Specific Latent Heat

Latent heat is the energy associated with changes of state. The word "latent" means "hidden" — the energy is hidden in the sense that it does not cause a temperature change. This topic is closely linked to internal energy and is frequently tested in AQA exams.

Spec mapping (AQA 7408 §3.6.2.2 — Specific latent heat): This lesson addresses the definition of specific latent heat L for both fusion (l_f) and vaporisation (l_v), the equation Q = mL, the use of this relationship in calculating mass, energy, or latent heat from experimental data, and an explanation of why latent heat values differ between fusion and vaporisation in terms of molecular bonds. Candidates are expected to apply Q = mL alongside Q = mcΔθ in multi-stage heating or cooling problems, and to interpret phase-change plateaus on temperature-time graphs. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Specific latent heat is the natural partner of specific heat capacity (§3.6.2.2): c handles temperature changes (kinetic energy), L handles state changes (potential energy). It links back to internal energy (§3.6.2.1) — the molecular explanation of why L_v >> L_f rests on which bonds are being broken (partial vs full). It also connects to electrical energy (§3.5) when L is determined by an electrical heater feeding a vapour-collection apparatus, and to uncertainty propagation (§3.1.3) where mass-of-steam-condensed and timing both contribute to combined uncertainty. Synoptically, latent heat reappears in evaporative cooling problems and in atmospheric physics (cloud formation, sweating) — themes worth signposting in extended-response questions.

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Definition and Equation

Key Definition: The specific latent heat (L) of a substance is the energy required to change the state of 1 kg of the substance at constant temperature.

Q = mL

where:

• Q is the energy transferred (J)
• m is the mass of the substance (kg)
• L is the specific latent heat (J kg⁻¹)

There are two types of specific latent heat:

• Specific latent heat of fusion (L_f) — the energy per kg to change between solid and liquid (melting or freezing).
• Specific latent heat of vaporisation (L_v) — the energy per kg to change between liquid and gas (boiling/evaporating or condensing).

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Why Is L_v Always Greater Than L_f?

During melting, the molecules are loosened from the rigid lattice structure but remain close together — intermolecular forces are only partially overcome. During vaporisation, the molecules must be completely separated from one another, overcoming all remaining intermolecular forces. Additionally, during vaporisation the substance expands greatly against atmospheric pressure, doing work on the surroundings (W = pΔV). This work also requires energy.

For water:

• L_f = 3.34 × 10⁵ J kg⁻¹ (334 kJ kg⁻¹)
• L_v = 2.26 × 10⁶ J kg⁻¹ (2260 kJ kg⁻¹)

The latent heat of vaporisation is about 6.8 times larger than the latent heat of fusion for water.

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Latent Heat Values for Common Substances

Substance	Melting Point (°C)	L_f (kJ kg⁻¹)	Boiling Point (°C)	L_v (kJ kg⁻¹)
Water	0	334	100	2260
Ethanol	−114	109	78	855
Lead	327	23	1750	858
Aluminium	660	397	2470	10 900
Copper	1083	205	2567	5070
Nitrogen	−210	25.7	−196	199
Oxygen	−219	13.9	−183	213

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Heating and Cooling Curves

A heating curve (temperature against time) for a substance heated at a constant rate shows a characteristic stepped shape:

1. Rising section (solid phase): Temperature increases linearly. The gradient depends on the specific heat capacity of the solid and the mass.
2. First flat section (melting): Temperature remains constant at the melting point. Energy is used to break bonds (latent heat of fusion). Duration depends on mL_f and the heating power.
3. Rising section (liquid phase): Temperature rises again, usually with a different gradient (different c value).
4. Second flat section (boiling): Temperature remains constant at the boiling point. This section is longer than the first because L_v > L_f.
5. Rising section (gas phase): Temperature increases once more.

Interpreting the Gradient

During a rising section, Q = mcΔθ and Q = Pt (where P is the constant heating power). Therefore:

Pt = mcΔθ → Δθ/t = P/(mc)

The gradient of the temperature-time graph equals P/(mc). A steeper gradient means either a smaller mass, a lower specific heat capacity, or a higher heating power.

Interpreting Flat Sections

During a flat section, Q = mL and Q = Pt. The duration of the flat section is:

t = mL/P

A longer flat section means a larger latent heat (for the same mass and power).

Exam Tip: Examiners often ask you to identify states, explain why sections are flat, or compare gradients. Always connect gradients to c and flat section lengths to L. If two substances are heated with the same power, the one with the longer flat section has the larger latent heat.

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Worked Examples

Worked Example 1 — Energy to Melt Ice

Question: Calculate the energy required to melt 0.50 kg of ice at 0 °C. (L_f = 3.34 × 10⁵ J kg⁻¹)

Solution:

Q = mL_f = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ J = 167 kJ

Worked Example 2 — Complete Heating Process

Question: A 0.20 kg block of ice at −15 °C is heated until it becomes steam at 100 °C. Calculate the total energy required. (c_ice = 2100 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹, L_f = 3.34 × 10⁵ J kg⁻¹, L_v = 2.26 × 10⁶ J kg⁻¹)

Solution:

Step 1: Heat ice from −15 °C to 0 °C. Q₁ = mcΔθ = 0.20 × 2100 × 15 = 6300 J

Step 2: Melt ice at 0 °C. Q₂ = mL_f = 0.20 × 3.34 × 10⁵ = 66 800 J

Step 3: Heat water from 0 °C to 100 °C. Q₃ = mcΔθ = 0.20 × 4200 × 100 = 84 000 J

Step 4: Boil water at 100 °C. Q₄ = mL_v = 0.20 × 2.26 × 10⁶ = 452 000 J

Step 5: Total energy. Q_total = Q₁ + Q₂ + Q₃ + Q₄ = 6300 + 66 800 + 84 000 + 452 000 = 609 100 J ≈ 609 kJ

Notice that the vaporisation step (452 kJ) accounts for about 74% of the total energy. This is typical and explains why steam burns are so dangerous — steam condensing on your skin releases enormous amounts of energy.

Worked Example 3 — Cooling a Hot Metal in Water

Question: A 0.15 kg copper block at 200 °C is dropped into 0.30 kg of water at 20 °C in an insulated container. Calculate the final temperature. (c_copper = 390 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹)

Solution:

Energy lost by copper = energy gained by water (assuming no losses).

m_Cu × c_Cu × (200 − T) = m_water × c_water × (T − 20)

0.15 × 390 × (200 − T) = 0.30 × 4200 × (T − 20)

58.5 × (200 − T) = 1260 × (T − 20)

11 700 − 58.5T = 1260T − 25 200

11 700 + 25 200 = 1260T + 58.5T

36 900 = 1318.5T

T = 36 900 / 1318.5 = 28.0 °C

The small temperature rise of the water (only 8 °C) despite the copper starting at 200 °C illustrates the large heat capacity of water compared to copper.

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Experimental Determination of Specific Latent Heat

Measuring L_f for Ice

Apparatus: Funnel, filter paper, beaker, electronic balance, heater, stopwatch.

Procedure:

1. Place crushed ice in a funnel with a filter paper and collect the meltwater in a beaker on a balance.
2. Without the heater, collect meltwater for a set time to measure the background melting rate.
3. Insert the heater into the ice, switch on, and collect meltwater for the same duration.
4. The extra mass of water collected (Δm) is due to the heater.
5. L_f = Pt / Δm = VIt / Δm

Sources of error:

• Background melting due to room temperature (controlled by subtracting a control run).
• Not all electrical energy goes into melting ice (some heats the meltwater after it melts).

Measuring L_v for Water

Apparatus: Electric kettle or immersion heater, electronic balance, stopwatch.

Procedure:

1. Bring water to a steady boil.
2. Record the mass of the kettle and water on the balance.
3. Continue boiling for a measured time and record the new mass.
4. The mass decrease (Δm) is the mass of water that boiled away.
5. L_v = Pt / Δm = VIt / Δm

Exam Tip: In latent heat experiments, the key challenge is accounting for background energy gains or losses. For fusion, subtract the control run. For vaporisation, ensure the water is already boiling steadily before you start timing (so the heater power goes entirely into producing steam, not into raising the temperature of the water).

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Evaporation vs Boiling

Feature	Evaporation	Boiling
Temperature	Below boiling point	At boiling point
Location	Surface only	Throughout the liquid
Bubbles	No bubbles	Bubbles form within the liquid
Rate	Slow, depends on conditions	Rapid, limited by energy supply
Temperature of liquid	Decreases (cooling effect)	Remains constant

Evaporation causes cooling because the fastest molecules (with the most kinetic energy) are the ones most likely to escape from the surface. This lowers the average kinetic energy of the remaining molecules, reducing the temperature.

Factors that increase the rate of evaporation:

• Higher temperature (more molecules have enough energy to escape)
• Larger surface area (more molecules near the surface)
• Lower humidity (fewer molecules returning to the liquid)
• Air movement across the surface (carries away vapour, maintaining a concentration gradient)

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Common Misconceptions

1. "You need to heat something to make it evaporate." Evaporation occurs at any temperature — some molecules always have enough energy to escape the surface.

2. "Boiling and evaporation are the same thing." They are different processes. Boiling occurs at a specific temperature throughout the liquid; evaporation occurs at any temperature from the surface only.

3. "The latent heat of fusion and vaporisation are similar in magnitude." For most substances, L_v is many times larger than L_f. For water, L_v is about 6.8 times larger.

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Additional Worked Example — Multi-Stage Heating of Ice to Steam

A 50.0 g block of ice at −20.0 °C is heated continuously until it becomes steam at 120.0 °C. Calculate the total energy required, working stage-by-stage. Use:

• c_ice = 2100 J kg⁻¹ K⁻¹
• c_water = 4180 J kg⁻¹ K⁻¹
• c_steam = 2010 J kg⁻¹ K⁻¹
• L_f (water) = 3.34 × 10⁵ J kg⁻¹
• L_v (water) = 2.26 × 10⁶ J kg⁻¹

This is a five-stage problem; do not collapse stages or miss any phase boundary.

Mass: m = 50.0 g = 0.0500 kg.

Stage 1 — Warm ice from −20 °C to 0 °C (Δθ = 20 K). Q₁ = m c_ice Δθ = 0.0500 × 2100 × 20 = 2 100 J.

Stage 2 — Melt ice at 0 °C (constant temperature). Q₂ = m L_f = 0.0500 × 3.34 × 10⁵ = 1.67 × 10⁴ J = 16 700 J.

Stage 3 — Warm water from 0 °C to 100 °C (Δθ = 100 K). Q₃ = m c_water Δθ = 0.0500 × 4180 × 100 = 20 900 J.

Stage 4 — Boil water at 100 °C (constant temperature). Q₄ = m L_v = 0.0500 × 2.26 × 10⁶ = 1.13 × 10⁵ J = 113 000 J.

Stage 5 — Warm steam from 100 °C to 120 °C (Δθ = 20 K). Q₅ = m c_steam Δθ = 0.0500 × 2010 × 20 = 2 010 J.