Specific Latent Heat

This lesson covers specific latent heat, the energy required to change the state of a substance without changing its temperature. This is a key concept in the AQA GCSE Combined Science Trilogy specification (8464, section 6.3.1).

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What Is Latent Heat?

Latent heat is the energy needed to change the state of a substance without changing its temperature. The word "latent" means "hidden" — the energy is hidden because there is no temperature change even though energy is being supplied (or released).

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The Equation

$E = m \times L$E=m×L

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

Exam Tip: The specific latent heat equation $E = mL$E=mL is given on the AQA equation sheet. You do not need to memorise it, but you must be able to use and rearrange it correctly.

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Two Types of Specific Latent Heat

Type	Change of state	Symbol
Specific latent heat of fusion ($L_f$Lf​)	Solid ↔ Liquid (melting or freezing)	$L_f$Lf​
Specific latent heat of vaporisation ($L_v$Lv​)	Liquid ↔ Gas (boiling/evaporating or condensing)	$L_v$Lv​

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

• During vaporisation, the bonds between particles must be completely broken (particles separate entirely).
• During fusion, bonds are only weakened (particles move from a rigid structure to being able to slide past each other).

Values for Water

Quantity	Value
Specific latent heat of fusion ($L_f$Lf​)	334 000 J/kg
Specific latent heat of vaporisation ($L_v$Lv​)	2 260 000 J/kg

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Worked Example 1

How much energy is needed to melt 2.0 kg of ice at 0 °C? ($L_f$Lf​ of water = 334 000 J/kg)

$E = m \times L = 2.0 \times 334\,000 = 668\,000 \text{ J} = 668 \text{ kJ}$E=m×L=2.0×334000=668000 J=668 kJ

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Worked Example 2

How much energy is needed to convert 0.5 kg of water at 100 °C to steam at 100 °C? ($L_v$Lv​ of water = 2 260 000 J/kg)

$E = m \times L = 0.5 \times 2\,260\,000 = 1\,130\,000 \text{ J} = 1130 \text{ kJ}$E=m×L=0.5×2260000=1130000 J=1130 kJ

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Worked Example 3

A substance releases 450 000 J of energy when 1.5 kg of it freezes. What is its specific latent heat of fusion?

$L = \frac{E}{m} = \frac{450\,000}{1.5} = 300\,000 \text{ J/kg}$L=mE​=1.5450000​=300000 J/kg

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Latent Heat on a Heating Curve

On a heating curve, the flat sections (constant temperature) correspond to changes of state where $E = mL$E=mL applies:

graph LR
    A["Solid<br/>(E = mcΔθ)"] --> B["MELTING<br/>(E = mL_f)<br/>Temp constant"]
    B --> C["Liquid<br/>(E = mcΔθ)"]
    C --> D["BOILING<br/>(E = mL_v)<br/>Temp constant"]
    D --> E["Gas<br/>(E = mcΔθ)"]

On the rising sections of the heating curve, the specific heat capacity equation $E = mc\Delta\theta$E=mcΔθ applies.

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Comparing Specific Latent Heat and Specific Heat Capacity

Feature	Specific heat capacity ($c$c)	Specific latent heat ($L$L)
What it measures	Energy to raise temperature of 1 kg by 1 °C	Energy to change state of 1 kg at constant temperature
Equation	$E = mc\Delta\theta$E=mcΔθ	$E = mL$E=mL
Unit	J/kg °C	J/kg
Temperature change?	Yes	No

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Common Misconceptions

Misconception	Correction
Latent heat increases the temperature	No — latent heat changes the state without changing the temperature
$L_f$Lf​ and $L_v$Lv​ are the same value	$L_v$Lv​ is always much greater than $L_f$Lf​ for the same substance
You lose energy during melting	Energy is absorbed during melting (and released during freezing)
Once you know the mass, you can calculate the energy without knowing which change of state occurs	You need to know which latent heat to use — fusion or vaporisation

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Summary

• Specific latent heat is the energy needed to change the state of 1 kg of a substance with no change in temperature.
• $E = mL$E=mL — this equation is provided on the AQA equation sheet.
• Latent heat of fusion ($L_f$Lf​) applies to melting/freezing.
• Latent heat of vaporisation ($L_v$Lv​) applies to boiling/condensing.
• $L_v > L_f$Lv​>Lf​ because completely separating particles requires more energy than just loosening their arrangement.
• On a heating curve, flat sections represent changes of state where the temperature is constant.

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Worked Example 4 — Combined Heating and Change of State

Calculate the total energy needed to convert 0.20 kg of ice at −10 °C into water at 20 °C. (Specific heat capacity of ice = 2100 J/kg °C; $L_f$Lf​ = 334 000 J/kg; specific heat capacity of water = 4200 J/kg °C.)

Step 1 — Warm the ice from −10 °C to 0 °C: $E_1 = mc\Delta\theta = 0.20 \times 2100 \times 10 = 4200 \text{ J}$E1​=mcΔθ=0.20×2100×10=4200 J

Step 2 — Melt the ice at 0 °C: $E_2 = mL = 0.20 \times 334\,000 = 66\,800 \text{ J}$E2​=mL=0.20×334000=66800 J

Step 3 — Warm the water from 0 °C to 20 °C: $E_3 = mc\Delta\theta = 0.20 \times 4200 \times 20 = 16\,800 \text{ J}$E3​=mcΔθ=0.20×4200×20=16800 J

Total: $E = 4200 + 66\,800 + 16\,800 = 87\,800$E=4200+66800+16800=87800 J ≈ 88 kJ.

Notice that most of the energy goes into melting the ice — nearly 76% of the total. This is why latent heat is so important: changes of state dominate the energy budget.

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Worked Example 5 — Finding Latent Heat Experimentally

A 100 W electric heater melts 0.060 kg of ice at 0 °C in 3.0 minutes (without raising the water temperature). Estimate the specific latent heat of fusion of water.

Energy supplied: $E = P \times t = 100 \times (3.0 \times 60) = 18\,000 \text{ J}$E=P×t=100×(3.0×60)=18000 J

Rearranging $E = mL$E=mL: $L = \frac{E}{m} = \frac{18\,000}{0.060} = 300\,000 \text{ J/kg}$L=mE​=0.06018000​=300000 J/kg

The accepted value is 334 000 J/kg. The lower experimental value suggests heat losses to the surroundings (some of the electrical energy was absorbed by the air or the beaker, not the ice) — a typical source of systematic error in this practical.

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Worked Example 6 — Condensation Releases Energy

A kettle produces 0.10 kg of steam at 100 °C. If all of this steam condenses on a person's hand, how much energy is released?