic energy of particles
Melting (flat section)	Solid changes to liquid	Constant (melting point)	Breaking intermolecular bonds (latent heat of fusion)
Liquid heating	Particles move faster	Rises	Increasing kinetic energy of particles
Boiling (flat section)	Liquid changes to gas	Constant (boiling point)	Breaking intermolecular bonds (latent heat of vaporisation)
Gas heating	Particles move even faster	Rises	Increasing kinetic energy of particles

How to Read a Heating Curve

• Flat sections = change of state. The temperature is constant because energy is being used to break intermolecular bonds, not to increase kinetic energy.
• Sloped sections = the substance is being heated within a single state. Temperature rises because particles are gaining kinetic energy.
• The length of the flat section for boiling is longer than for melting (if the rate of energy supply is constant), because $L_v > L_f$Lv​>Lf​.

Exam Tip: If you are asked to explain a flat section of a heating curve, make sure you state: (1) a change of state is occurring, (2) energy is being used to break (or form) intermolecular bonds, and (3) the kinetic energy of the particles does not increase, so the temperature stays constant.

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Multi-Step Problems

Some questions require you to combine the SHC equation with the latent heat equation.

Example: Heating Ice to Steam

Question: Calculate the total energy needed to heat 0.1 kg of ice at 0 °C to steam at 100 °C.

Data: $L_f$Lf​ = 334,000 J/kg, c (water) = 4200 J/kg°C, $L_v$Lv​ = 2,260,000 J/kg.

Solution:

Step 1: Melt the ice at 0 °C (latent heat of fusion): $E_1 = m L_f = 0.1 \times 334{,}000 = 33{,}400 \text{ J}$E1​=mLf​=0.1×334,000=33,400 J

Step 2: Heat the water from 0 °C to 100 °C (SHC): $E_2 = m c \Delta\theta = 0.1 \times 4200 \times 100 = 42{,}000 \text{ J}$E2​=mcΔθ=0.1×4200×100=42,000 J