During melting or boiling, heat is supplied but the temperature stays constant. The energy changes the arrangement of the particles, and it is calculated with Q = ml.
This lesson is part of heat. It pairs with using specific heat capacity and leads to interpreting temperature-time graphs.
Where does the energy go?
In a solid, particles are held in place by forces between them. To melt it, energy must be used to loosen those forces.
That energy raises the potential energy of the particles and leaves the kinetic energy unchanged. Temperature measures the average kinetic energy, so it does not change during the change of state.
Worked example: warming then boiling
How much energy changes 0.10 kg of water at 20°C into steam at 100°C? Use c = 4200 J kg⁻¹ °C⁻¹ and l = 2.26 × 10⁶ J kg⁻¹ for boiling.
- Warm the water: Q₁ = mcθ = 0.10 × 4200 × 80 = 33 600 J.
- Boil it: Q₂ = ml = 0.10 × 2.26 × 10⁶ = 226 000 J.
- Total = 33 600 + 226 000 = 259 600 J, which is 2.60 × 10⁵ J.
Notice that the boiling stage needs about seven times the energy of the warming stage. A large share of the total goes into changing state.
Using a heater
A 1000 W heater melts 0.20 kg of ice at 0°C. Take l = 3.34 × 10⁵ J kg⁻¹ for melting. Energy needed is 0.20 × 3.34 × 10⁵ = 66 800 J, so the time is 66 800 ÷ 1000 = 66.8 s.
The mistake that costs marks
The slip is to use Q = mcθ during a change of state, where θ is zero, or to forget the warming stage before boiling.
| Slip | Result | Fix |
|---|---|---|
| Q = mcθ for boiling at 100°C | θ = 0 gives Q = 0 | Use Q = ml |
| Only ml for 20°C to steam | Misses 33 600 J | Add the warming stage |
| Saying the energy “raises the temperature” during melting | Temperature is constant | Energy raises potential energy |
Split any long question into stages, and label each stage with its formula. The units and significant figure checker helps you keep J and kJ apart.
Check yourself
0.050 kg of ice at 0°C is changed into water at 20°C. Use l = 3.34 × 10⁵ J kg⁻¹ and c = 4200 J kg⁻¹ °C⁻¹. Find the energy needed.
Answer
Melting: Q₁ = 0.050 × 3.34 × 10⁵ = 16 700 J.
Warming the water from 0°C to 20°C: Q₂ = 0.050 × 4200 × 20 = 4200 J.
Total = 16 700 + 4200 = 20 900 J.
What to study next
See both formulas at work on one graph in interpreting temperature-time graphs. Then try the heat practice set.
For a teacher to check your staged calculations, see online one-to-one Physics tuition.