Specific Latent Heat

This lesson explains specific latent heat, the difference between latent heat of fusion and latent heat of vaporisation, and how these concepts relate to heating curves. This is a required equation for the Edexcel GCSE Combined Science specification (1SC0).

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What Is Latent Heat?

When a substance changes state (e.g. melting or boiling), energy is transferred to break or form bonds between particles. During this process:

• The temperature stays constant.
• The energy goes into changing the potential energy of the particles (not their kinetic energy).

This energy is called latent heat (from the Latin latere, meaning "to hide") because the energy is "hidden" — it does not cause a temperature change.

$E = m L$E=mL

Symbol	Quantity	Unit
$E$E	Energy transferred	joules (J)
$m$m	Mass	kilograms (kg)
$L$L	Specific latent heat	joules per kilogram (J/kg)

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Two Types of Specific Latent Heat

Type	Symbol	Change of state	Definition
Specific latent heat of fusion	$L_f$Lf​	Solid ⇌ Liquid	Energy needed to change 1 kg of a substance from solid to liquid (or released when liquid freezes) at its melting point
Specific latent heat of vaporisation	$L_v$Lv​	Liquid ⇌ Gas	Energy needed to change 1 kg of a substance from liquid to gas (or released when gas condenses) at its boiling point

graph LR
    A["Solid"] -->|"Latent heat of fusion (L_f)"| B["Liquid"]
    B -->|"Latent heat of vaporisation (L_v)"| C["Gas"]
    C -->|"Energy released (L_v)"| B
    B -->|"Energy released (L_f)"| A

Why Is L_v Always Greater Than L_f?

• During vaporisation, the particles must completely separate and overcome all the intermolecular forces. The spacing between particles increases enormously.
• During fusion (melting), the particles only need to overcome some of the forces — they remain close together but can move past each other.
• Therefore, $L_v > L_f$Lv​>Lf​ for the same substance.

Exam Tip: The latent heat of vaporisation is always larger than the latent heat of fusion for the same substance. This is a very common comparison question.

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Typical Values

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

Notice that for water, $L_v$Lv​ is about 6.8 times larger than $L_f$Lf​.

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Rearranging the Equation

To find	Rearrangement
Energy	$E = m L$E=mL
Mass	$m = \frac{E}{L}$m=LE​
Specific latent heat	$L = \frac{E}{m}$L=mE​

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Worked Example 1

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

$E = m L_f = 0.5 \times 334\,000 = 167\,000 \text{ J} = 167 \text{ kJ}$E=mLf​=0.5×334000=167000 J=167 kJ

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Worked Example 2

How much energy is needed to boil 2.0 kg of water at 100 °C? ($L_v$Lv​ for water = 2 260 000 J/kg)

$E = m L_v = 2.0 \times 2\,260\,000 = 4\,520\,000 \text{ J} = 4520 \text{ kJ}$E=mLv​=2.0×2260000=4520000 J=4520 kJ

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Worked Example 3

A student supplies 50 000 J of energy to a block of metal at its melting point. 0.25 kg of the metal melts. Calculate the specific latent heat of fusion of the metal.

$L_f = \frac{E}{m} = \frac{50\,000}{0.25} = 200\,000 \text{ J/kg}$Lf​=mE​=0.2550000​=200000 J/kg

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Latent Heat and Heating Curves

Recall the heating curve from the previous lesson. The flat sections correspond to changes of state where the energy supplied equals the latent heat.

graph LR
    A["Solid: E = mcΔθ<br/>(temp rises)"] --> B["Melting: E = mL_f<br/>(temp constant)"]
    B --> C["Liquid: E = mcΔθ<br/>(temp rises)"]
    C --> D["Boiling: E = mL_v<br/>(temp constant)"]
    D --> E["Gas: E = mcΔθ<br/>(temp rises)"]

Section	Equation used	What changes
Sloped (single state)	$E = m c \Delta\theta$E=mcΔθ	Temperature increases; kinetic energy increases
Flat (change of state)	$E = m L$E=mL	State changes; potential energy increases; temperature constant

Exam Tip: When asked to calculate the total energy to heat a substance from below its melting point to above its boiling point, you may need to combine three calculations using $E = mc\Delta\theta$E=mcΔθ and two using $E = mL$E=mL.

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Combining Both Equations

How much energy is needed to heat 0.5 kg of ice at −10 °C to steam at 100 °C?

Data: $c_{\text{ice}} = 2100$cice​=2100 J/kg °C, $L_f = 334\,000$Lf​=334000 J/kg, $c_{\text{water}} = 4200$cwater​=4200 J/kg °C, $L_v = 2\,260\,000$Lv​=2260000 J/kg.

Stage	Calculation	Energy (J)
1. Heat ice from −10 °C to 0 °C	$0.5 \times 2100 \times 10$0.5×2100×10	10 500
2. Melt ice at 0 °C	$0.5 \times 334\,000$0.5×334000	167 000
3. Heat water from 0 °C to 100 °C	$0.5 \times 4200 \times 100$0.5×4200×100	210 000
4. Boil water at 100 °C	$0.5 \times 2\,260\,000$0.5×2260000	1 130 000
Total		1 517 500 J ≈ 1518 kJ

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Common Misconceptions

Misconception	Correction
The temperature rises during a change of state	The temperature is constant during melting or boiling
Latent heat of fusion and vaporisation are the same	$L_v$Lv​ is always greater than $L_f$Lf​ for the same substance
Energy is not needed if the temperature doesn't change	Energy is supplied — it goes into breaking intermolecular bonds (increasing potential energy)

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Summary