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A heating curve packs a lot of physics into a few straight lines. Exam questions hand you one and ask for a latent heat, a specific heat or a ratio, and the numbers are all in the slopes and the flat parts. This post shows how to read each part, works through an example with ice, and lists the misreadings that cost marks.
What does each part of a heating curve mean?
The NCERT Class 11 Physics chapter on thermal properties of matter describes heating ice from −10 °C: the temperature rises to 0 °C, then stays constant while the ice melts, then rises again once all the ice has melted. The same happens at the boiling point. NCERT also notes that the slopes of the lines differ because the specific heats of the solid, liquid and gas are not equal.
If heat is supplied at a constant power P to a mass m:
Sloping part: P = mc × (rate of temperature rise). So the slope equals P/(mc). A steeper line means a smaller heat capacity mc, not a larger one.
Flat part: temperature is constant while the state changes. The heat taken is mL, so the length of the flat part in time is mL/P.
Change of state and latent heat are in both the JEE (Main) 2026 syllabus and the CBSE Physics curriculum for 2026-27, which also lists a Class XI activity to "observe change of state and plot a cooling curve for molten wax".
Worked example: heating ice to steam
These are sample numbers for practice. A 0.20 kg block of ice at −20 °C is heated by a heater that delivers a steady 420 W, all of it absorbed. Use NCERT's values: specific heat of ice 2060 J/(kg K), water 4186 J/(kg K), latent heat of fusion 3.33 × 10⁵ J/kg and of vaporisation 22.6 × 10⁵ J/kg.
Ice from −20 °C to 0 °C: Q = 0.20 × 2060 × 20 = 8240 J. Time = 8240/420 ≈ 19.6 s. Slope = 420/(0.20 × 2060) ≈ 1.02 °C per second.
Melting at 0 °C: Q = 0.20 × 3.33 × 10⁵ = 66,600 J. Time ≈ 158.6 s. Flat.
Water from 0 °C to 100 °C: Q = 0.20 × 4186 × 100 = 83,720 J. Time ≈ 199.3 s. Slope ≈ 0.50 °C per second.
Boiling at 100 °C: Q = 0.20 × 22.6 × 10⁵ = 452,000 J. Time ≈ 1076 s. Flat.
Three things to read off. The ice line is about twice as steep as the water line, because 4186/2060 ≈ 2.03. The boiling plateau is about 6.8 times longer than the melting plateau, because 22.6/3.33 ≈ 6.79. And the longest stretch by far is boiling, which matches NCERT's point that steam at 100 °C carries much more heat than water at 100 °C.
How do you find latent heat from a graph alone?
Often you do not know the power or the mass. Use the slope of a sloping part to remove them.
Sample graph: a solid is heated at a steady rate. It melts at 80 °C and the flat part lasts 120 s. Just after melting, the liquid warms at 0.5 °C per second. The liquid's specific heat is 2000 J/(kg K).
From the liquid line, P = m × 2000 × 0.5, so P/m = 1000 W/kg. From the flat part, L = P × 120/m = 1000 × 120 = 1.2 × 10⁵ J/kg.
The mass and power cancel. In one line: L = c × (slope of the liquid line) × (length of the flat part).
Which misreadings cost marks?
"Flat means no heat is being absorbed." Heat is still flowing in; it goes into changing the state, not raising the temperature.
Steeper means larger specific heat. It is the reverse at constant power.
Mixing axes. On a temperature-heat graph, the flat lengths are mL directly. On a temperature-time graph they are mL/P.
Reading a cooling curve as a heating curve. On cooling, the flat part is where latent heat is released.
These are classic graph errors. McDermott, Rosenquist and van Zee (1987) described two kinds of difficulty with physics graphs: connecting graphs to physical concepts and connecting them to the real world. Beichner (1994), testing 895 high school and college students on kinematics graphs, listed slope-height confusion and reading graphs as pictures among the common errors. Both show up on heating curves. Our post on reading physics graphs gives a general method, and the calorimetry post covers mixing problems.
When a slope or a plateau does not make sense, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Sketch the ice-to-steam curve from memory, labelling each part with "P/(mc)" or "mL/P".
Redo the worked example with a different heater power and check that only the time axis changes, not the shape.
Solve two "find L from the graph" problems with made-up numbers, then swap: find c from a known L.
If your school does the molten wax activity, sketch the cooling curve you expect before you record it.
Draw the sample graph above on squared paper tonight and check that you get 1.2 × 10⁵ J/kg.
