A temperature-time graph shows how temperature changes while energy is supplied or removed steadily. Sloping sections show the temperature changing within one state. Flat sections show a change of state, when energy is still moving but the temperature is not changing.
It appears in thermal questions on melting, boiling and cooling, and it needs the ideas in distinguishing temperature from thermal energy. A tool such as the rate and energy graph interpreter lets you practise choosing what a gradient means.
What does each part of the graph show?
A rising slope: the substance is gaining energy and its particles are speeding up, so the temperature rises. The state stays the same.
A flat section during heating: the substance is melting or boiling. The energy supplied is used to overcome the bonds between particles, so the particles gain potential energy without a change in average kinetic energy.
A falling slope: the substance is losing energy to its surroundings.
A flat section during cooling: the substance is condensing or freezing. Energy is released as bonds form, which balances the energy lost.
The melting point or boiling point is the temperature of the flat section.
Worked example
A solid is heated at a steady power of 300 W. The table shows invented readings.
| Time (min) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 26 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Temperature (°C) | 20 | 40 | 60 | 80 | 80 | 80 | 80 | 80 | 90 | 100 | 110 | 140 |
(a) Melting point: the temperature stays at 80 °C, so the melting point is 80 °C.
(b) Time spent melting: flat from 6 min to 14 min, so 14 − 6 = 8 min.
(c) Energy supplied while melting: 8 min = 8 × 60 = 480 s. Energy = power × time = 300 × 480 = 144 000 J. This assumes all the power reaches the sample.
(d) Rate of rise as a solid: (80 − 20) ÷ 6 = 10 °C/min.
(e) Rate of rise as a liquid: from 14 to 26 min the temperature rises 80 to 140 °C, which is 60 °C in 12 min, so 5 °C/min.
(f) Comparison: the liquid warms at half the rate of the solid under the same power, so for the same mass it needs twice as much energy per degree. Its specific heat capacity is about twice as large.
The mistake to watch for
Mistaken answer: “From 6 to 14 min the temperature is constant, so no energy is being supplied.”
The heater is still on at 300 W. The flat section tells you the energy is doing something other than raising temperature.
The correction is to say that the energy is used to break bonds between particles, so the temperature stays constant. Always link a flat section to a change of state and to the energy still going in.
Check yourself
A liquid is cooled at a steady rate. Invented readings: 90 °C at 0 min, 60 °C at 6 min, 60 °C until 15 min, then 40 °C at 20 min.
1. What is the freezing point, and how long does freezing take?
Show answer
The flat section is at 60 °C, so that is the freezing point. It lasts from 6 to 15 min, which is 15 − 6 = 9 min.
2. Find the rate of cooling of the liquid before freezing.
Show answer
(90 − 60) ÷ 6 = 30 ÷ 6 = 5 °C/min.
3. Explain why the temperature stays constant during freezing although energy is still leaving.
Show answer
As the particles form bonds, they release energy. This released energy balances the energy lost to the surroundings, so the average kinetic energy of the particles, and the temperature, stays constant until all the liquid has frozen.
Where this leads next
After this, explaining insulation from a supplied structure uses cooling curves to compare designs, and the calculations on energy for each section continue in heat calculations. Graph skills in other topics are covered in the graph-model and residual explorer.
Some students can read each point on a graph but struggle to explain what is happening to the particles. That link is exactly what our teachers build in online one-to-one Physics tuition.