Heating Curve of Water: Heat Needed to Turn Ice into Steam
How much heat does it take to turn a block of ice from −10 °C into steam at 110 °C? Students often use one formula for the whole journey and get a wrong answer. The correct method is to split the journey into five steps: three heating steps where the temperature changes, and two steps (melting and boiling) where it does not. This is a favourite numerical in Class 11 thermodynamics and in the “states of matter” chapter.
Two different formulas for two different situations
When the temperature of a substance changes and its state stays the same, the heat needed is q = m × c × ΔT. When the state changes at a fixed temperature, the heat needed is q = m × L. Never use one formula in the place of the other.
State changes (fixed temperature): q = m × L
Total heat = q1 + q2 + q3 + q4 + q5
- m = mass of the sample in grams.
- c = specific heat capacity, the heat needed to raise 1 g of the substance by 1 °C. It is different for ice, liquid water and steam.
- ΔT = final temperature minus initial temperature (same size in °C and in K).
- L = latent heat per gram: Lfus for melting, Lvap for boiling.
Values used in this article
Textbooks round these values slightly differently, so always use the numbers printed in your own book or in the question. We use the following common values:
| Quantity | Value used here |
|---|---|
| c of ice | 2.09 J g−1 °C−1 |
| c of liquid water | 4.18 J g−1 °C−1 |
| c of steam | 2.01 J g−1 °C−1 |
| Latent heat of fusion of ice | 334 J g−1 (6.01 kJ mol−1) |
| Latent heat of vaporisation of water | 2260 J g−1 (about 40.7 kJ mol−1) |
Reading the heating curve
If you plot temperature against heat added, you get a curve with two flat parts. The sloping parts are the heating steps. The flat parts are melting at 0 °C and boiling at 100 °C, where all the heat goes into breaking intermolecular attractions and the temperature does not rise.
| Step | What happens | Formula |
|---|---|---|
| 1 | Ice warms from −10 to 0 °C | m cice ΔT |
| 2 | Ice melts at 0 °C | m Lfus |
| 3 | Water warms from 0 to 100 °C | m cwater ΔT |
| 4 | Water boils at 100 °C | m Lvap |
| 5 | Steam warms from 100 to 110 °C | m csteam ΔT |
Worked examples
Example 1. Find the total heat needed to convert 50 g of ice at −10 °C into steam at 110 °C.
Step 1: q1 = 50 × 2.09 × 10 = 1045 J
Step 2: q2 = 50 × 334 = 16 700 J
Step 3: q3 = 50 × 4.18 × 100 = 20 900 J
Step 4: q4 = 50 × 2260 = 113 000 J
Step 5: q5 = 50 × 2.01 × 10 = 1005 J
Total = 1045 + 16 700 + 20 900 + 113 000 + 1005 = 152 650 J ≈ 152.65 kJ.
Notice that boiling alone takes 113 000 J, which is about 74% of the total. This is why steam burns are so much more serious than hot-water burns.
Example 2. How much heat is needed to change 18 g of ice at 0 °C into water at 25 °C?
Melting: 18 × 334 = 6012 J
Warming the water: 18 × 4.18 × 25 = 1881 J
Total = 6012 + 1881 = 7893 J ≈ 7.89 kJ.
Here we start at exactly 0 °C, so steps 1, 4 and 5 do not exist. Always begin by deciding which steps your question really contains.
Example 3. 200 g of water at 40 °C is poured over a large amount of ice at 0 °C. How much ice can melt if the water cools to 0 °C? (Ignore heat lost to the surroundings.)
Heat given out by water = 200 × 4.18 × 40 = 33 440 J.
This heat melts ice: mass melted = 33 440 ÷ 334 = 100.1 g.
Check: 100.1 g × 334 J g−1 = 33 433 J, which matches the heat released (the tiny difference comes from rounding the mass to one decimal).
Cross-check using moles
Many Class 11 books give the latent heats per mole. Melting 18 g (about 1 mol, since the molar mass of water is 18.015 g mol−1) needs 6.01 kJ. Heating 1 mol of liquid water by 100 °C needs 18.015 × 4.184 × 100 ÷ 1000 ≈ 7.54 kJ. Both routes (per gram or per mole) must give the same answer if you convert units correctly.
Common mistakes
- Using q = m c ΔT for the melting or boiling step. The temperature does not change there, so ΔT = 0 would wrongly give zero heat.
- Using c of liquid water (4.18) for ice or steam. Their specific heats are about half as large.
- Forgetting a step, usually the boiling step, which is the largest term.
- Mixing joules and kilojoules, or grams and kilograms, in the same line.
- Writing ΔT as initial minus final. It is always final minus initial for the heat absorbed.
Where this appears in exams
| Topic | How a heating curve is used |
|---|---|
| Class 11 states of matter | Latent heat, change of state, intermolecular forces |
| Class 11 thermodynamics | Heat q, enthalpy of fusion and vaporisation per mole |
| Class 9–10 physics overlap | Calorimetry and the principle of mixtures |
| Board numericals | Check the current official syllabus and sample papers for the expected depth |
Do the multiplication and addition for each step carefully on a calculator, and keep every step on its own line.
Open the Chemistry Calculator Suite →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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