A substance being heated either changes temperature or changes state, and on a heating curve the two look completely different. A sloping section means the temperature is changing; a flat section means a change of state is happening. Questions ask you to identify each section, explain it in terms of particles and use the correct equation for the stage you are in.
This lesson follows interpreting an energy-temperature slope and sets up calculating latent energy.
What is happening at each stage of a heating curve?
During a sloping section, energy supplied increases the kinetic energy of the particles. They vibrate or move faster, so the temperature rises. The equation Q = m × c × Δθ applies here.
During a flat section, the energy supplied increases the potential energy of the particles as they are pulled apart or freed from their neighbours. The average kinetic energy stays the same, so the temperature is constant. Q = m × c × Δθ gives zero here, which is a clue that it is the wrong equation for this stage.
- Look at the graph shape and label each section as sloping or flat.
- Name the stage: solid warming, melting, liquid warming, boiling or gas warming.
- Choose the equation by stage: m × c × Δθ for sloping sections, the latent energy equation for flat sections.
- Use the time as a measure of energy if the power is steady: energy = power × time.
Worked example
A solid is heated by a steady 300 W heater. (Invented example data.) The temperature is recorded every minute:
| Time (min) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| Temperature (°C) | 30 | 45 | 60 | 60 | 60 | 60 | 70 | 80 | 90 |
Describe each stage and find the energy supplied during the flat section.
Step 1, stages: 0 to 2 min, the solid warms from 30 °C to 60 °C. From 2 to 5 min the temperature stays at 60 °C, so the substance is melting at 60 °C. From 5 to 8 min the liquid warms from 60 °C to 90 °C.
Step 2, compare the slopes: the solid rises 15 °C per minute and the liquid only 10 °C per minute. The same energy arrives each minute, so the liquid needs more energy per degree. Its m × c is larger.
Step 3, energy in the flat section: duration = 3 min = 180 s. E = P × t = 300 × 180 = 54 000 J.
Step 4, check: 300 W is 300 × 60 = 18 000 J per minute. Three minutes gives 3 × 18 000 = 54 000 J, which matches.
The mistake to watch for
A common slip is to say that nothing is happening during the flat section.
Mistaken answer: “The temperature is constant, so no energy is being transferred to the substance.”
The heater is still on, and 54 000 J is supplied in this example.
The correction is to say where the energy goes: it increases the potential energy of the particles while the substance changes state, so the temperature stays constant. The energy is transferred, but it is not used to raise the temperature.
Check yourself
1. In the example, what is the energy supplied in the first two minutes?
Show answer
2 min = 120 s. E = 300 × 120 = 36 000 J. Check: 2 × 18 000 J = 36 000 J.
2. During which time intervals is the average kinetic energy of the particles increasing?
Show answer
0 to 2 min and 5 to 8 min. The temperature is rising only in those intervals. In the flat section the average kinetic energy is constant.
3. Explain in one or two sentences why the temperature stays constant while the substance melts.
Show answer
The energy supplied is used to overcome the forces holding the particles in place, increasing their potential energy. The average kinetic energy does not change, so the temperature stays constant until melting is complete.
Where this leads next
The flat section can be calculated: continue with calculating latent energy, if your syllabus includes it. The heat calculations practice set includes heating-curve questions.
A flat line is easy to spot but harder to explain in words that earn marks. A teacher in online one-to-one Physics tuition can read your explanation and tighten each sentence.