CP13 – Specific Latent Heat | Edexcel A Level Physics

Theory — Specific Latent Heat

Energy is required to change the state of a substance without changing its temperature.

What is Specific Latent Heat?

The specific latent heat L of a substance is the energy required per unit mass to change its state at constant temperature. There are two types:

Specific latent heat of fusion L_f — energy to melt (solid → liquid) or freeze (liquid → solid) 1 kg
Specific latent heat of vaporisation L_v — energy to vaporise (liquid → gas) or condense (gas → liquid) 1 kg

Q = mL

Q = energy transferred (J) · m = mass undergoing phase change (kg) · L = specific latent heat (J kg⁻¹)

Energy Balance — Ice Melting in Warm Water

When ice at 0°C is added to warm water, the energy balance is:

Heat lost by warm water = Heat to melt ice + Heat to warm meltwater m_w · c · (θ₁ − θ₂) = m_i · L_f + m_i · c · θ₂ Solving for L_f: L_f = [m_w · c · (θ₁ − θ₂) − m_i · c · θ₂] / m_i

m_w = mass of warm water (kg) · m_i = mass of ice (kg) · c = 4186 J kg⁻¹ K⁻¹ · θ₁ = initial water temperature · θ₂ = final mixture temperature

Note the ice must be at exactly 0°C (let it drip-dry before adding). If ice is below 0°C there is an additional term m_i·c_ice·(0 − T_ice) on the right side.

Energy Balance — Steam Condensing in Cool Water

Heat gained by cool water = Heat released condensing steam + Heat released cooling condensate m_w · c · (θ₂ − θ₁) = m_s · L_v + m_s · c · (100 − θ₂) Solving for L_v: L_v = [m_w · c · (θ₂ − θ₁) − m_s · c · (100 − θ₂)] / m_s

m_s = mass of steam (kg) · θ₁ = initial water temperature · θ₂ = final mixture temperature

Accepted Values

Quantity	Value
L_f (water, fusion)	3.34 × 10⁵ J kg⁻¹
L_v (water, vaporisation)	2.26 × 10⁶ J kg⁻¹
c (water, specific heat)	4186 J kg⁻¹ K⁻¹

Procedure

Equipment

Ice (≈50g) · Funnel · Two beakers · Electronic balance · Thermometer · Stirring rod · Paper towels

1

Warm ice to 0°C

Place ice in a funnel and allow it to warm to 0°C — water dripping from the funnel indicates the ice is at 0°C. This ensures no extra energy is needed to warm the ice before melting.

💡 Do not skip this step — ice below 0°C would absorb extra heat to reach 0°C, making L_f appear too large.

2

Measure m₀ — empty beaker

Weigh the dry, empty beaker on the balance. Record m₀ in grams.

💡 Dry the beaker thoroughly — any water already present will affect m_water.

3

Add warm water — measure m₁ and θ₁

Add approximately 100 cm³ of water to the beaker. Weigh to get m₁. Measure the temperature θ₁. Mass of water m_w = m₁ − m₀.

💡 Use warm water (not hot) — around 30°C is ideal. This gives a larger temperature drop and reduces heat exchange with surroundings.

4

Add ice and stir

Quickly add approximately 20g of 0°C ice to the beaker. Stir gently and continuously until all ice has melted.

💡 Stir gently — vigorous stirring adds energy to the system. Keep the beaker insulated with polystyrene if possible.

5

Record θ₂ — lowest temperature

Record the lowest temperature reached. This occurs as the last piece of ice melts. Do not record the temperature after the ice has all melted — by then the mixture is warming again.

💡 Watch the thermometer carefully — the temperature continues to fall as long as ice is present, then rises once all ice has melted.

6

Weigh again — measure m₂

Weigh the beaker + water + melted ice. Record m₂. Mass of ice m_i = m₂ − m₁.

💡 Some ice may stick to the thermometer or stirrer. Shake gently to ensure all melted water is in the beaker before weighing.

7

Calculate L_f

L_f = [m_w · c · (θ₁ − θ₂) − m_i · c · θ₂] / m_i where c = 4186 J kg⁻¹ K⁻¹.

💡 Check: Q_lost (heat from warm water) should roughly equal Q_melt + Q_warm (heat absorbed by ice). If they differ by more than 15%, check for heat loss to surroundings or ice not at 0°C.

      🧊 Set up conditions using the sliders, then press ▶ Run Experiment to watch the simulation.
    

Phase Change

Mass of Warm Water m_w

m_w 100 g

50 g200 g

Mass of Ice m_i

m_i 20 g

5 g60 g

Initial Water Temperature θ₁

θ₁ 30 °C

10°C60°C

Expected Result

Calculated L

—

J kg⁻¹

Real Experimental Values

Enter your own lab measurements here to calculate L from real data.

—

J kg⁻¹ — from your real data

Results Data Table

Each row = one experiment. Simulation and real values both recorded here.

#	Type	Source	m_w / g	m_change / g	θ₁ / °C	θ₂ / °C	Q_lost/gained / J	Q_phase / J	L measured / J kg⁻¹	L accepted / J kg⁻¹	% error
No data yet.

Energy Balance Chart

Compares Q_lost (heat from water) with Q_gained (heat to phase change + warming). A good experiment shows these nearly equal.

Energy audit

—% energy accounted for

Last experiment

Q_water (lost/gained)—

Q_phase change—

Q_warming/cooling—

Imbalance—

Common sources of imbalance

Run an experiment to see the energy balance.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Explain why the temperature of a pure substance remains constant during a phase change, even though thermal energy is still being supplied.

During a phase change, the energy supplied (latent heat) is used to break the intermolecular bonds that hold the substance in its current state — not to increase the average kinetic energy of the molecules. Since temperature is a measure of the mean kinetic energy of the molecules, and this is not increasing, the temperature remains constant. For melting: energy breaks the ordered lattice structure of the solid. For vaporisation: energy completely separates molecules against the intermolecular attractive forces. Once all bonds are broken (all ice melted or all liquid vaporised), further energy input again raises kinetic energy and therefore temperature.

Q2

In the ice experiment, a student forgot to let the ice warm to 0°C first. Explain how this would affect the calculated value of L_f (would it be too high or too low?).

If the ice is below 0°C (say T_ice = −5°C), it requires extra energy to warm it to 0°C before it can melt: Q_extra = m_i × c_ice × (0 − T_ice). This extra energy comes from the warm water, making the water cool more than expected. The measured θ₂ is lower than it would be if the ice were at 0°C. The student uses the formula L_f = [m_w·c·(θ₁−θ₂) − m_i·c·θ₂]/m_i. The numerator (Q_lost − Q_warming) is larger because θ₂ is too low, but this extra energy was actually used to warm the ice to 0°C, not to melt it. So the student attributes energy to L_f that was actually used to warm the ice — the calculated L_f is too high.

Q3

Why is L_v (vaporisation) approximately seven times larger than L_f (fusion) for water?

During melting (solid → liquid), the molecules gain enough energy to break free from their fixed lattice positions, but they remain in close contact with neighbouring molecules — the intermolecular forces are not fully overcome. Only about 15% of the intermolecular bonds are broken on average. During vaporisation (liquid → gas), the molecules must overcome all intermolecular attractive forces completely and move very far apart. The amount of energy required to fully separate molecules is much greater than just disrupting the ordered lattice. For water, hydrogen bonds are exceptionally strong, making L_v particularly large (2.26 MJ/kg) compared to L_f (0.334 MJ/kg) — a ratio of approximately 6.8.

Q4

Suggest three sources of systematic error in the ice experiment and explain how each would affect the calculated value of L_f.

1. Heat exchange with the surroundings — the beaker and water lose heat to the cooler surroundings during the experiment. This means Q_lost (from water to ice) is overestimated because some cooling is due to the environment, not just the ice. Calculated L_f appears too high. Minimise by using polystyrene insulation. 2. Wet ice — if ice has surface water that is not at 0°C or brings extra mass, the measured m_i is incorrect. Dry ice with paper towels to remove surface water. 3. Heat capacity of the beaker — the beaker also loses heat to the ice but this is ignored in the formula. The beaker cools from θ₁ to θ₂, releasing energy m_beaker × c_beaker × (θ₁ − θ₂). This energy is attributed to L_f but actually came from the beaker, making L_f appear too high. Correct by using a thin polystyrene cup instead of a glass beaker, or by measuring the heat capacity of the beaker and including it in the calculation.

Q5

A student adds 22.4 g of ice at 0°C to 105 g of water at 28°C. The final temperature is 8°C. Calculate L_f and find the percentage difference from the accepted value of 3.34 × 10⁵ J kg⁻¹.

m_w = 0.1050 kg, m_i = 0.0224 kg, θ₁ = 28°C, θ₂ = 8°C, c = 4186 J kg⁻¹ K⁻¹. Q_lost by water = m_w × c × (θ₁−θ₂) = 0.1050 × 4186 × 20 = 8790.6 J. Q_warm meltwater = m_i × c × θ₂ = 0.0224 × 4186 × 8 = 750.6 J. Q_melt = Q_lost − Q_warm = 8790.6 − 750.6 = 8040 J. L_f = Q_melt / m_i = 8040 / 0.0224 = 3.589 × 10⁵ J kg⁻¹. % difference = |(3.589 − 3.34)/3.34| × 100 = 7.5%. This is within typical experimental uncertainty for this method.