Latent Heat ICSE Class 10: Change of State, Heating Curve and Latent Heat of Fusion

What Latent Heat is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Latent Heat is in the 5. Heat (ii) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Change of state at constant temperature

A substance changes state at a fixed temperature for a given pressure: ice melts (and water freezes) at 0 °C; water boils at 100 °C at normal pressure.

During the change, heat is absorbed (melting, boiling) or released (freezing, condensing), but the temperature stays constant. This heat changes the potential energy of the molecules (breaking or forming the bonds that hold them in place); the average kinetic energy, and so the temperature, does not change.

Worked example

During the melting of ice at 0 °C, is heat absorbed or released? What happens to the temperature and to the average kinetic energy of the molecules?

Try it first, then show the solutionSolution

1. Heat is absorbed to break the rigid arrangement of molecules.
2. The temperature stays at 0 °C, so the average kinetic energy is unchanged; the potential energy increases.

Answer: Absorbed; temperature and average KE constant

Latent heat and specific latent heat of fusion

Latent heat is the heat absorbed or released during a change of state at constant temperature.

The specific latent heat of fusion $L$ of a solid is the heat needed to change unit mass of it from solid to liquid at its melting point, without change in temperature:
$Q=mL$.
SI unit: J kg⁻¹ (also J g⁻¹). For ice, $L=336$ J g⁻¹ $=3.36\times 10^5$ J kg⁻¹ (about 80 cal g⁻¹).

The same amount of heat is released when the same mass of liquid freezes. Heat during a change of state depends on the mass (and the substance), not on temperature change or specific heat.

Worked example

How much heat is needed to change 200 g of ice at 0 °C into water at 0 °C? ($L=336$ J g⁻¹)

Try it first, then show the solutionSolution

1. $Q=mL=200\times 336$.
2. $Q=67200$ J.

Answer: 67 200 J

The heating curve of water

If ice at, say, −10 °C is heated steadily, a graph of temperature against time shows:
1. AB: temperature of ice rises from −10 °C to 0 °C ($m{c}_{\text{ice}}\Delta T$).
2. BC: flat at 0 °C while ice melts ($m{L}_{\text{fusion}}$); ice and water coexist.
3. CD: temperature of water rises from 0 °C to 100 °C ($m{c}_{\text{water}}\Delta T$).
4. DE: flat at 100 °C while water boils.

The flat parts show latent heat. The slope of CD is less steep than AB because water's specific heat (4.2 J g⁻¹ °C⁻¹) is about twice that of ice (2.1 J g⁻¹ °C⁻¹).

Worked example

A 500 W heater supplies heat steadily to 200 g of ice at 0 °C. How long is the flat part of the heating curve at 0 °C (no losses)?

Try it first, then show the solutionSolution

1. Heat to melt $=200\times 336=67200$ J.
2. Time $=\frac{67200}{500}=134.4$ s.

Answer: About 134 s

Heating ice into warm water

Add the heat for each stage separately:

• warm the ice to 0 °C: $m{c}_{\text{ice}}(0-t_1)$;
• melt it: $mL$;
• warm the melted water: $m{c}_{\text{water}}(t_2-0)$.

The melted ice must also be warmed: do not leave out this term.

Worked example

Find the heat needed to change 50 g of ice at −10 °C into water at 20 °C. ($c_{\text{ice}}=2.1$, $c_{\text{water}}=4.2$ J g⁻¹ °C⁻¹, $L=336$ J g⁻¹)

Try it first, then show the solutionSolution

1. Warm ice: $50\times 2.1\times 10=1050$ J.
2. Melt: $50\times 336=16800$ J.
3. Warm water: $50\times 4.2\times 20=4200$ J.
4. Total $=22050$ J.

Answer: 22 050 J

Mixtures of ice and water

Use the principle of method of mixtures:
heat lost by the warm water = heat gained by the ice (to melt) + heat gained by the melted ice (to warm up).
$m_w{c}_w(t_w-t)=m_{i}L+m_i{c}_w(t-0)$ for ice at 0 °C.

Check first whether there is enough heat to melt all the ice: if the warm water would have to cool below 0 °C, only part of the ice melts and the final temperature is 0 °C.

Worked example

What mass of ice at 0 °C is needed to cool 150 g of water from 40 °C to 10 °C?

Try it first, then show the solutionSolution

1. Heat lost by water $=150\times 4.2\times 30=18900$ J.
2. Heat gained by $m$ g of ice $=m\times 336+m\times 4.2\times 10=378m$.
3. $378m=18900$, so $m=50$ g.

Answer: 50 g

Everyday effects of latent heat of fusion

• Ice cools better than ice-cold water: 1 g of ice at 0 °C absorbs 336 J just to melt, before it starts warming; water at 0 °C does not. So water at 0 °C holds more heat than ice at 0 °C.
• Freezing lakes warm the weather: as water freezes, it releases a large amount of latent heat to the surroundings, so the air feels pleasantly warm.
• Snow on mountains melts slowly, keeping streams cool and releasing water gradually.
• Ice-cream and cold drinks are kept cold by ice or ice–salt mixtures.
• Frozen food stays cold because thawing absorbs a lot of heat.

Worked example

Which contains more heat energy: 1 g of ice at 0 °C or 1 g of water at 0 °C? By how much?

Try it first, then show the solutionSolution

1. Water at 0 °C has absorbed latent heat to melt from ice.
2. It has 336 J more than 1 g of ice at 0 °C.

Answer: Water at 0 °C, by 336 J

Physics in real life: Latent Heat

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.

Topics in this chapter

• Specific Latent Heat of Fusion
• Heating Curve of Water
• Mixtures of Ice and Water

Latent Heat formula sheet

Latent heat

$Q=mL$

No temperature change during melting.

Specific latent heat of fusion of ice

$L=336{\text{J g}}^{-1}=3.36\times 10^5{\text{J kg}}^{-1}$

About 80 cal g⁻¹.

Ice at t₁ (< 0 °C) to water at t₂

$Q=m{c}_{\text{ice}}(0-t_1)+mL+m{c}_{\text{water}}{t}_2$

c_ice ≈ 2.1, c_water = 4.2 J g⁻¹ °C⁻¹.

Ice at 0 °C in warm water

$m_w{c}_w(t_w-t)=m_{i}L+m_i{c}_{w}t$

Heat lost = heat gained.

Heat in 1 g of water vs ice at 0 °C

$Q_{\text{water}}-Q_{\text{ice}}=336\text{J}$

Water at 0 °C holds more heat.

Latent Heat: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Latent Heat

The question patterns to expect, and the method that works for each.

Tricks to remember Latent Heat

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

• Flat line = latent

  On a heating curve, flat lines are where latent heat goes in.

  Use it when: Reading heating curves.

• 336 = 80 × 4.2

  L of ice is 80 cal/g = 336 J/g; 4.2 is a factor of 336, which simplifies sums.

  Use it when: Quick arithmetic in mixtures.

• Melt, then warm

  Ice in water: first mL to melt, then mcΔT to warm the meltwater.

  Use it when: Avoiding the missing term.

• Same temperature, more heat

  Water at 0 °C has 336 J/g more than ice at 0 °C.

  Use it when: Comparing ice and water.

Latent Heat: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Latent Heat

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

• ✗Mistake: Forgetting to warm the melted ice

  CISCE examiners' comments, 2024: in finding the mass of ice to cool water to 5 °C, candidates took only mL for melting and left out mcΔT for warming the melted ice water to 5 °C; many did not attempt it.

  ✓Do this: Heat gained by ice = mL + mc(t − 0). Build the equation from the heating curve.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Thinking temperature changes during melting

  CISCE examiners' comments, 2024: candidates reasoned that absorbing heat must raise temperature, and were unsure whether ice at 0 °C absorbs or releases heat while melting.

  ✓Do this: During a change of state heat changes potential energy only; temperature stays constant.

  Usually costs: 1 mark.

• ✗Mistake: Ice holds more heat than water at 0 °C

  CISCE examiners' comments, 2024: some candidates said ice at 0 °C contains more heat, mixing up temperature and heat; others said average KE increases or decreases on freezing, and confused heat released on freezing with heat absorbed on melting.

  ✓Do this: Water at 0 °C has 336 J g⁻¹ more than ice; average KE is constant during freezing; freezing releases latent heat.

  Usually costs: 1 mark each.

• ✗Mistake: Wrong units and wrong data

  CISCE examiners' comments, 2024 and 2025: candidates converted mass to kg but kept L in J g⁻¹, used 336 J g⁻¹ when 330 J g⁻¹ was given, omitted units, or converted only one quantity.

  ✓Do this: Use the values given in the question and one unit system throughout; write J.

  Usually costs: 1 mark.

• ✗Mistake: Latent heat linked to specific heat

  CISCE examiners' comments, 2025: asked what heat absorbed during a change of phase depends on, candidates chose specific heat capacity instead of mass.

  ✓Do this: Q = mL: depends on mass (and the substance), not on specific heat or temperature change.

  Usually costs: 1 mark.

Practice questions on Latent Heat

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

1. Q1 · 1 mark · MCQBoard level

   The SI unit of specific latent heat is:
   (a) J
   (b) J kg⁻¹
   (c) J kg⁻¹ K⁻¹
   (d) J K⁻¹

   Hint

   Per kg, no temperature.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The heat absorbed during the melting of a solid at its melting point depends on its:
   (a) mass
   (b) change in temperature
   (c) specific heat capacity
   (d) initial temperature

   Hint

   Q = mL.

   Answer

   (a)

3. Q3 · 1 mark · MCQHarder board level

   When water at 0 °C freezes into ice at 0 °C, the average kinetic energy of its molecules:
   (a) increases
   (b) decreases
   (c) remains the same
   (d) becomes zero

   Hint

   Temperature is constant.

   Answer

   (c)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): The weather becomes pleasantly warm when lakes begin to freeze.
   Reason (R): Water releases its latent heat of fusion to the surroundings as it freezes.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Freezing releases heat.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   Find the heat absorbed when 300 g of a wax melts at its melting point, if its specific latent heat of fusion is 150 J g⁻¹.

   Hint

   Q = mL.

   Answer

   $45000$ J

6. Q6 · 2 marks · Very short answerHarder board level

   A 1000 W heater is placed in a large block of ice at 0 °C. How much ice does it melt in 2 minutes, if all its heat goes into the ice? ($L=336$ J g⁻¹)

   Hint

   Heat = Pt; mass = heat ÷ L.

   Answer

   Heat $=1000\times 120=120000$ J; mass $=\frac{120000}{336}\approx 357$ g.

7. Q7 · 3 marks · Long answerBoard level

   (a) Define specific latent heat of fusion. (b) Explain, in terms of molecular energy, why temperature does not rise while ice melts.

   Hint

   PE vs KE.

   Answer

   (a) Heat needed to change unit mass of a solid into liquid at its melting point without change in temperature. (b) The heat is used to increase the potential energy of molecules (breaking the solid structure); the average kinetic energy, and hence the temperature, stays the same.

8. Q8 · 3 marks · Long answerHarder board level

   Find the heat needed to change 20 g of ice at −20 °C into water at 50 °C. ($c_{\text{ice}}=2.1$, $c_{\text{water}}=4.2$ J g⁻¹ °C⁻¹, $L=336$ J g⁻¹)

   Hint

   Three stages.

   Answer

   $20\times 2.1\times 20+20\times 336+20\times 4.2\times 50=840+6720+4200=11760$ J

9. Q9 · 3 marks · Long answerHOTS / competency

   100 g of ice at 0 °C is added to 100 g of water at 20 °C in an insulated flask. (a) Show that not all the ice melts. (b) Find the mass of ice that melts and the final temperature.

   Hint

   Compare heat available with heat needed.

   Answer

   (a) Water can give 8400 J cooling to 0 °C; melting all the ice needs 33 600 J. (b) $\frac{8400}{336}=25$ g melts; final temperature 0 °C (75 g ice remains).

10. Q10 · 3 marks · Long answerHarder board level

    What mass of ice at 0 °C must be added to 252 g of water at 40 °C to cool it to 0 °C, with all the ice just melting?

    Hint

    mL = heat lost by water.

    Answer

    Heat lost $=252\times 4.2\times 40=42336$ J; $m=\frac{42336}{336}=126$ g.

11. Q11 · 1 mark · MCQHOTS / competency

    On the heating curve of water, the flat portion at 0 °C represents:
    (a) warming of ice
    (b) melting of ice at constant temperature
    (c) warming of water
    (d) boiling of water

    Hint

    Flat = change of state.

    Answer

    (b)

12. Q12 · 4 marks · Long answerHOTS / competency

    50 g of ice at 0 °C is dropped into 250 g of water at 60 °C. (a) Show that all the ice melts. (b) Find the final temperature. (c) If the container were a copper calorimeter at 60 °C, would the final temperature be higher or lower? Explain.

    Hint

    Heat lost = mL + mcΔT.

    Answer

    (a) Melting needs $50\times 336=16800$ J; the water can give $250\times 4.2\times 60=63000$ J in cooling to 0 °C. (b) $250\times 4.2\times (60-t)=16800+50\times 4.2\times t$, so $63000-1050t=16800+210t$, $t\approx 36.7$ °C. (c) Higher: the warm calorimeter would also give heat to the ice.

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. 1.Define latent heat.Show answer

   Heat absorbed or released during a change of state at constant temperature.

2. 2.Specific latent heat of fusion of ice?Show answer

   336 J g⁻¹ (3.36 × 10⁵ J kg⁻¹).

3. 3.Formula for latent heat?Show answer

   Q = mL

4. 4.What stays constant during melting?Show answer

   Temperature (average kinetic energy).

5. 5.What changes during melting?Show answer

   Potential energy of the molecules.

6. 6.Which has more heat: ice or water at 0 °C?Show answer

   Water, by 336 J per gram.

7. 7.What do the flat parts of a heating curve show?Show answer

   Changes of state (latent heat).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why temperature is constant while ice melts?
• ☐Can you draw and label the heating curve?
• ☐Do you add mL and mcΔT for melted ice?
• ☐Do you check whether all the ice melts?
• ☐Can you explain freezing lakes and ice-cold drinks?

Latent Heat: frequently asked questions