PhysicsICSE Class 10 · Latent Heat

Heating Curve of Water: Stages, Flat Portions, Slopes and the Cooling Curve (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

The heating curve of water is a graph of temperature against time when ice below 0 °C is heated steadily. It rises while the ice warms, stays flat at 0 °C while the ice melts, rises again while the water warms, and stays flat at 100 °C while it boils. The flat parts show latent heat.

Understand Heating Curve of Water

The experiment. Crushed ice at, say, −20 °C is heated in a beaker by a heater of constant power, with steady stirring, and its temperature is recorded at regular time intervals. Since the heater supplies equal heat in equal times, the time axis is also a measure of the heat supplied ( $Q = P t$ ).

The stages.

Part	Temperature	State	What the heat does	Heat
AB	Rises from below 0 °C to 0 °C	Ice	Raises KE: ice warms	$m c_{i ce} Δ T$
BC	Constant at 0 °C	Ice + water	Raises PE: ice melts	$m L$ (fusion)
CD	Rises from 0 °C to 100 °C	Water	Raises KE: water warms	$m c_{w a t er} Δ T$
DE	Constant at 100 °C	Water + steam	Water boils	latent heat of vaporisation

KE = average kinetic energy of the molecules; PE = their potential energy.

Reading the flat parts. On BC and DE the temperature does not change although heat is still being supplied. This heat is latent heat: it changes the state, not the temperature. The length of a flat part (time) is proportional to the latent heat absorbed: with power $P$ , $m L = P \times t_{B C}$ . A larger mass makes BC longer; a more powerful heater makes it shorter. The melting point (0 °C) does not change with mass or heater power.

Reading the slopes. On a sloping part, $P t = m c Δ T$ , so the slope is

$\frac{Δ T}{t} = \frac{P}{m c} .$

For the same heater and mass, the slope is inversely proportional to the specific heat. Water's specific heat (4.2 J g⁻¹ °C⁻¹) is twice that of ice (2.1 J g⁻¹ °C⁻¹), so CD is less steep than AB (half the slope).

Why BC is so long. Melting 1 g of ice needs 336 J, enough to warm 1 g of water by 80 °C. So the flat part at 0 °C lasts much longer than the time to warm the ice by a few degrees.

The cooling curve. If water is cooled steadily (for example in a freezer that removes heat at a constant rate), the curve runs the other way: the temperature falls to 0 °C, stays at 0 °C while the water freezes (latent heat is released), and then falls again as the ice cools. The freezing point equals the melting point.

Using the curve in numericals. With a heater of power $P$ (all heat going into the sample):

time to warm ice: $t = \frac{m c _{i ce} Δ T}{P}$ ;
time to melt: $t = \frac{m L}{P}$ ;
time to warm water: $t = \frac{m c _{w a t er} Δ T}{P}$ .

Working backwards, a measured time on a part of the curve gives $c$ or $L$ .

Heating Curve of Water: formulas and key facts

Heat supplied in time t: $Q = P t$; Constant-power heater; time is a measure of heat.
Sloping parts: $P t = m c Δ T, slope = \frac{Δ T}{t} = \frac{P}{m c}$; Larger c, gentler slope.
Flat part at 0 °C: $P t_{B C} = m L$; Length ∝ mass × latent heat.
Values: $c_{i} ce \approx 2.1 J g^{- 1} ° C^{- 1} ∣ c_{w} a t er = 4.2 J g^{- 1} ° C^{- 1} ∣ L_{i} ce = 336 J g^{- 1}$; Slope of CD = half the slope of AB.

How to solve Heating Curve of Water questions

Treat the curve as a sequence of separate stages; each needs its own heat formula.

1Label the stages: ice warming, melting, water warming (and boiling, if shown).
2For each sloping stage, use $Q = m c Δ T$ with the correct $c$ (ice or water).
3For each flat stage, use $Q = m L$ ; there is no $Δ T$ .
4With a constant-power heater, convert each heat to time with $t = \frac{Q}{P}$ , or each time to heat with $Q = P t$ .
5For slope comparisons, use slope $= \frac{P}{m c}$ : same $P$ and $m$ , slope $\propto \frac{1}{c}$ .

Heating Curve of Water in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ice cubes in a glass

Suppose you leave a glass of ice cubes on the table and check it with a kitchen thermometer every few minutes.

What will the thermometer show while ice remains in the glass, and why?

Try it first, then show the solutionSolution

Once the ice has warmed to 0 °C, the heat flowing in from the room is used to melt the ice.
This is the flat part of the heating curve: the temperature stays at 0 °C until the last piece of ice melts.
Only then does the water start warming towards room temperature.

Answer: It stays at about 0 °C as long as any ice is left.

Memory hook: Ice in the glass, zero on the dial.

Making ice in the freezer

Imagine you put a tray of tap water in the freezer and note how long each stage takes.

Why does the water take much longer to freeze than to cool from room temperature to 0 °C?

Try it first, then show the solutionSolution

Cooling 1 g of water by 20 °C removes only $4.2 \times 20 = 84$ J.
Freezing 1 g at 0 °C removes 336 J of latent heat, four times as much.
With the freezer removing heat at a steady rate, the freezing stage lasts about four times longer.

Answer: Freezing must remove far more heat (latent heat) than cooling does.

Memory hook: The flat part is the long part.

Tricks to remember Heating Curve of Water

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slope, flat, slope, flat
Ice warms, ice melts, water warms, water boils: up, flat, up, flat.
Use it when: Drawing the curve.
Flat = phase change
Whenever the line is flat, two states are present and latent heat is at work.
Use it when: Identifying parts of the curve.
Steep for ice
Ice (c = 2.1) warms twice as fast as water (c = 4.2): AB steeper than CD.
Use it when: Slope questions.
Length means latent
Longer flat part = more mass or larger latent heat (or a weaker heater).
Use it when: Comparing curves.

Heating Curve of Water: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: The flat part at the top · 1 mark · MCQBoard level

When ice is heated steadily to steam, the temperature–time graph has a horizontal section at 100 °C. This section shows:
(a) warming of water
(b) melting of ice
(c) boiling of water at constant temperature
(d) cooling of steam

Try it first, then show the solutionSolution

At 100 °C the water changes to steam; heat is absorbed as latent heat with no rise in temperature.

Answer: (c) boiling of water at constant temperature

Example 2: Describing the curve · 3 marks · Short answerBoard level

Ice at −10 °C is heated steadily until the water reaches 50 °C. (a) Sketch or describe the temperature–time graph and name the state(s) in each part. (b) In which part is latent heat absorbed?

Try it first, then show the solutionSolution

(a) AB: rising line from −10 °C to 0 °C, ice only. BC: horizontal line at 0 °C, ice and water together. CD: rising line from 0 °C to 50 °C, water only, less steep than AB. (2)
(b) Along BC, while the ice melts at 0 °C. (1)

Answer: (a) Rise (ice), flat at 0 °C (ice + water), gentler rise (water) (b) BC

Example 3: Timing each stage · 3 marks · Short answerHarder board level

100 g of ice at −20 °C is heated by an 84 W heater; all the heat goes into the ice or water. ( $c_{i ce} = 2.1$ , $c_{w a t er} = 4.2$ J g⁻¹ °C⁻¹, $L = 336$ J g⁻¹) Find the time taken (a) to warm the ice to 0 °C, (b) to melt it, (c) to warm the water to 100 °C.

Try it first, then show the solutionSolution

(a) $Q = 100 \times 2.1 \times 20 = 4200$ J; $t = \frac{4200}{84} = 50$ s. (1)
(b) $Q = 100 \times 336 = 33600$ J; $t = \frac{33 600}{84} = 400$ s. (1)
(c) $Q = 100 \times 4.2 \times 100 = 42000$ J; $t = \frac{42 000}{84} = 500$ s. (1)

Answer: (a) 50 s (b) 400 s (c) 500 s

Example 4: Comparing the slopes · 3 marks · Short answerHarder board level

In the example above, find the rate of rise of temperature (in °C s⁻¹) on AB and on CD, and explain the difference.

Try it first, then show the solutionSolution

AB: $\frac{20}{50} = 0.4$ °C s⁻¹. (1)
CD: $\frac{100}{500} = 0.2$ °C s⁻¹. (1)
Slope $= \frac{P}{m c}$ ; water's specific heat is twice that of ice, so CD has half the slope of AB. (1)

Answer: AB 0.4 °C s⁻¹, CD 0.2 °C s⁻¹; water's c is twice that of ice

Example 5: Heat with no temperature rise · 2 marks · Short answerHarder board level

Along the flat part BC of the heating curve, heat is supplied but the temperature stays at 0 °C. Explain in terms of molecular energy.

Try it first, then show the solutionSolution

The heat is used to free the molecules from their fixed positions in the ice, increasing their potential energy. (1)
The average kinetic energy of the molecules, which decides the temperature, stays the same. (1)

Answer: Heat raises molecular potential energy, not average kinetic energy.

Example 6: Reading c and L from a curve · 3 marks · Short answerHOTS / competency

A 50 W heater is used on 200 g of ice at −10 °C. The curve shows the ice reaching 0 °C after 84 s, and the temperature then stays at 0 °C for a further 1344 s. Assuming no heat losses, find (a) the specific heat capacity of ice, (b) the specific latent heat of fusion of ice.

Try it first, then show the solutionSolution

(a) $Q = 50 \times 84 = 4200$ J; $c = \frac{4200}{200 \times 10} = 2.1$ J g⁻¹ °C⁻¹. (1½)
(b) $Q = 50 \times 1344 = 67200$ J; $L = \frac{67 200}{200} = 336$ J g⁻¹. (1½)

Answer: (a) 2.1 J g⁻¹ °C⁻¹ (b) 336 J g⁻¹

Example 7: Case study: the cooling curve · 4 marks · Case-basedHOTS / competency

250 g of water at 20 °C is placed in a freezer that removes heat at a steady 35 W, until it becomes ice at −10 °C. ( $c_{w a t er} = 4.2$ , $c_{i ce} = 2.1$ J g⁻¹ °C⁻¹, $L = 336$ J g⁻¹)
(a) Describe the shape of the temperature–time graph.
(b) Find the time for each stage.
(c) Which stage takes longest, and why?
(d) At what temperature is the flat part? Is heat absorbed or released there?

Try it first, then show the solutionSolution

(a) A falling line from 20 °C to 0 °C, a horizontal line at 0 °C, then a falling line to −10 °C (steeper, since ice has the lower specific heat). (1)
(b) Cooling water: $\frac{250 \times 4.2 \times 20}{35} = 600$ s; freezing: $\frac{250 \times 336}{35} = 2400$ s; cooling ice: $\frac{250 \times 2.1 \times 10}{35} = 150$ s. (1)
(c) Freezing (2400 s): the latent heat to be removed (84 000 J) is far more than the heat removed in either cooling stage. (1)
(d) At 0 °C; latent heat is released by the freezing water (and removed by the freezer). (1)

Answer: (a) Fall, flat at 0 °C, steeper fall (b) 600 s, 2400 s, 150 s (c) Freezing (d) 0 °C; released

Where students lose marks in Heating Curve of Water

The slips examiners see most often, and what to write instead.

Mistake: Temperature drawn rising during melting
During a change of state the temperature stays constant.
Wrong working
A continuous sloping line from −10 °C to 50 °C.
Right working
A horizontal section at 0 °C between the two sloping parts.
Do this: Mark 0 °C on the axis and draw a flat line there.
Usually costs: Reported by CISCE examiners for 2024 Q1(xi).
Mistake: Leaving out a stage
Each stage needs its own heat; the examiners advise building the equation from the heating curve.
Wrong working
Ice at −10 °C to water at 30 °C: Q = mL only.
Right working
Q = mc_ice(10) + mL + mc_water(30).
Do this: Sketch the curve and write one term for each part.
Usually costs: Reported by CISCE examiners for 2024 Q9(i).
Mistake: Wrong slope comparison
The slope is P/(mc): larger c gives a gentler slope.
Wrong working
'The water part is steeper because water has a higher specific heat.'
Right working
'The water part is less steep because water has a higher specific heat.'
Do this: Write ΔT = Pt/(mc) and read it.
Usually costs: Usually costs the mark.
Mistake: Unlabelled axes
A graph without quantities and units cannot be read.
Wrong working
A curve with no axis labels or temperature values.
Right working
Axes: temperature (°C) against time (s or min); 0 °C and 100 °C marked.
Do this: Label both axes before drawing the curve.
Usually costs: Usually costs the diagram mark.

Exam tips

Label each part of the curve with the state(s) present.
Write 'temperature remains constant' and 'latent heat is absorbed' for the flat portions.
Draw the water-warming part less steep than the ice-warming part.
In numericals, give time in seconds and convert minutes before using Q = Pt.

Practice questions on Heating Curve of Water

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Along the flat portion at 0 °C of the heating curve of water, the substance is:
(a) ice only
(b) water only
(c) a mixture of ice and water
(d) water and steam
Hint
Melting is in progress.
Answer
(c) a mixture of ice and water
Q2 · 2 marks · Short answerBoard level
On the heating curve of water, why is the part for warming water less steep than the part for warming ice?
Hint
Compare specific heats.
Answer
Water has a higher specific heat (4.2 J g⁻¹ °C⁻¹) than ice (2.1 J g⁻¹ °C⁻¹), so the same heat per second raises its temperature more slowly.
Q3 · 1 mark · Very short answerBoard level
State whether heat is absorbed or released along the flat part of (a) the heating curve, (b) the cooling curve of water at 0 °C.
Hint
Melting versus freezing.
Answer
(a) Absorbed (melting). (b) Released (freezing).
Q4 · 2 marks · Short answerHarder board level
50 g of ice at 0 °C is heated by a 42 W heater. How long does it take to melt completely? ( $L = 336$ J g⁻¹)
Hint
$t = \frac{m L}{P}$ .
Answer
$t = \frac{50 \times 336}{42} = 400$ s.
Q5 · 1 mark · Very short answerHarder board level
Continuing, how long does the same heater take to warm the 50 g of melted ice from 0 °C to 40 °C? ( $c = 4.2$ J g⁻¹ °C⁻¹)
Hint
$t = \frac{m c Δ T}{P}$ .
Answer
$t = \frac{50 \times 4.2 \times 40}{42} = 200$ s.
Q6 · 3 marks · Short answerHarder board level
The mass of ice in a heating-curve experiment is doubled, using the same heater. State the effect on (a) the length of the flat part at 0 °C, (b) the temperature of that flat part, (c) the slope of the ice-warming part.
Hint
$P t = m L$ ; slope $= \frac{P}{m c}$ .
Answer
(a) It doubles. (b) No change: still 0 °C. (c) The slope halves.
Q7 · 2 marks · Short answerHarder board level
Describe the cooling curve of water from 30 °C to ice at −5 °C.
Hint
Reverse of the heating curve.
Answer
The temperature falls from 30 °C to 0 °C; it stays constant at 0 °C while the water freezes (releasing latent heat); then it falls again, more steeply, from 0 °C to −5 °C as the ice cools.
Q8 · 1 mark · Very short answerHarder board level
A 100 W heater warms 300 g of ice from −10 °C to 0 °C. How long does this take? ( $c_{i ce} = 2.1$ J g⁻¹ °C⁻¹)
Hint
$Q = m c Δ T$ .
Answer
$t = \frac{300 \times 2.1 \times 10}{100} = 63$ s.
Q9 · 3 marks · Short answerHOTS / competency
Two students heat equal masses of ice from −10 °C, one with a 50 W heater and the other with a 100 W heater. Compare their curves: the melting temperature, the length of the flat part and the slope of the ice-warming part.
Hint
Only P changes.
Answer
The melting temperature is 0 °C for both. With 100 W the flat part lasts half as long ( $t = \frac{m L}{P}$ ) and the ice-warming part is twice as steep (slope $= \frac{P}{m c}$ ).
Q10 · 2 marks · Short answerHOTS / competency
On a heating curve made with a 56 W heater, the temperature stays at 0 °C for 5 minutes. Find the mass of ice that melted. ( $L = 336$ J g⁻¹)
Hint
Heat = Pt with t in seconds.
Answer
$Q = 56 \times 300 = 16800$ J; $m = \frac{16 800}{336} = 50$ g.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Temperature of the first flat part?Show answer
0 °C.
2.States present on the flat part at 0 °C?Show answer
Ice and water.
3.Which part is steeper: ice warming or water warming?Show answer
Ice warming.
4.Effect of a stronger heater on the melting point?Show answer
None.
5.Heat on the flat part of a cooling curve?Show answer
Released.

Heating Curve of Water: frequently asked questions

What is the heating curve of water?

It is a graph of temperature against time when ice below 0 °C is heated steadily. It shows ice warming, a flat part at 0 °C while the ice melts, water warming, and a flat part at 100 °C while the water boils.

Why is the heating curve flat at 0 °C?

The heat supplied is used as latent heat to melt the ice, increasing the potential energy of the molecules. Their average kinetic energy, and so the temperature, stays constant until all the ice has melted.

Why is the water part of the curve less steep than the ice part?

Water's specific heat capacity (4.2 J g⁻¹ °C⁻¹) is about twice that of ice (2.1 J g⁻¹ °C⁻¹), so the same heat raises its temperature about half as fast.