Specific Latent Heat

When you boil a kettle, the water reaches 100 °C and then continues to absorb energy — but the temperature stops rising. The energy is still going somewhere: it is breaking the intermolecular bonds that hold the liquid water together, converting it into steam. The energy required for this change of state is called latent heat, from the Latin word "latens" meaning hidden, because the energy is absorbed without any temperature change.

Definition and Equation

The specific latent heat (L) of a substance is the energy required to change the state of 1 kg of that substance without a change in temperature.

Q = mL

where:

Q = energy transferred (J)
m = mass of substance changing state (kg)
L = specific latent heat (J kg⁻¹)

There are two types:

Specific latent heat of fusion (L_f): the energy to change 1 kg of solid to liquid (or liquid to solid) at the melting point.
Specific latent heat of vaporisation (L_v): the energy to change 1 kg of liquid to gas (or gas to liquid) at the boiling point.

graph LR
    A["Solid"] -->|"Melting: Q = mL_f\n(energy absorbed)"| B["Liquid"]
    B -->|"Freezing: Q = mL_f\n(energy released)"| A
    B -->|"Boiling: Q = mL_v\n(energy absorbed)"| C["Gas"]
    C -->|"Condensing: Q = mL_v\n(energy released)"| B
    A -->|"Sublimation\n(energy absorbed)"| C
    C -->|"Deposition\n(energy released)"| A

Key Values

Substance	Melting point / °C	L_f / kJ kg⁻¹	Boiling point / °C	L_v / kJ kg⁻¹	L_v / L_f ratio
Water	0	334	100	2260	6.8
Ethanol	−114	108	78	855	7.9
Lead	327	23	1749	871	37.9
Aluminium	660	397	2519	10 900	27.5
Oxygen	−219	13.8	−183	213	15.4
Nitrogen	−210	25.5	−196	199	7.8

For water: L_f = 334 000 J kg⁻¹ and L_v = 2 260 000 J kg⁻¹. Notice that L_v is always significantly larger than L_f for any given substance.

Why is L_v Much Greater Than L_f?

This is a commonly examined question. During melting (solid → liquid):

Only some of the intermolecular bonds are broken — the particles gain enough energy to move past each other but remain close together.
The change in separation between particles is relatively small.

During boiling (liquid → gas):

Nearly all remaining intermolecular bonds must be broken — the particles must gain enough energy to completely separate and move independently.
The change in separation is large.
Work must also be done against atmospheric pressure as the substance expands significantly (liquid water expands by a factor of about 1700 when it becomes steam at 100 °C and 1 atm).

Because more bonds are broken and work is done against the atmosphere during vaporisation, L_v is always significantly larger than L_f for the same substance.

Worked Example 1: Simple Latent Heat

How much energy is needed to convert 0.50 kg of ice at 0 °C to water at 0 °C? (L_f = 334 000 J kg⁻¹)

Q = mL_f = 0.50 × 334 000 = 167 000 J = 167 kJ

Worked Example 2: Vaporisation

How much energy is needed to convert 0.20 kg of water at 100 °C to steam at 100 °C? (L_v = 2 260 000 J kg⁻¹)

Q = mL_v = 0.20 × 2 260 000 = 452 000 J = 452 kJ

Notice that despite the mass being smaller, the energy required for vaporisation is much larger than for fusion — reflecting the much larger value of L_v.

Worked Example 3: Multi-Step Calculation (SHC + Latent Heat)

Many exam problems require combining Q = mcΔθ and Q = mL. For example:

How much energy is needed to convert 0.30 kg of ice at −10 °C to water at 40 °C? (c_ice = 2100 J kg⁻¹ K⁻¹, L_f = 334 000 J kg⁻¹, c_water = 4200 J kg⁻¹ K⁻¹)

Step 1: Heat ice from −10 °C to 0 °C Q₁ = mc_ice × Δθ = 0.30 × 2100 × 10 = 6300 J

Step 2: Melt the ice at 0 °C Q₂ = mL_f = 0.30 × 334 000 = 100 200 J

Step 3: Heat water from 0 °C to 40 °C Q₃ = mc_water × Δθ = 0.30 × 4200 × 40 = 50 400 J

Total: Q = Q₁ + Q₂ + Q₃ = 6300 + 100 200 + 50 400 = 156 900 J ≈ 157 kJ

Notice that the melting step (100.2 kJ) accounts for 64% of the total energy, even though it produces no temperature change. This is a common pattern — latent heat steps typically dominate.

Worked Example 4: Mass of Steam Produced

A 2.0 kW kettle is left boiling for 3 minutes after the water reaches 100 °C. What mass of water boils away? (L_v = 2 260 000 J kg⁻¹)

Energy supplied: Q = Pt = 2000 × (3 × 60) = 2000 × 180 = 360 000 J

Mass boiled: m = Q / L_v = 360 000 / 2 260 000 = 0.159 kg ≈ 0.16 kg

This is about 160 mL — roughly a cupful. It would take about 14 minutes at this power to boil away a full 1.0 kg of water.

Worked Example 5: Thermal Equilibrium with Phase Change

50 g of steam at 100 °C is passed into 0.40 kg of water at 20 °C. Find the final temperature. (L_v = 2 260 000 J kg⁻¹, c_water = 4200 J kg⁻¹ K⁻¹)

Energy released by steam condensing: Q₁ = mL_v = 0.050 × 2 260 000 = 113 000 J

The condensed steam (now water at 100 °C) then cools to final temperature T: Q₂ = 0.050 × 4200 × (100 − T) = 210(100 − T)

Energy gained by the 0.40 kg of water heating from 20 °C to T: Q₃ = 0.40 × 4200 × (T − 20) = 1680(T − 20)

Energy balance: Q₁ + Q₂ = Q₃

113 000 + 210(100 − T) = 1680(T − 20)

113 000 + 21 000 − 210T = 1680T − 33 600

167 600 = 1890T

T = 167 600 / 1890 = 88.7 °C

The high final temperature demonstrates the enormous amount of energy carried by steam — this is why steam burns are so dangerous.

Measuring Specific Latent Heat Experimentally

Latent heat of fusion of ice:

Place an electrical heater in a funnel containing crushed ice over a beaker.
Measure the current (I) and potential difference (V) across the heater.
Switch on the heater for a measured time (t).
Collect and weigh the melted water.
Run a control experiment with no heater for the same time, to measure the mass of ice that melts due to the surroundings.
Subtract the control mass: net mass melted = m_heater − m_control.
L_f = VIt / m_net.

Latent heat of vaporisation of water:

Bring water to a steady boil.
Measure the mass of the water + container.
Switch on a heater of known power (or measure V and I).
After a measured time, re-weigh to find the mass of water that boiled away (m).
L_v = VIt / m.

Heating and Cooling Curves

A heating curve plots temperature against time (or energy supplied) as a substance is heated steadily.

The key features are:

Rising sections: temperature increases, energy is increasing kinetic energy (Q = mcΔθ applies).
Flat sections: temperature is constant during a change of state, energy is increasing potential energy (Q = mL applies).
The gradient of the rising sections depends on the specific heat capacity — a steeper gradient means a smaller heat capacity (same power input causes a faster temperature rise).
The length of the flat sections depends on the latent heat — a longer flat section means a larger latent heat (more energy needed at constant temperature).

Feature of heating curve	What it tells you	Equation used
Gradient of sloped section	Inversely proportional to mc (steeper = lower heat capacity)	Q = mcΔθ → Δθ/t = P/(mc)
Length of flat section	Proportional to mL (longer = larger latent heat)	Q = mL → t = mL/P
Different gradients for solid vs liquid	Different specific heat capacities	Compare c values

A cooling curve is the reverse: flat sections appear at the freezing and condensation points as latent heat is released.

Common Exam Mistakes

Mistake	Correct approach
Using Q = mcΔθ during a change of state	Use Q = mL — temperature is constant during phase changes
Forgetting the control experiment for L_f	Essential to subtract ice melted by surroundings
Omitting the latent heat step in multi-stage problems	Draw a heating curve first to identify all stages
Confusing L_f and L_v	L_f = fusion (melting/freezing); L_v = vaporisation (boiling/condensing)
Ignoring work done against atmosphere in L_v	L_v includes energy to break bonds AND to expand against atmospheric pressure

Summary

Q = mL gives the energy for a change of state at constant temperature.
L_f (fusion) is the latent heat for solid ↔ liquid; L_v (vaporisation) is for liquid ↔ gas.
L_v > L_f because vaporisation breaks more bonds and does work against atmospheric pressure.
Multi-step calculations often combine Q = mcΔθ and Q = mL.
Heating curves have flat regions at changes of state and sloped regions during temperature changes.
The gradient of sloped regions reveals relative SHC values; the length of flat regions reveals relative latent heats.

A-Level Deep Dive: Specific Latent Heat

Spec mapping

Edexcel 9PH0 specification Topic 9 — Thermodynamics covers specific latent heat as part of the thermal physics core, sitting alongside specific heat capacity, internal energy and the kinetic theory of gases (refer to the official specification document for exact wording). The defining equation $Q = mL$ is examined on Paper 2 (Topics 6–8 advanced material) and on Paper 3 (Practical Skills in Physics II), where the experimental determination of $L_f$ and $L_v$ provides standard required-practical material. Latent heat is not in the Pearson formula booklet under its own line, but $Q = mL$ is treated as an assumed working equation alongside $Q = mc\Delta\theta$ . Synoptically the topic links forward to internal energy (the latent-heat input increases the molecular potential energy with no kinetic-energy change) and backward to AS-level energy transfer (electrical work $E = VIt$ is the standard input route in the lab).

Worked example with full mark scheme

Question (8 marks):

A 50 W electrical heater is used to determine the specific latent heat of fusion of ice. After the heater has reached steady operation, 38 g of ice is collected as melt-water in 240 s. A control run (heater off) gives 4 g of melt-water in the same time, attributed to heat gained from the surroundings.

(a) Use the data to calculate an experimental value for the specific latent heat of fusion of ice. (5)

(b) Comment on whether your value is likely to be an over- or under-estimate of the accepted value of $3.34 \times 10^5$ J kg⁻¹, giving one reason. (3)

Solution with mark scheme:

(a) Step 1 — net mass melted by the heater.

$m_{\mathrm{net}} = m_{\mathrm{heater}} - m_{\mathrm{control}} = 38 - 4 = 34 \text{ g} = 0.034 \text{ kg}$

M1 — subtracting the control mass to remove the contribution from background heat transfer. Candidates who use 38 g uncorrected lose this method mark immediately.

Step 2 — energy delivered by the heater.

$E = Pt = 50 \times 240 = 12\,000 \text{ J}$

M1 — correct use of $E = Pt$ for an electrical heater. Equivalent route via $E = VIt$ is acceptable.

Step 3 — apply $Q = mL_f$ .

$L_f = \dfrac{E}{m_{\mathrm{net}}} = \dfrac{12\,000}{0.034}$

M1 — correct rearrangement of $Q = mL$ for $L$ .

Step 4 — evaluate.

$L_f = 3.53 \times 10^5 \text{ J kg}^{-1}$

A1 — numerically correct to 3 s.f.

A1 — units stated as J kg⁻¹ (or equivalent J/kg). Bare numbers without units cap the answer at A0.

(b) Likely over-estimate. B1

Reason: the control run only corrects for heat gained when the heater is off. With the heater running, the funnel and apparatus are slightly warmer than the surroundings (or thermal contact between heater and ice is imperfect), so additional heat may be lost from the apparatus rather than melting ice — meaning more energy was delivered than actually went into the phase change. B1 B1 — one mark for identifying a specific energy-loss route, one for linking that loss to an inflated $L_f$ .

Total: 8 marks (M3 A2 B3 as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A domestic kettle is rated at 2.2 kW. It contains 0.80 kg of water already at the boiling point of 100 °C. The kettle is left on for 90 s after the water first boils. Assume no heat is lost to the surroundings.

(a) Calculate the mass of water that boils away in this time. Take $L_v = 2.26 \times 10^6$ J kg⁻¹. (3)

(b) State and explain one reason why the actual mass of water lost would be smaller than your calculated value if heat losses are taken into account. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — recognising that all electrical energy delivered after boiling goes into vaporisation: $E = Pt = 2200 \times 90 = 1.98 \times 10^5$ J.
M1 (AO1.2) — applying $m = E / L_v$ .
A1 (AO1.1) — $m = 1.98 \times 10^5 / 2.26 \times 10^6 = 0.0876$ kg ≈ 0.088 kg (3 s.f.).

(b)

B1 (AO1.1) — identifies a specific loss route, e.g. heat conducted through the kettle walls to the surroundings, or convective loss through the open spout.
B1 (AO2.1) — explains that some of the supplied 1.98 × 10⁵ J does not reach the water, so less energy is available for vaporisation.
B1 (AO3.2a) — concludes that less mass actually boils away because the available energy for the phase change is reduced.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. Specific latent heat questions on Paper 2 typically combine routine AO1 calculation with an AO2/AO3 commentary on heat loss or experimental validity.

Synoptic links

Connects to:

Specific heat capacity (Topic 9, prior lesson): the parallel structure $Q = mc\Delta\theta$ versus $Q = mL$ underpins every multi-stage problem (e.g. ice → steam crossing two phase boundaries). Strong A* candidates separate the heating curve into discrete sections and apply the appropriate equation in each.
Internal energy (Topic 9, kinetic-model lesson): $U = U_K + U_P$ . During a phase change at constant temperature, $U_K$ is unchanged (since temperature is the macroscopic measure of mean molecular kinetic energy), so the latent-heat input goes entirely into $U_P$ — increasing the molecular potential energy by separating particles against intermolecular forces.
Kinetic theory of gases (Topic 9): $pV = \tfrac{1}{3}Nm\overline{c^2}$ assumes no intermolecular forces. Latent heat is precisely the energy needed to overcome those forces — the very thing the ideal-gas model neglects. This is why real gases deviate from ideal behaviour near condensation.