All subjects · Physics · WPH15 (Unit 5)

Core practical 13: specific latent heat

Core practical 13 measures a specific latent heat, usually of fusion of ice or vaporisation of water, using an electrical heater.

Latent heat of fusion of ice

Put an immersion heater in a funnel packed with crushed ice at 0 °C, with a beaker underneath. Set up an identical control funnel with no heater (or the heater switched off).
Switch on the heater for a measured time t and record its power P (from a joulemeter, or P = IV).
Collect and weigh the meltwater from each funnel.
The control shows how much ice melts from the room's energy, so L = Pt / (m_heater - m_control).

Latent heat of vaporisation of water

Put a beaker of water with an immersion heater on a top-pan balance and bring it to the boil.
Once boiling steadily, record the fall in mass Δm over a time t.
L = Pt / Δm. Energy lost to the surroundings makes this value too large, because not all of Pt evaporates water.
Better: repeat at two powers. The losses are about the same, so L = (P_1 - P_2) / (rate_1 - rate_2), where 'rate' is the mass lost per second.

Good practice

Crushed ice gives good contact with the heater; the heater must be fully covered.
Use ice at 0 °C (melting), so no energy goes into warming it.
Wait until water boils steadily before timing.

Worked example

A 40 W heater runs for 300 s. 0.050 kg of meltwater is collected from its funnel and 0.015 kg from the control. Find L.

Energy = 40 x 300 = 12000 J
Ice melted by the heater = 0.050 - 0.015 = 0.035 kg
L = 12000 / 0.035 = 342857 J/kg

Answer: L = 3.4 x 10⁵ J/kg

Key idea: L = Pt / Δm; a control funnel removes melting caused by the room; two powers cancel heat losses in vaporisation.

Check yourself

A 40 W heater runs for 300 s. Its funnel collects 0.050 kg of meltwater; the control funnel collects 0.015 kg. Find L, in units of 10⁵ J/kg, to 3 s.f.
Show answer
3.43 x 10⁵ J/kg. L = 12000 / 0.035 = 342857 J/kg = 3.43 x 10⁵ J/kg.
A 60 W heater keeps water boiling. The mass falls by 4.5 g in 180 s. Find L, in units of 10⁶ J/kg.
Show answer
2.4 x 10⁶ J/kg. L = Pt / Δm = 10800 / 0.0045 = 2.4 x 10⁶ J/kg.
Water boils away at 1.20 x 10^-5 kg/s with a 50 W heater and at 0.31 x 10^-5 kg/s with a 30 W heater. Assuming equal heat losses, find L, in units of 10⁶ J/kg, to 3 s.f.
Show answer
2.25 x 10⁶ J/kg. L = (50 - 30) / (0.89 x 10^-5) = 2.25 x 10⁶ J/kg.
What is the purpose of the control funnel with no heater?
- To measure how much ice melts due to energy from the surroundings
- To check the heater's power
- To keep the ice cold
- To collect extra water
Show answer
To measure how much ice melts due to energy from the surroundings. Subtracting the control's meltwater leaves only the ice melted by the heater.
In the vaporisation experiment with a single heater power, why is the calculated L usually too large?
- The balance reads too high
- Water boils at less than 100 °C
- Some of the heater's energy is lost to the surroundings rather than evaporating water
- The heater's power increases over time
Show answer
Some of the heater's energy is lost to the surroundings rather than evaporating water. Pt includes energy lost, so dividing by Δm overestimates L.
Why should the ice be melting (at 0 °C) at the start?
- So the heater's energy only melts the ice and does not warm it first
- So it melts more slowly
- So the meltwater is easier to weigh
- So the heater does not overheat
Show answer
So the heater's energy only melts the ice and does not warm it first. Ice below 0 °C would need mcΔθ before it could melt.
Why must the heater be fully surrounded by crushed ice?
- To stop the heater moving
- To make the ice melt evenly in the control
- So that all the energy from the heater goes into the ice
- To increase the heater's power
Show answer
So that all the energy from the heater goes into the ice. Any exposed part would heat the air instead.

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Last updated 7 November 2026. Spotted a mistake? Tell us.