K323 / 2027
Thermal properties of matter overview

Lesson 5 of 6

Latent heat

  • I can define latent heat and specific latent heat.Syllabus

    Syllabus K323, 9(f). Define latent heat and specific latent heat.

  • I can apply energy transfer for a change of state = mass x specific latent heat.Syllabus

    Syllabus K323, 9(g). Apply energy transfer for a change of state = mass x specific latent heat.

  • I can explain latent heat in terms of particle behaviour.Syllabus

    Syllabus K323, 9(h). Explain latent heat in terms of particle behaviour.

Make a guess. You are not marked.

You want to cool a drink as much as possible. You can add 10 g of ice at 0 degrees C or 10 g of water at 0 degrees C. Which cools the drink more?

  1. The water, because it mixes in at once.
  2. The ice, because melting takes in extra energy.
  3. Both cool it equally, because both start at 0 degrees C.
Show the answer

The ice, because melting takes in extra energy.

The ice must first take in latent heat from the drink to melt, about 3340 J for 10 g, before the meltwater warms up as the water would. That extra energy comes out of the drink.

An energy transfer can change the state of a sample while its temperature stays constant. Latent heat measures that transfer.

For a pure substance at the stated transition and fixed pressure, energy changes the particle arrangement and potential energy during the transition. Average random kinetic energy, and therefore temperature, remains unchanged.

Latent heat, L
The energy transferred when the specified sample changes state without a temperature change. It is a total energy, measured in joules.
Specific latent heat, l
The energy transferred per unit mass for that change of state without a temperature change. The lowercase letter l has units J/kg or J/g, depending on the stated mass unit.
E = mlUse the mass that changes state and the specific latent heat for that transition. For this sample's transition, its total latent heat L is ml.

Specific latent heat of fusion concerns solid becoming liquid; the same magnitude is released per unit mass in the reverse solidification process at the same conditions. Specific latent heat of vaporisation concerns liquid becoming gas; condensation transfers that energy out.

During melting, the arrangement changes so that particles can move past one another while remaining close together. During boiling, particles become widely separated against their attractions. Both processes increase the particle potential-energy contribution. During solidification or condensation, it decreases as energy is transferred out. The particles continue moving; latent heat is not energy used to stop them or split the molecules into smaller particles.

Worked example

Ice already at its melting point

A 0.050 kg sample of ice is already at 0°C under normal atmospheric pressure. Use the supplied specific latent heat of fusion lf = 334000 J/kg.

  1. Identify the stage: only melting is required; there is no warming interval in this question.
  2. Use the mass that melts: E = mlf = 0.050 x 334000 = 16700 J.
  3. Describe the result: after this ideal transfer, the sample is liquid water at 0°C, not water at a higher temperature.

Equivalently, 334000 J/kg is 334 J/g, and 50 g x 334 J/g gives the same 16700 J. If 0.050 kg of water freezes at the same conditions, 16700 J is transferred out.

Try it step by step

Work it out in steps: Energy to warm the water to 100 degrees C in J, then Energy to boil the water away in J, then Total energy in J.

  1. Problem 1. 0.20 kg of water at 60 degrees C is heated until it has all boiled away at 100 degrees C. Take c of water = 4200 J/(kg degrees C) and the specific latent heat of vaporisation = 2260000 J/kg. Find the total energy supplied.
  2. Problem 2. 0.10 kg of water at 80 degrees C is heated until it has all boiled away at 100 degrees C. Take c = 4200 J/(kg degrees C) and lv = 2260000 J/kg. Find the total energy supplied.
  3. Problem 3. 0.50 kg of water at 90 degrees C is heated until it has all boiled away at 100 degrees C. Take c = 4200 J/(kg degrees C) and lv = 2260000 J/kg. Find the total energy supplied.
Show the answer

Problem 1.

Energy to warm the water to 100 degrees C in J: E1 = mc x change in T = 0.20 x 4200 x 40 = 33600 J

Energy to boil the water away in J: E2 = m lv = 0.20 x 2260000 = 452000 J

Total energy in J: E = 33600 + 452000 = 485600 J, which is 486000 J to 3 sf

Problem 2.

Energy to warm the water to 100 degrees C in J: E1 = 0.10 x 4200 x 20 = 8400 J

Energy to boil the water away in J: E2 = 0.10 x 2260000 = 226000 J

Total energy in J: E = 8400 + 226000 = 234400 J, which is 234000 J to 3 sf

Problem 3.

Energy to warm the water to 100 degrees C in J: E1 = 0.50 x 4200 x 10 = 21000 J

Energy to boil the water away in J: E2 = 0.50 x 2260000 = 1130000 J

Total energy in J: E = 21000 + 1130000 = 1151000 J, which is 1150000 J to 3 sf

Calculate each stage separately

Now start with 0.10 kg of ice at -10°C and finish with water at 20°C, at normal atmospheric pressure. First warm the ice to 0°C, then melt it, then warm the water. The supplied heat capacities apply to their respective states.

Follow the same 0.10 kg sample through three stages

Ice at -10 °C becomes water at 20 °C. Use the supplied material values and a melting point of 0 °C at the stated pressure. The energy amounts below all reach the sample; warming the apparatus and transfers to the surroundings are excluded.

1. Warm the ice

1. Warm the ice: 2100 joules transferred into the sampleThe same 0.10 kg sample remains ice as its temperature rises from minus 10 to zero degrees Celsius. Its temperature rise is 10 degrees Celsius. Using the supplied ice specific heat capacity of 2100 joules per kilogram per degree Celsius, the energy transferred in is 2100 joules. The state boxes and transition arrow are schematic. Their equal sizes do not represent equal energy amounts or equal durations.Ice at -10 °C2100 Jtransferred inIce at 0 °C

cice = 2100 J/(kg °C); ΔT = 0 - (-10) = 10 °C.

E = mciceΔT = 0.10 × 2100 × 10 = 2100 J.

2. Melt the ice

2. Melt the ice: 33400 joules transferred into the sampleThe 0.10 kg sample changes from ice to water at zero degrees Celsius, at the stated pressure. Its temperature does not rise during melting. Using the supplied specific latent heat of fusion of 334000 joules per kilogram, the energy transferred in is 33400 joules. The symbol for specific latent heat is lowercase l. The state boxes and transition arrow are schematic. Their equal sizes do not represent equal energy amounts or equal durations.Ice at 0 °C33400 Jtransferred inWater at 0 °C

Specific latent heat of fusion: lf = 334000 J/kg.

E = mlf = 0.10 × 334000 = 33400 J.

The lowercase lf is energy per unit mass. The sample's temperature stays at 0 °C during melting.

3. Warm the water

3. Warm the water: 8400 joules transferred into the sampleThe 0.10 kg sample remains liquid water as its temperature rises from zero to 20 degrees Celsius. Its temperature rise is 20 degrees Celsius. Using the supplied water specific heat capacity of 4200 joules per kilogram per degree Celsius, the energy transferred in is 8400 joules. The state boxes and transition arrow are schematic. Their equal sizes do not represent equal energy amounts or equal durations.Water at 0 °C8400 Jtransferred inWater at 20 °C

cwater = 4200 J/(kg °C); ΔT = 20 - 0 = 20 °C.

E = mcwaterΔT = 0.10 × 4200 × 20 = 8400 J.

Total energy transferred in = 2100 + 33400 + 8400 = 43900 J.

The diagrams show the sequence. Box sizes and arrow lengths do not measure energy or time.

Choose the equation for each stage before substituting. The warming stages use mcΔT; melting uses ml. The cards are schematic and are not sized in proportion to their energy values. The total 43900 J is energy reaching the sample.

The melting contribution is 33400 J even though the temperature does not rise during that stage. The two warming contributions are 2100 J and 8400 J. Adding them gives 43900 J; neither one mcΔT calculation across -10°C to 20°C nor ml alone accounts for the complete change.

This total excludes energy warming the apparatus or escaping to the surroundings. An actual heater may need to supply more. The supplied material values are data for this example, not values that can be applied to every substance.

Determine specific latent heat of fusion

Use ice already at its melting point, with a heater transferring energy at a known constant rate. Collect and measure the mass of liquid produced over a measured time. Melting can also occur because the surroundings are warmer than the ice, so not all collected meltwater is necessarily due to the heater.

  1. Establish the transition: let the ice reach its melting point and drain away water present before the timed collection. Keep some ice present throughout the measurement interval so the intended process remains melting.
  2. Measure meltwater mass: collect the water in a dry weighed container; subtract its empty mass from its final mass. A balance reading that includes the container is not the mass that melted.
  3. Estimate background melting: use a matched unpowered ice setup over the same time. Match the surroundings, exposed geometry, initial conditions and collection method as closely as possible.
  4. Separate the heater contribution: estimate the additional mass melted by the heater as mheated - mcontrol. Use its mass unit consistently with the intended l unit.
  5. Relate energy and mass: if apparatus warming and other differences are negligible or corrected, lf = Pt / (mheated - mcontrol).

Supplied practical data

Separate heater melting from background melting

A 16.7 W heater operates for 200 s. The powered setup produces 14.0 g of collected meltwater; a matched unpowered control produces 4.0 g. Both contain ice at its melting point throughout. Assume equal background heating, negligible apparatus warming during collection and negligible energy loss from the heater or leads outside the melting region.

  1. Heater energy: E = 16.7 x 200 = 3340 J.
  2. Additional melted mass: 14.0 - 4.0 = 10.0 g = 0.0100 kg.
  3. Specific latent heat: lf = 3340 / 0.0100 = 334000 J/kg.

The 4.0 g correction estimates melting due to the surroundings. It does not, by itself, account for energy that leaves the heater through its leads or warms another part of the apparatus.

Ignoring background melting makes too much mass appear to have melted because of the heater. Dividing its energy by this excessive mass gives an estimate of lf that is too low.

The control is an estimate, not an automatic correction for every transfer. It must experience comparable background heating. Meltwater left behind instead of collected, initial water included by mistake, apparatus warming and changing heater output can also affect the result. Repetition helps reveal variation but does not remove an unmatched control or a wrong energy account.

Choose the stage, then the equation. Use mcΔT for a temperature change within one state. Use ml for the mass that changes state at its transition temperature. Capital L is total energy; lowercase l is energy per mass.

Check your understanding

A 0.10 kg sample of ice at -10 degrees C becomes water at 20 degrees C. Supplied values are c_ice = 2100 J/(kg degree C), l_f = 334000 J/kg and c_water = 4200 J/(kg degree C). Which total energy reaches the sample at the stated normal-pressure transition?

  1. 6300 J, using the ice specific heat capacity through the full 30 degrees C interval.
  2. 33400 J, because the energy goes into melting while the temperature is steady.
  3. 43900 J, adding warming the ice, melting it and warming the water.
Show the answer

43900 J, adding warming the ice, melting it and warming the water.

Choose the equation for each physical stage: 0.10 x 2100 x 10, then 0.10 x 334000, then 0.10 x 4200 x 20. Adding the three transfers gives 43900 J. It excludes energy warming the apparatus or escaping to the surroundings.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

Define specific latent heat of vaporisation. [2]

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Exam-style question

A 2.0 kW kettle keeps water boiling. In 3.0 minutes, the mass of water in the kettle falls by 0.15 kg. Calculate the specific latent heat of vaporisation of water from these results, and state one assumption you made. [4]

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