🧊 Explain what latent heat is and why temperature stays constant during a change of state
📐 Use the equation E = mL to calculate energy transferred during changes of state
🔥 Distinguish between latent heat of fusion and latent heat of vaporisation
📈 Interpret heating and cooling curves, identifying plateau regions
💧 Explain condensation in terms of energy transfer and particle behaviour
📊 Select and use appropriate SI units for energy, mass and specific latent heat
🧊 What is Latent Heat?
When a substance changes state — for example, melting from solid to liquid or boiling from liquid to gas — energy is transferred without any change in temperature. This energy is called latent heat (from the Latin word meaning "hidden").
Specific Latent Heat (L) is the amount of energy required to change the state of 1 kg of a substance at constant temperature, without changing its temperature.
The word "specific" means "per kilogram". So the specific latent heat tells us exactly how much energy is needed for each kilogram of material to undergo a change of state.
During a change of state, the energy supplied is used to break or form intermolecular bonds — not to increase the kinetic energy of particles. Because temperature is a measure of average kinetic energy, the temperature stays the same while bonds are being broken or formed.
💡 Temperature does NOT change during a change of state. All the energy goes into changing the arrangement of particles, not speeding them up.
📐 The Equation: E = mL
E = m × L
Symbol
Quantity
SI Unit
E
Energy transferred
Joules (J)
m
Mass of substance
Kilograms (kg)
L
Specific latent heat
Joules per kilogram (J/kg)
This equation can be rearranged to find any unknown:
m = E ÷ L L = E ÷ m
Specific latent heat values are often very large. For example, the specific latent heat of vaporisation of water is approximately 2,260,000 J/kg (2.26 × 10⁶ J/kg). This explains why steam burns are so much more severe than boiling water burns — the steam releases this enormous amount of energy as it condenses on your skin.
⚠️ Always convert mass to kilograms (kg) before using the equation. 1 g = 0.001 kg.
🔥 Fusion vs Vaporisation
There are two main types of specific latent heat, depending on which change of state is occurring:
Specific Latent Heat of Fusion (L_f) — the energy needed to melt (or release when freezing) 1 kg of a substance at its melting point, at constant temperature.
Specific Latent Heat of Vaporisation (L_v) — the energy needed to vaporise (or release when condensing) 1 kg of a substance at its boiling point, at constant temperature.
Substance
L_f (J/kg)
L_v (J/kg)
Water
334,000
2,260,000
Ethanol
108,000
841,000
Aluminium
397,000
10,900,000
Lead
24,700
871,000
Notice that the latent heat of vaporisation is always much larger than the latent heat of fusion. This is because turning a liquid into a gas requires completely separating the particles from each other, which needs far more energy than simply loosening the regular structure of a solid.
The same equation E = mL applies to both, and the same amount of energy is released when the reverse process occurs (condensing or freezing).
📈 Heating Curves and Plateaus
When a substance is heated at a constant rate and its temperature is plotted against time, we get a characteristic heating curve. The key features are:
Rising sections: Temperature increases as the substance is heated. Energy increases the kinetic energy of particles.
Flat sections (plateaus): Temperature stays constant during a change of state. Energy breaks intermolecular bonds rather than increasing kinetic energy.
A typical heating curve for water shows:
Temperature rising as ice is heated (solid phase)
A plateau at 0°C — ice melting to water (latent heat of fusion)
Temperature rising as water is heated (liquid phase)
A plateau at 100°C — water boiling to steam (latent heat of vaporisation)
Temperature rising as steam is heated (gas phase)
📊 The longer the plateau, the greater the specific latent heat. The steeper the slope, the smaller the specific heat capacity of that phase.
A cooling curve looks like the reverse: the plateaus still appear at the same temperatures, but now energy is being released as the substance freezes or condenses.
💧 Condensation and Energy Release
Condensation is the change of state from gas to liquid. It is the reverse of vaporisation. When a gas condenses, it releases energy equal to the latent heat of vaporisation:
E released = m × L_v
This is why:
Steam at 100°C causes worse burns than boiling water at 100°C — the steam releases 2,260,000 J/kg of extra energy as it condenses on skin.
Clouds form when warm, moist air rises, cools, and the water vapour condenses into tiny droplets, releasing latent heat that can fuel storm systems.
Radiators in central heating systems work efficiently because steam condenses inside them, releasing large amounts of energy.
Similarly, freezing releases the latent heat of fusion. The formation of ice on a pond releases energy to the surrounding water, which is why large bodies of water freeze slowly even in very cold conditions.
🔄 Energy changes are symmetric: the energy needed to melt a substance equals the energy released when it freezes. The energy needed to vaporise equals the energy released when it condenses.
Example 1: How much energy is needed to completely melt 2.5 kg of ice at 0°C?
(Specific latent heat of fusion of water = 334,000 J/kg)
1Identify the equation and values:
E = m × L
m = 2.5 kg, L = 334,000 J/kg
2Substitute into the equation:
E = 2.5 × 334,000
3Calculate:
E = 835,000 J
✅ E = 835,000 J (835 kJ)
Example 2: A student supplies 678,000 J of energy to convert liquid water at 100°C into steam. What mass of water was vaporised?
(Specific latent heat of vaporisation of water = 2,260,000 J/kg)
1Write the equation and rearrange for mass:
E = m × L → m = E ÷ L
2Substitute values:
m = 678,000 ÷ 2,260,000
3Calculate:
m = 0.3 kg
✅ m = 0.3 kg (300 g)
Example 3: 500 g of ethanol at its boiling point releases 420,500 J of energy as it condenses. Calculate the specific latent heat of vaporisation of ethanol.
1Convert mass to kg:
m = 500 g = 500 ÷ 1000 = 0.5 kg
2Rearrange E = mL for L:
L = E ÷ m
3Substitute and calculate:
L = 420,500 ÷ 0.5 = 841,000 J/kg
✅ L = 841,000 J/kg (8.41 × 10⁵ J/kg)
Example 4: A heating curve shows a plateau at 0°C lasting 4 minutes and a plateau at 100°C lasting 27 minutes. A heater supplies 500 J per minute. Use this to estimate the ratio of the latent heat of vaporisation to the latent heat of fusion for water.
1Calculate energy supplied during each plateau:
Fusion plateau: E_f = 500 × 4 = 2,000 J
Vaporisation plateau: E_v = 500 × 27 = 13,500 J
2Since the same mass is used, L is proportional to E:
L_v ÷ L_f = E_v ÷ E_f = 13,500 ÷ 2,000
3Calculate the ratio:
Ratio = 6.75
✅ L_v is approximately 6.75 times greater than L_f (the actual ratio is about 6.76, showing excellent agreement)
Question 1: During a change of state, what happens to the temperature of a substance?
Question 2: The specific latent heat of fusion of water is 334,000 J/kg. How much energy is needed to melt 3 kg of ice at 0°C?
Question 3: What is the SI unit of specific latent heat? Type your answer below.
Question 4: Why is the specific latent heat of vaporisation always larger than the specific latent heat of fusion for the same substance?
Question 5: A heater provides 1,130,000 J of energy to water at 100°C. Using L_v = 2,260,000 J/kg, calculate the mass of water vaporised. Give your answer in kg.
Challenge 1: A student is heating 0.2 kg of a substance. They record the following data from a heating curve: the fusion plateau lasts 5 minutes and the heater supplies energy at a rate of 60 W (60 J/s). Calculate the specific latent heat of fusion of the substance. Show all your working.
Step 1: Find total energy supplied during plateau.
Time = 5 minutes = 5 × 60 = 300 s
E = P × t = 60 × 300 = 18,000 J
Step 2: Rearrange E = mL for L.
L = E ÷ m = 18,000 ÷ 0.2
Answer: L = 90,000 J/kg
Challenge 2: Steam at 100°C condenses onto a patient's skin during an accident. If 50 g of steam condenses, calculate the energy released. Explain why this is more dangerous than 50 g of boiling water at 100°C spilling on the same area of skin. (L_v of water = 2,260,000 J/kg)
Step 1: Convert mass. m = 50 g = 0.05 kg
Step 2: E = m × L = 0.05 × 2,260,000 = 113,000 J
Answer: 113,000 J (113 kJ) is released.
Explanation: When steam condenses, it releases 113,000 J of latent heat energy in addition to any energy released as both the condensed water AND the original water subsequently cool from 100°C. Boiling water at 100°C only releases energy as it cools from 100°C to body temperature — no latent heat is released. This means steam deposits a far greater amount of energy into the skin, causing much more severe burns.
Challenge 3: A heating curve experiment uses a 500 W heater on a 1 kg sample of a pure substance. The fusion plateau lasts 668 s and the vaporisation plateau lasts 4,520 s. Calculate (a) the specific latent heat of fusion and (b) the specific latent heat of vaporisation. Comment on whether these values match water.
(a) Latent heat of fusion:
E_f = P × t = 500 × 668 = 334,000 J
L_f = E_f ÷ m = 334,000 ÷ 1 = 334,000 J/kg ✅
(b) Latent heat of vaporisation:
E_v = P × t = 500 × 4,520 = 2,260,000 J
L_v = E_v ÷ m = 2,260,000 ÷ 1 = 2,260,000 J/kg ✅
Comment: Both values match the accepted values for water exactly (L_f = 334,000 J/kg and L_v = 2,260,000 J/kg), confirming the substance is water. The latent heat of vaporisation is approximately 6.76 times larger than the latent heat of fusion, which reflects the much greater energy needed to fully separate water molecules in the gas phase.
Challenge 4 (Extended): Explain, using ideas about particles and energy, why there is a temperature plateau on a heating curve during melting. In your answer, refer to: bonds between particles, kinetic energy, and what the energy supplied is used for.
Model answer:
In a solid, particles are held in fixed positions by strong intermolecular bonds/forces. When energy is supplied at the melting point, rather than increasing the speed of the particles, it is used entirely to break the bonds between particles, allowing them to move more freely as a liquid.
Because temperature is a measure of the average kinetic energy of particles, and the kinetic energy is not increasing during this time, the temperature remains constant — forming the plateau on the heating curve.
Once all the bonds in the solid have been broken and the substance has fully melted, any further energy supplied does increase the kinetic energy of particles, so the temperature rises again — ending the plateau.
The amount of energy needed for this process per kilogram is the specific latent heat of fusion.