P5: Specific Latent Heat
Understand specific latent heat of fusion and vaporisation, interpret heating and cooling curves, and learn how internal energy changes during state changes.
What is Specific Latent Heat?
When a substance changes state (e.g. solid to liquid, or liquid to gas), energy must be transferred to break or form bonds between particles. During this process, the temperature of the substance does not change, even though energy is still being supplied or removed. The energy needed for a state change is called latent heat.
Definition
Specific latent heat is the amount of energy required to change the state of 1 kg of a substance without changing its temperature. It is measured in joules per kilogram (J/kg).
Latent Heat vs Specific Heat Capacity
Specific heat capacity involves a temperature change with no state change. Specific latent heat involves a state change with no temperature change. These are two different properties and must not be confused. SHC tells you how much energy is needed to change the temperature. SLH tells you how much energy is needed to change the state.
Specific Latent Heat of Fusion
Definition
Specific latent heat of fusion is the energy required to change 1 kg of a substance from solid to liquid (or liquid to solid) at its melting point, with no change in temperature.
The word "fusion" refers to melting or freezing. When a solid melts, energy must be supplied to break the bonds holding the particles in a rigid structure. When a liquid freezes, energy must be removed so that bonds can form between particles.
Worked Example 1
Calculate the energy required to melt 0.5 kg of ice at 0 degrees C. The specific latent heat of fusion of water is 334 000 J/kg.
E = mL
E = 0.5 × 334 000
E = 167 000 J
E = 167 kJ
Worked Example 2
A freezer removes energy from 2 kg of water at 0 degrees C to turn it into ice at 0 degrees C. The specific latent heat of fusion of water is 334 000 J/kg. Calculate the energy removed.
E = mL
E = 2 × 334 000
E = 668 000 J
E = 668 kJ
Specific Latent Heat of Vaporisation
Definition
Specific latent heat of vaporisation is the energy required to change 1 kg of a substance from liquid to gas (or gas to liquid) at its boiling point, with no change in temperature.
The specific latent heat of vaporisation is always much larger than the specific latent heat of fusion for the same substance. This is because much more energy is needed to completely separate particles (turn liquid to gas) than to weaken the bonds enough for them to slide past each other (turn solid to liquid).
Worked Example 3
Calculate the energy required to boil 0.2 kg of water at 100 degrees C. The specific latent heat of vaporisation of water is 2 260 000 J/kg.
E = mL
E = 0.2 × 2 260 000
E = 452 000 J
E = 452 kJ
Worked Example 4
Steam at 100 degrees C condenses on a cold surface. If 0.01 kg of steam condenses, calculate the energy released. The specific latent heat of vaporisation of water is 2 260 000 J/kg.
E = mL
E = 0.01 × 2 260 000
E = 22 600 J
E = 22.6 kJ
The Specific Latent Heat Equation
Specific Latent Heat Equation
E = mL
E = energy transferred (J)
m = mass (kg)
L = specific latent heat (J/kg)
For fusion: Lf = specific latent heat of fusion
For vaporisation: Lv = specific latent heat of vaporisation
Rearranging the Equation
To find the mass: m = E / L
To find the specific latent heat: L = E / m
Worked Example 5
45 000 J of energy is supplied to melt a block of ice at 0 degrees C. The specific latent heat of fusion of ice is 334 000 J/kg. Calculate the mass of ice that melts.
m = E / L
m = 45 000 / 334 000
m = 0.135 kg
m = 0.13 kg (2 s.f.)
Worked Example 6
100 000 J of energy is supplied to vaporise 0.04 kg of a liquid at its boiling point. Calculate the specific latent heat of vaporisation.
L = E / m
L = 100 000 / 0.04
L = 2 500 000 J/kg
Heating and Cooling Curves
A heating curve shows how the temperature of a substance changes as energy is supplied at a constant rate. A cooling curve shows how the temperature changes as energy is removed at a constant rate.
Features of a Heating Curve
The temperature rises as the solid is heated (SHC of the solid applies). At the melting point, the temperature stays constant while the substance melts. This flat section represents the energy being used to break bonds (latent heat of fusion). The temperature then rises as the liquid is heated (SHC of the liquid applies). At the boiling point, the temperature stays constant again while the substance boils. This second flat section represents the energy being used to completely separate particles (latent heat of vaporisation). The temperature then rises as the gas is heated (SHC of the gas applies).
Features of a Cooling Curve
The temperature falls as the gas cools. At the condensation point, the temperature stays constant while the gas condenses into a liquid. The temperature then falls as the liquid cools. At the freezing point, the temperature stays constant while the liquid freezes into a solid. The temperature then falls as the solid cools. The flat sections on a cooling curve occur at the same temperatures as on a heating curve.
Interpreting the Flat Sections
The flat sections of a heating or cooling curve represent state changes. During these sections, all the energy is being used to break or form bonds between particles. The temperature does not change because the energy is going into changing the internal energy (specifically the potential energy of the particles), not their kinetic energy. The longer the flat section, the more energy is needed for the state change. The flat section at the boiling point is always longer than at the melting point because the latent heat of vaporisation is greater than the latent heat of fusion.
You may be asked to calculate the total energy needed to heat a substance through a state change. This requires two steps: first calculate the energy to heat the substance to its melting/boiling point using E = mcΔT, then calculate the energy for the state change using E = mL. Add both values together for the total energy.
Worked Example 7: Multi-Step Calculation
Calculate the total energy required to turn 0.3 kg of ice at -10 degrees C into water at 30 degrees C.
SHC of ice = 2100 J/kg degrees C
SHC of water = 4200 J/kg degrees C
Specific latent heat of fusion of water = 334 000 J/kg
Step 1: Heat ice from -10 degrees C to 0 degrees C
E1 = mcΔT = 0.3 × 2100 × 10 = 6300 J
Step 2: Melt ice at 0 degrees C
E2 = mL = 0.3 × 334 000 = 100 200 J
Step 3: Heat water from 0 degrees C to 30 degrees C
E3 = mcΔT = 0.3 × 4200 × 30 = 37 800 J
Total energy = 6300 + 100 200 + 37 800 = 144 300 J = 144.3 kJ
Worked Example 8: Multi-Step Calculation
Calculate the total energy needed to convert 0.1 kg of water at 25 degrees C into steam at 100 degrees C.
SHC of water = 4200 J/kg degrees C
Specific latent heat of vaporisation of water = 2 260 000 J/kg
Step 1: Heat water from 25 degrees C to 100 degrees C
E1 = mcΔT = 0.1 × 4200 × 75 = 31 500 J
Step 2: Boil water at 100 degrees C
E2 = mL = 0.1 × 2 260 000 = 226 000 J
Total energy = 31 500 + 226 000 = 257 500 J = 257.5 kJ
Internal Energy During State Changes
What is Internal Energy?
Internal energy is the total energy stored by the particles that make up a substance. It is the sum of the kinetic energy of the particles (due to their random motion) and the potential energy of the particles (due to their positions relative to each other and the bonds between them).
Heating Without State Change
When a substance is heated and its temperature rises (but no state change occurs), the kinetic energy of the particles increases. The particles move faster (in a liquid or gas) or vibrate more vigorously (in a solid). The potential energy stays roughly the same because the arrangement of particles does not change significantly. Temperature is a measure of the average kinetic energy of the particles.
Heating During a State Change
When a substance is changing state, the temperature stays constant. This means the average kinetic energy of the particles does not change. Instead, the energy being supplied goes into increasing the potential energy of the particles. The bonds between particles are being broken (during melting or boiling), which requires energy. The particles become more spread out and less strongly bonded. This is why energy is still needed even though the temperature is not rising.
Internal Energy and State
A gas has more internal energy than a liquid of the same substance at the same temperature, because the particles in a gas have more potential energy (the bonds have been completely broken so the particles are free). Similarly, a liquid has more internal energy than a solid of the same substance at the same temperature, because the bonds have been partially broken.
Temperature measures the average kinetic energy of particles. During a state change, kinetic energy does not change (temperature stays constant), but potential energy does change (bonds are being broken or formed). This is why the temperature stays flat on a heating curve even though energy is still being supplied.
Comparison of Latent Heat Values
| Substance | Lf (J/kg) | Lv (J/kg) | Melting Point (degrees C) | Boiling Point (degrees C) |
|---|---|---|---|---|
| Water | 334 000 | 2 260 000 | 0 | 100 |
| Aluminium | 397 000 | 10 900 000 | 660 | 2470 |
| Iron | 247 000 | 6 300 000 | 1538 | 2862 |
| Lead | 23 000 | 860 000 | 327 | 1749 |
| Ethanol | 108 000 | 855 000 | -114 | 78 |
| Oxygen | 14 000 | 210 000 | -219 | -183 |
| Nitrogen | 26 000 | 200 000 | -210 | -196 |
| Mercury | 11 700 | 296 000 | -39 | 357 |
Key Observations
The latent heat of vaporisation is always much larger than the latent heat of fusion for any substance. For water, Lv is nearly 7 times larger than Lf. This is because completely separating particles (vaporisation) requires much more energy than just weakening bonds enough for them to slide past each other (fusion). Substances with strong bonds between particles tend to have higher latent heat values.
Why Steam Burns Are Worse Than Boiling Water Burns
When steam at 100 degrees C touches your skin, it first condenses (releasing the latent heat of vaporisation, 2 260 000 J/kg) and then the resulting water cools down. The energy released during condensation is enormous compared to the energy released by water at 100 degrees C simply cooling on your skin. This is why steam burns are much more severe than boiling water burns, even though both are at the same temperature.
Worked Example 9: Steam vs Boiling Water
Compare the energy released when 0.001 kg (1 g) of steam at 100 degrees C condenses and cools to body temperature (37 degrees C), versus 0.001 kg of boiling water at 100 degrees C cooling to 37 degrees C.
Steam condensing: E = mL = 0.001 × 2 260 000 = 2260 J
Water cooling: E = mcΔT = 0.001 × 4200 × 63 = 264.6 J
The condensed steam then also cools: E = 0.001 × 4200 × 63 = 264.6 J
Total from steam: 2260 + 264.6 = 2524.6 J
Total from boiling water: 264.6 J
The steam releases nearly 10 times more energy than the boiling water.
Practice Questions
1. Calculate the energy required to melt 1.5 kg of ice at 0 degrees C. The specific latent heat of fusion of water is 334 000 J/kg.
E = mL = 1.5 × 334 000 = 501 000 J = 501 kJ
2. 0.05 kg of ethanol is vaporised at its boiling point. The specific latent heat of vaporisation of ethanol is 855 000 J/kg. Calculate the energy required.
E = mL = 0.05 × 855 000 = 42 750 J = 42.75 kJ
3. 150 000 J of energy is supplied to melt a solid at its melting point. The specific latent heat of fusion is 75 000 J/kg. Calculate the mass that melts.
m = E / L = 150 000 / 75 000 = 2 kg
4. Explain why the temperature of a substance does not change while it is melting, even though energy is still being supplied.
During melting, the energy supplied is used to break bonds between particles (increasing their potential energy) rather than increasing their kinetic energy. Since temperature is a measure of the average kinetic energy of the particles, and kinetic energy does not change, the temperature stays constant.
5. Calculate the total energy needed to convert 0.5 kg of ice at -20 degrees C to steam at 100 degrees C.
Step 1: Heat ice from -20 to 0: E = 0.5 × 2100 × 20 = 21 000 J. Step 2: Melt ice: E = 0.5 × 334 000 = 167 000 J. Step 3: Heat water from 0 to 100: E = 0.5 × 4200 × 100 = 210 000 J. Step 4: Boil water: E = 0.5 × 2 260 000 = 1 130 000 J. Total = 21 000 + 167 000 + 210 000 + 1 130 000 = 1 528 000 J = 1528 kJ
6. Explain why steam at 100 degrees C causes more severe burns than boiling water at 100 degrees C.
When steam condenses on the skin, it releases the latent heat of vaporisation (2 260 000 J/kg for water) in addition to the energy released as the resulting water cools. Boiling water only releases energy as it cools. The extra energy from condensation makes steam burns much more severe.
🔢 Maths Skills
Mathematical Skills for this Topic
Rearranging E = mL: To find the mass: m = E / L. To find the specific latent heat: L = E / m. This equation is simpler than the SHC equation because there is no squared term. However, you must be careful to use the correct latent heat value — Lf for fusion (melting/freezing) and Lv for vaporisation (boiling/condensing). These are different values for the same substance and must not be mixed up.
Interpreting heating and cooling curves: A heating curve plots temperature on the y-axis against time (or energy) on the x-axis. Diagonal sections show temperature changing (SHC applies). Horizontal flat sections show state changes (SLH applies). The gradient of a diagonal section is related to 1/(mc) — steeper gradients mean lower SHC. The length of a flat section is proportional to the mass × latent heat — longer flat sections mean more energy is needed for the state change. The flat section at the boiling point is always longer than at the melting point because Lv is greater than Lf.
Reading values from graphs: When reading a heating curve, identify the melting point and boiling point from the temperatures of the flat sections. Read the initial and final temperatures from the diagonal sections. You may need to calculate the energy during a flat section if the power is known: energy = power × time (where time is read from the x-axis). Multi-step calculations require you to calculate energy for each section separately and add them together.
Standard form for large latent heat values: Latent heat values are often very large. For water, Lv = 2 260 000 J/kg = 2.26 × 10⁶ J/kg. Writing in standard form reduces errors when multiplying or dividing. When using these values, keep track of the powers of ten carefully.
⚠️ Common Misconceptions
Watch Out!
Students often think that the temperature changes during melting or boiling. Wrong: The temperature gradually increases during melting or boiling as more energy is supplied. Correct: During a state change, the temperature stays constant. All the energy supplied is used to break bonds between particles (increasing potential energy), not to increase the kinetic energy of the particles. Since temperature depends on average kinetic energy, the temperature does not change during melting or boiling.
Students often think that latent heat is the same thing as specific heat capacity. Wrong: Latent heat and specific heat capacity are the same property — they both measure how much energy is needed to heat a substance. Correct: Specific heat capacity (SHC) is the energy needed to change the temperature of 1 kg of a substance by 1°C WITHOUT a state change. Specific latent heat (SLH) is the energy needed to change the state of 1 kg of a substance WITH NO temperature change. SHC involves a temperature change; SLH involves a state change. They are completely different properties.
✍️ 6-Mark Question
Extended Answer Question
6 marks: Explain why the temperature of a boiling kettle remains at 100°C until all the water has turned to steam.
When water reaches 100°C at standard atmospheric pressure, it begins to boil. During boiling, the water is changing state from a liquid to a gas (steam). The energy being supplied by the kettle's heating element is not increasing the kinetic energy of the water molecules. Instead, all the energy is being used to overcome the intermolecular forces between the water molecules, allowing them to separate completely and escape as a gas. This energy increases the potential energy of the molecules — they move further apart and the bonds between them are fully broken.
Since temperature is a measure of the average kinetic energy of the particles, and the kinetic energy is not increasing during the state change, the temperature remains constant at 100°C. The energy being supplied is described as the specific latent heat of vaporisation. For water, this is 2 260 000 J per kilogram, which is a very large amount of energy. The temperature will not begin to rise above 100°C until all the liquid water has been converted to steam. This is why a boiling kettle does not get hotter than 100°C no matter how long it is left on, as long as there is still liquid water present.
Mark scheme: 2 marks for explaining that energy goes into breaking bonds not increasing kinetic energy, 2 marks for linking temperature to average kinetic energy and explaining why it stays constant, 1 mark for mentioning potential energy increases, 1 mark for referencing latent heat of vaporisation
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student heated a sample of a pure substance at a constant rate and recorded the temperature every minute. The graph of temperature against time shows: the temperature rises from 20°C to 80°C over 4 minutes, then stays at 80°C for 8 minutes, then rises from 80°C to 150°C over 5 minutes, then stays at 150°C for 20 minutes.
The power of the heater is 50 W and the mass of the substance is 0.2 kg. Calculate the specific latent heat of vaporisation and explain why the flat section at 150°C is much longer than the one at 80°C.
Answer: The flat section at 150°C represents boiling (vaporisation). Energy supplied during this section = power × time = 50 × (20 × 60) = 50 × 1200 = 60 000 J. Specific latent heat of vaporisation Lv = E / m = 60 000 / 0.2 = 300 000 J/kg. The flat section at 150°C is much longer (20 minutes) than the flat section at 80°C (8 minutes) because the latent heat of vaporisation is much greater than the latent heat of fusion. Vaporisation requires completely separating particles (breaking all intermolecular bonds), while melting only requires weakening bonds enough for particles to slide past each other. More energy per kilogram is needed, so the flat section lasts longer when heating at a constant rate.
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