Specific Latent Heat

This lesson covers specific latent heat as required by the AQA GCSE Physics specification (4.3.2). Specific latent heat is the quantity of energy needed to change the state of a substance without changing its temperature. This concept explains why it takes a lot of energy to melt ice or boil water, even though the temperature does not change during these processes.

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What Is Latent Heat?

Latent heat is the energy required to change the state of a substance without changing its temperature. The word "latent" means "hidden" — the energy is "hidden" because it does not cause a temperature rise. Instead, the energy is used to overcome the forces of attraction (bonds) between particles.

There are two types of latent heat:

Type	Change of State	Definition
Specific latent heat of fusion (Lf)	Solid to liquid (melting) or liquid to solid (freezing)	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 (Lv)	Liquid to gas (boiling) or gas to liquid (condensation)	Energy needed to change 1 kg of a substance from liquid to gas (or vice versa) at its boiling point

Exam Tip: The word "specific" means "per unit mass" (per kilogram). The word "latent" means "hidden" — the energy does not cause a temperature change. "Fusion" refers to melting (fusing from solid to liquid), and "vaporisation" refers to boiling (turning from liquid to vapour). Learn these definitions precisely.

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The Specific Latent Heat Equation

The equation for calculating the energy involved in a change of state is:

energy for a change of state = mass x specific latent heat

E = m x L

Where:

• E = energy transferred, measured in joules (J)
• m = mass of the substance, measured in kilograms (kg)
• L = specific latent heat, measured in joules per kilogram (J/kg)

Rearranging the Equation

To Find	Formula
Energy	E = m x L
Mass	m = E / L
Specific latent heat	L = E / m

Exam Tip: The equation E = m x L is given on the equation sheet in the exam, so you do not need to memorise it. However, you MUST know how to use it and rearrange it. Make sure you know the difference between Lf (fusion) and Lv (vaporisation) and when to use each one.

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Values of Specific Latent Heat

Substance	Specific Latent Heat of Fusion Lf (J/kg)	Specific Latent Heat of Vaporisation Lv (J/kg)
Water	334,000 (3.34 x 10^5)	2,260,000 (2.26 x 10^6)
Ethanol	108,000 (1.08 x 10^5)	855,000 (8.55 x 10^5)
Lead	23,000 (2.3 x 10^4)	871,000 (8.71 x 10^5)
Aluminium	397,000 (3.97 x 10^5)	10,900,000 (1.09 x 10^7)
Iron	247,000 (2.47 x 10^5)	6,090,000 (6.09 x 10^6)

Important Observations

• The specific latent heat of vaporisation is always much greater than the specific latent heat of fusion for the same substance.
• This is because during vaporisation, all the intermolecular forces must be completely overcome to separate the particles entirely. During fusion, only some of the forces are overcome — the particles move from a regular arrangement to a close but irregular arrangement.

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Worked Examples

Worked Example 1: Melting Ice

How much energy is needed to melt 0.5 kg of ice at 0 degrees C? (Lf for water = 334,000 J/kg)

Step 1: Write down the known values: m = 0.5 kg, L = 334,000 J/kg

Step 2: Use the equation: E = m x L

Step 3: Substitute: E = 0.5 x 334,000

Step 4: Calculate: E = 167,000 J = 167 kJ

Worked Example 2: Boiling Water

How much energy is needed to boil 2 kg of water at 100 degrees C? (Lv for water = 2,260,000 J/kg)

Step 1: Write down the known values: m = 2 kg, L = 2,260,000 J/kg

Step 2: Use the equation: E = m x L

Step 3: Substitute: E = 2 x 2,260,000

Step 4: Calculate: E = 4,520,000 J = 4,520 kJ = 4.52 MJ

Worked Example 3: Finding Mass

A total of 668,000 J of energy is supplied to melt a block of ice at 0 degrees C. Calculate the mass of the ice. (Lf for water = 334,000 J/kg)

Step 1: Write down the known values: E = 668,000 J, L = 334,000 J/kg

Step 2: Rearrange: m = E / L

Step 3: Substitute: m = 668,000 / 334,000

Step 4: Calculate: m = 2 kg

Exam Tip: In calculations, always check that your mass is in kilograms (not grams). If the question gives mass in grams, convert to kilograms by dividing by 1000 before substituting into the equation. For example, 500 g = 0.5 kg.

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Latent Heat and the Heating Curve

The concept of latent heat explains the flat sections on a heating curve:

graph TD
    subgraph Heating_Curve["Heating Curve for Water"]
        A["Below 0C:<br/>Solid ice heating<br/>Temperature rises"] --> B["At 0C:<br/>MELTING<br/>Temperature constant<br/>Energy = m x Lf"]
        B --> C["0C to 100C:<br/>Liquid water heating<br/>Temperature rises"]
        C --> D["At 100C:<br/>BOILING<br/>Temperature constant<br/>Energy = m x Lv"]
        D --> E["Above 100C:<br/>Steam heating<br/>Temperature rises"]
    end

    style A fill:#3498db,color:#fff
    style B fill:#9b59b6,color:#fff
    style C fill:#2ecc71,color:#fff
    style D fill:#9b59b6,color:#fff
    style E fill:#e74c3c,color:#fff

• At the melting point (0 degrees C for water): the flat section represents the energy needed to melt the substance. The total energy absorbed is E = m x Lf.
• At the boiling point (100 degrees C for water): the flat section represents the energy needed to boil the substance. The total energy absorbed is E = m x Lv.
• The flat section at the boiling point is much longer than at the melting point because Lv is much greater than Lf.

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Cooling and Latent Heat

When a substance condenses (gas to liquid) or freezes (liquid to solid), it releases latent heat to the surroundings.

• The energy released during condensation = m x Lv
• The energy released during freezing = m x Lf

This is why:

• Steam burns are worse than hot water burns — when steam condenses on your skin, it releases a large amount of latent heat (2,260,000 J per kg).
• Freezing water releases energy to the surroundings — this is why fruit growers sometimes spray water on crops on cold nights; as the water freezes, it releases latent heat that helps protect the crops from frost.

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Linking Latent Heat to the Particle Model