Latent heat and specific latent heat
Key idea: Changing state takes energy without changing the temperature. The energy needed equals the mass that changes state multiplied by the specific latent heat.
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Guided steps
- Read the units and identify the mass that actually changes state.
- Compare particle-energy changes and the two different heating curves.
- Predict melting temperature, particle-energy changes and the effect of doubled power.
- Compare melting times, energy changes and temperature at both powers.
- Follow the phase choice, changed mass and energy units.
- Calculate how much ice melts from the energy input.
- Find the specific latent heat by dividing energy by changed mass.
- Read when heater input can be assigned entirely to changing state.
- Convert time to seconds, calculate heater energy, then find the melted mass.
- Calculate the energies for warming and boiling the same mass.
- Calculate energy retained, ice melted, ice remaining and final temperature.
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Melting ice and boiling water both take in energy without getting any hotter. That energy is called latent heat, and how much is needed depends on the substance and on how much of it changes state. With one equation you can work out how long a kettle takes to boil dry or how much ice a drink can melt.
Latent heat and specific latent heat
The latent heat, L, of a body is the energy absorbed or released when it changes state without a change in temperature. Its unit is the joule.
The specific latent heat, l, of a substance is the energy needed to change the state of 1 kg of it without a change in temperature. Its unit is J kg⁻¹.
For a mass m changing state, the energy transferred is
At the same pressure, melting a given mass needs as much energy as freezing it releases; boiling and condensing are also reverse changes. The mass in Q = ml is the amount that actually changes state, not automatically all the material present.
There are two values for each substance:
- the specific latent heat of fusion, lf, for melting and freezing (solid ↔ liquid);
- the specific latent heat of vaporisation, lᵥ, for boiling and condensing (liquid ↔ gas).
For water, lf = 3.34 × 10⁵ J kg⁻¹ and lᵥ = 2.26 × 10⁶ J kg⁻¹.
What the latent heat does to the particles
For a pure substance changing state at fixed pressure, latent heat changes particle arrangement and potential energy while average random kinetic energy and temperature stay constant. Melting lets particles move past one another; boiling gives a much more widely separated gas. Particle potential energy increases. Separation need not increase during melting: ice becomes denser as water forms. Freezing and condensation reverse the potential-energy change and release energy; the water molecules themselves stay the same size.
For water, this explains why lᵥ is much larger than lf. Melting only loosens the particles enough to slide past one another; boiling has to separate them completely, which takes much more energy.
Heating curve for a pure substance
Temperature rises as solid warms to a melting plateau at 20 °C, then liquid warms to a boiling plateau at 80 °C, then gas warms. The temperatures and times are illustrative, not measured.
If the figure is wider than its frame, scroll across to read all labels.
View figure data
This table is included in print while it is open.
| Illustrative time (min) | Temperature |
|---|---|
| 0 | 0 |
| 2 | 20 |
| 4 | 20 |
| 6 | 80 |
| 8 | 80 |
| 10 | 120 |
Check your understanding 1 Predict the model’s energy stages
The model starts with 0.10 kg of ice at −20 °C and heats it at atmospheric pressure. Before running it, predict what happens to temperature and particle energy while the ice melts. With all heater energy reaching the sample, what changes if Heater power doubles?
Show answer
The temperature rises to 0 °C, stays there while ice melts, rises to 100 °C, then stays there while water boils. During melting, latent heat increases particle potential energy; average random kinetic energy stays constant.
Doubling Heater power from 500 to 1000 W halves each stage’s duration. The same mass still needs the same energy, and the phase-change temperatures stay the same. Compare Temperature, State and Energy supplied as the model runs. Its time graph uses minutes. The earlier schematic illustrates a different substance; water melts at 0 °C and boils at 100 °C at the stated pressure.
A 500 watt heater is heating 100 grams of ice that started at −20 degrees Celsius. It is now ice at −20.0 degrees Celsius after 0 kilojoules.
- Temperature
- −20.0 °C
- State
- ice
- Energy supplied, E = P t
- 0 kJ
- Dull black can
- — °C
- Shiny silver can
- — °C
Try this
0 of 4 doneHeat the ice until the water boils, and compare the two flat parts of the graph. (not done yet)
While ice melts or water boils, latent energy changes the arrangement and interparticle potential energy instead of raising temperature. Ice melting does not require increased particle separation. For the same mass of water, boiling takes far more energy than melting.
Record a cooling curve for stearic acid from above 72 °C to below 66 °C. (not done yet)
The flat part of your curve is the melting point, 69 °C: as the liquid solidifies it gives out latent heat, which balances the heat lost.
Find out which rod conducts heat best. (not done yet)
Copper's pins drop first. In metals, free electrons carry energy along the rod quickly; glass relies only on vibrating particles.
Compare how quickly the dull black and shiny silver cans warm up. (not done yet)
The dull black can warms faster: dull black surfaces are good absorbers (and good emitters) of infrared radiation; shiny surfaces reflect it.
You’ve tried every task.
Your readings
| # | t / min | θ / °C | Remove |
|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | |||
Check your understanding 2 Compare the melting plateau
Run Heating curve of ice at 500 W. During the 0 °C plateau, record the times and Energy supplied when melting starts and ends. Find the increases in time and energy. Reset and run at 1000 W; compare the plateau duration, energy change and temperature with your predictions. Use State to identify melting. Explain the particle-energy change. You may use the written readings below if interaction is unavailable.
Show answer
At 500 W, melting starts at about 0.14 min with 4.2 kJ received and ends at 1.25 min with 37.6 kJ received. The differences are about 1.11 min (66.8 s) and 33.4 kJ, matching 0.10 × 3.34 × 10⁵ = 33400 J. At 1000 W the duration halves to about 0.56 min, while the energy change is still 33.4 kJ and temperature stays at 0 °C. Total energy received at the end includes the earlier warming of ice; subtract the start value to get only the melting energy. Particle potential energy increases while average random kinetic energy stays constant.
Worked example 1
Energy to melt ice
Problem
Show full solution
Choose the equation
Method
Use Q = mlf.Reason
The ice changes state from solid to liquid with no change in temperature.Working
Answer
Q = mlfSubstitute and calculate
Method
Multiply the mass by the specific latent heat of fusion.Reason
The mass is already in kilograms, which matches J kg⁻¹.Working
Answer
Q = 0.50 kg × 3.34 × 10⁵ J kg⁻¹ = 1.67 × 10⁵ J
Quick check
Guided practice 1
How much ice melts?
Problem
Make the mass the subject
Hints
Hint 1
start from the equation
Use Q = mlf.
Hint 2
divide by the latent heat
m = Q/lf.
Show solution step by step
Rearrange for mass
Method
Divide the energy by the specific latent heat.Reason
In Q = mlf the mass is multiplied by lf.Working
Answer
m = Q/(lf)Calculate the mass
Method
Substitute and divide.Reason
Joules divided by joules per kilogram leaves kilograms.Working
Answer
m = (6.68 × 10⁵ J)/(3.34 × 10⁵ J kg⁻¹) = 2.0 kg
Quick check
Spot the mistake 1
Finding the specific latent heat of vaporisation
Learner response
9.04 × 10⁵ J turns 0.40 kg of water at 100°C into steam at 100°C. To find lᵥ, a student multiplies Q by m. Find the error and calculate lᵥ.
Diagnose, then calculate
Show solution step by step
Rearrange for the specific latent heat
Method
Divide the energy by the mass.Reason
“Specific” means per kilogram, so you need energy ÷ mass.Working
Answer
lᵥ = Q/mCalculate
Method
Substitute the values.Reason
The water changes from liquid to gas, so this is the latent heat of vaporisation.Working
Answer
lᵥ = (9.04 × 10⁵ J)/(0.40 kg) = 2.26 × 10⁶ J kg⁻¹
Common mistake
Supplying the energy with a heater
A heater of power P supplies energy E = Pt in time t. If all of it goes into changing the state of the substance,
In a real experiment some energy warms the container or is lost to the surroundings, and some ice can melt from the warmth of the room. Insulation reduces these effects. Questions usually tell you to assume that no energy is lost.
Exam-style question 1
Melting ice with a heater
Examination question
Write your answer before viewing the mark scheme
Show the steps Optional working
Show solution step by step
Find the energy supplied
2 marksMethod
Convert minutes to seconds, then use E = Pt.Reason
A watt is a joule per second, so the time must be in seconds.Working
Answer
t = 8.0 × 60 = 480 s, E = 500 W × 480 s = 2.40 × 10⁵ JFind the mass melted
3 marksMethod
Divide the energy by lf.Reason
All the energy melts ice, so Pt = mlf.Working
Answer
m = (2.40 × 10⁵ J)/(3.34 × 10⁵ J kg⁻¹) = 0.72 kg
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark each point separately.
Common mistake
Using 8.0 minutes without converting gives 0.012 kg and loses the first two marks.
Check your understanding 3 Warming water or boiling it away?
Calculate the energy for warming 1.0 kg of water from 0°C to 100°C, and for boiling the same water away at 100°C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and lᵥ = 2.26 × 10⁶ J kg⁻¹.
Show answer
Warming, with a temperature change, uses Q = mcΔθ:
Boiling, with a change of state, uses Q = mlᵥ:
Boiling it away needs over five times as much energy. That is why a kettle takes a few minutes to boil but much longer to boil dry.
Check your understanding 4 Account for losses and partial melting
A mixture contains 0.10 kg each of ice and water at 0 °C. A 200 W heater runs for 120 s, losing 4000 J to the surroundings. The rest melts ice. Using fusion 334 000 J/kg, find the retained energy, melted mass and remaining ice. What is the final temperature?
Show the steps Optional working
Show the steps Optional working
Show answer
Input energy = 200 × 120 = 24000 J. Retained energy = 24000-4000 = 20000 J. Mass melted = 20000/334000 ≈ 0.060 kg; remaining ice = 0.10-0.060 ≈ 0.040 kg. The mixture stays at 0 °C because ice and water are still present.
Use only the mass changing state in Q = ml. The total sample remains 0.20 kg, but some water was already liquid and some ice has not melted.
Common mistakes
- Using Q = mcΔθ during a change of state. When the temperature is constant, use Q = ml.
- Forgetting to convert grams to kilograms.
- Using lf for boiling or lᵥ for melting.
- Forgetting to convert minutes to seconds before using E = Pt.
- Saying latent heat makes the particles move faster. It changes their potential energy.
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