9478 / 2027
Thermodynamic Systems overview

Lesson 4 of 4

Phase change and latent heat

  • I can define and apply specific heat capacity and specific latent heat under the stated thermal and process conditions.Syllabus

    Syllabus 9478, 13(g). Define and apply specific heat capacity and specific latent heat under the stated thermal and process conditions.

Make a guess. You are not marked.

Ice at 0 degrees C is heated until it has all become water at 0 degrees C. What happens to its internal energy?

  1. It falls, because the molecules end up closer together in water.
  2. It stays the same, because the temperature has not changed.
  3. It increases, as the potential energy of the molecules rises.
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It increases, as the potential energy of the molecules rises.

Energy is supplied to break the rigid structure. The mean kinetic energy stays the same, but the molecular potential energy rises, so the internal energy increases.

Decide whether the energy changes temperature or changes phase. At a phase change, energy can enter while temperature stays constant under the stated pressure and conditions.

Define specific latent heat

Specific latent heat l is the energy required by heating per unit mass for a stated change of phase at constant temperature under stated conditions. Its unit is J/kg.

Q = ml

Specific latent heat of fusion concerns melting; specific latent heat of vaporisation concerns liquid becoming vapour. Their values need not be equal. Freezing and condensation transfer energy out under the corresponding reversed conditions.

Here lowercase l denotes the specific quantity. Some references use L for it; check the definition and unit. A value in J/kg is not the total energy for an arbitrary sample.

During a phase change, energy can change microscopic potential energy and supply expansion work without raising temperature. Q = mcΔT describes warming within a phase, so setting ΔT = 0 in that equation does not prove that no energy entered during a phase change.

Split a multi-stage change

Before calculating precisely, roughly 0.1 kg multiplied by a supplied fusion latent heat of order 3 × 105 J/kg needs energy of order 3 × 104 J. Melting should therefore dominate the following account compared with its short warming stages.

Worked warming and melting

Ice at -5.0°C to water at 25.0°C

A 0.0800 kg ice sample ends as liquid water. Use cice = 2100 J/(kg K), lf = 3.34 × 105 J/kg and cwater = 4200 J/(kg K). The stated melting temperature is 0°C. Assume constant properties over each interval, negligible losses and container heat capacity, and negligible volume work for these condensed phases.

Energy entering each stage of the supplied ice-to-water model
StageCalculationEnergy / J
Warm ice from -5 to 0°C0.0800 × 2100 × 5.0840
Melt all ice at 0°C0.0800 × 3.34 × 10526720
Warm water from 0 to 25°C0.0800 × 4200 × 25.08400
Total input = 840 + 26720 + 8400
= 35960 J ≈ 36.0 kJ

Use cumulative energy to separate warming from melting

For the stated 0.0800 kg ice/water sample, assume the supplied heat capacities and fusion latent heat, negligible loss and vessel capacity, and negligible volume work. The horizontal coordinate is energy supplied to the sample, not time.

A true energy scale shows that melting dominates the supplied ice-to-water accountTemperature in degrees Celsius is plotted against cumulative input energy in kilojoules. The exact endpoints are A, zero kilojoules and minus five degrees; B, 0.840 kilojoules and zero degrees; C, 27.560 kilojoules and zero degrees; and D, 35.960 kilojoules and 25 degrees. One linear energy scale is used throughout: 250 drawing units for 36 kilojoules. Consequently the initial warming interval is only 5.8333 drawing units wide and has not been widened. Leaders and a separate point key identify the closely spaced initial energy coordinates. The long horizontal B-to-C segment requires 26.720 kilojoules of melting energy at constant temperature. The final water-warming stage uses 8.400 kilojoules. The arrowed horizontal energy axis is at zero degrees Celsius, and A is below it. The graph gives no elapsed duration or heating-rate assumption.010203036-50102025ABCDTemperature / °CCumulative input energy / kJ

A: 0 kJ, -5 °C. B: 0.840 kJ, 0 °C. C: 27.560 kJ, 0 °C. D: 35.960 kJ, 25 °C.

A-B warms ice; B-C melts it; C-D warms water. The initial stage is genuinely narrow on this energy scale. A flat temperature segment does not mean that energy input has stopped.

The horizontal axis is cumulative energy entering the sample, not time. The four model endpoints are 0, 0.840, 27.560 and 35.960 kJ. The long horizontal interval represents melting with continued energy input at constant temperature.

The first 840 J warms the ice. The next 26720 J melts it, so complete melting occurs at cumulative input 27560 J. Only after this can further input warm an entirely liquid sample in this model. The graph's energy widths do not by themselves specify stage durations.

Check whether all the material changes phase

Suppose a separate 0.0200 kg ice sample is already at its melting point and receives 5000 J, with negligible other transfers. Complete melting would require:

Qall = 0.0200(3.34 × 105) = 6680 J
Mass melted = 5000/(3.34 × 105)
= 0.01497 kg ≈ 15.0 g

About 5.03 g of ice remains. Ice and water coexist at the melting temperature under the stated conditions; there is insufficient energy to melt everything and then warm the liquid.

Try it step by step

Work it out in steps: Energy for the change of state, Q = mL, in J, then Time at the heater power, t = Q / P, in s.

  1. Problem 1. A 100 W heater melts 0.20 kg of ice at 0 degrees C. The specific latent heat of fusion of ice is 3.34 x 105 J/kg. Assume all the heater's energy goes to the ice. How long does it take?
  2. Problem 2. A 50 W heater melts 0.050 kg of ice at 0 degrees C. Use the same latent heat and assumption. How long does it take?
  3. Problem 3. A 2.0 kW kettle boils away 0.30 kg of water at 100 degrees C. The specific latent heat of vaporisation of water is 2.26 x 106 J/kg. Assume all the energy goes to the water. Find the energy needed and the time taken.
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Problem 1.

Energy for the change of state, Q = mL, in J: Q = 0.20 x 3.34 x 105 = 6.68 x 104 J

Time at the heater power, t = Q / P, in s: t = 6.68 x 104 / 100 = 668 s

Problem 2.

Energy for the change of state, Q = mL, in J: Q = 0.050 x 3.34 x 105 = 1.67 x 104 J

Time at the heater power, t = Q / P, in s: t = 1.67 x 104 / 50 = 334 s

Problem 3.

Energy for the change of state, Q = mL, in J: Q = 0.30 x 2.26 x 106 = 6.78 x 105 J

Time at the heater power, t = Q / P, in s: t = 6.78 x 105 / 2000 = 339 s

Check your understanding

A 20.0 g ice sample is already at its melting point. It receives 5000 J with negligible other transfers. Use l_f = 3.34 x 105 J/kg. What happens?

  1. The whole of the ice melts, and the water then rises above the melting temperature.
  2. About 15.0 g melts and the temperature stays at the melting point.
  3. No ice melts because the supplied energy must first raise its temperature.
Show the answer

About 15.0 g melts and the temperature stays at the melting point.

Melted mass = 5000/(3.34 x 105) = 0.01497 kg. The energy is insufficient to melt all 20.0 g, so ice and water remain together at the melting temperature.

Account for work during vaporisation

Latent heating need not all become an internal-energy increase when the sample expands significantly. Choose a fixed sample as the system and use ΔU = Q + Won.

Worked phase-change energy account

Separate heating from expansion work

In a separate supplied model, 0.0100 kg vaporises at constant temperature against external pressure 2.0 × 105 Pa. Use lv = 1.80 × 106 J/kg and a volume increase of 0.0060 m3. These are supplied model properties, not stated values for water.

Q = mlv = 0.0100(1.80 × 106) = 18000 J
Wby = pextΔV = 1200 J
Won = -1200 J
ΔU = 18000 - 1200 = 16800 J

The heating supplies both the internal-energy increase and the work output. Constant temperature does not mean zero ΔU, and Q = ml does not automatically make ΔU equal to Q.

Measure fusion latent heat with a matched control

A melting investigation can compare water collected with a measured heater input against a comparable no-heater control. Begin with ice and apparatus at the melting temperature, drain pre-existing water before timed collection, and use the same collection geometry and duration. Measure collected mass with a tared collector and record the input and timing resolutions.

Compare collected melt with a matched no-input control

Begin with ice and apparatus at the melting temperature, and drain pre-existing water before each timed collection. The two runs use the same arrangement and 300 s interval. The supplied model assumes equal background heating and complete collection of the relevant melt.

Heater-on collection

A 16.7-watt heater run collects twenty grams of melt in 300 secondsBoth panels have the same draining ice holder, immersed heater, perforated support, outlet and tared collector on a balance. Ice and apparatus start at the melting temperature. Two wires connect the heater to its input device. The measured heater input is 16.7 watts for 300 seconds, or 5010 joules. The supplied final collected-water mass is 20.00 grams. An inward orange arrow identifies the assumed matched background heating in each run. The perforated support retains ice while water drains through the open outlet into the collector; a droplet illustrates collection during the run. The balance label gives the supplied final water mass after taring the container. Water levels are schematic, not calibrated volume measurements. No boundary labels this draining holder as a closed fixed-mass system. Subtracting the two collected masses isolates the heater contribution only under the stated matching and collection assumptions.Collect for 300 sHeater on16.7 WIce at 0 °CMatchedbackgroundSupportTaredcollector20.00 g

Matched control: no heater input

The matching no-heater-input control collects five grams in 300 secondsBoth panels have the same draining ice holder, immersed heater, perforated support, outlet and tared collector on a balance. Ice and apparatus start at the melting temperature. Two wires connect the heater to its input device. The same heater is unpowered during the 300-second control, and the supplied final collected-water mass is 5.00 grams. An inward orange arrow identifies the assumed matched background heating in each run. The perforated support retains ice while water drains through the open outlet into the collector; a droplet illustrates collection during the run. The balance label gives the supplied final water mass after taring the container. Water levels are schematic, not calibrated volume measurements. No boundary labels this draining holder as a closed fixed-mass system. Subtracting the two collected masses isolates the heater contribution only under the stated matching and collection assumptions.Collect for 300 sHeater off0 WIce at 0 °CMatchedbackgroundSupportTaredcollector5.00 g

The extra collected mass is 20.00 - 5.00 = 15.00 g. The measured heater input is 16.7 × 300 = 5010 J, giving lf = 5010/0.01500 = 3.34 × 105 J/kg under the stated assumptions.

This is an input-versus-extra-melt comparison. Initial warming, trapped water or a changed background transfer would undermine the simple subtraction. Because meltwater leaves the holder, a closed fixed-mass account for the holder would omit the outflow.

The heater-on and control arrangements collect meltwater over the same 300 s interval. Background heating is assumed comparable. This is an input-versus-extra-melt comparison; the draining vessel alone is not a closed fixed-mass system.

In the supplied comparison, measured heater power is 16.7 W = 16.7 J/s for 300 s. The heated run collects 20.00 g of water; the comparable control collects 5.00 g.

Heater input = 16.7 × 300 = 5010 J
Extra melted mass = (20.00 - 5.00) g = 0.01500 kg
lf = 5010/0.01500
= 3.34 × 105 J/kg

This calculation assumes complete collection, equal background transfer in the two runs, and that the measured heater input accounts for the additional melting rather than appreciable apparatus warming or an unmatched loss. The control is a conditional correction, not a guarantee that every error has disappeared.

  • Ice initially below its melting point: part of the input first warms it. Treating all that input as melting energy can make the inferred l too high.
  • Water retained in the apparatus: the collected extra mass can be smaller than the mass actually melted. Dividing input by that underestimated mass can also give too large an l.
  • Different background conditions: a heater can change temperature gradients or losses. A control that no longer matches these conditions cannot simply be subtracted as an equal background.

Water leaving the vessel crosses its boundary. Do not apply a closed, fixed-mass first-law account to that vessel while ignoring the outflow. The matched input-and-mass calculation above states the required assumptions without treating the draining vessel as such a system.

Practise an exam question

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

Exam-style question

The specific latent heat of vaporisation of water is much larger than its specific latent heat of fusion. Explain why. [3]

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