PhysicsISC Class 11 · Thermal Properties of Matter

Calorimetry and Latent Heat: Mixtures, Melting, Boiling and Heating Curves

Updated 17 October 2026 · Fact-checked

Heat gained or lost without change of state is Q = mcΔT; during a change of state at constant temperature it is Q = mL. The principle of calorimetry says that heat lost by the hot bodies equals heat gained by the cold ones, if no heat escapes to the surroundings.

Sloping parts: temperature rises, Q = mcΔT. Flat parts: temperature stays fixed while heat changes the state, Q = mL; boiling needs far more heat.

Illustration drawn for QuestionBuddy.

Understand Calorimetry and Latent Heat

Specific heat and heat capacity. The specific heat capacity c is the heat needed to raise the temperature of 1 kg of a substance by 1 K (unit J kg⁻¹ K⁻¹). The heat capacity of a body is C = mc (J K⁻¹). The water equivalent of a body is the mass of water with the same heat capacity: m_w = mc/c_water. Water has an unusually high specific heat (about 4200 J kg⁻¹ K⁻¹), so it takes in or gives out much heat for a small temperature change.

Principle of calorimetry. In an insulated calorimeter, heat lost = heat gained. Write each term as m c (T_hotter − T_final) or m c (T_final − T_colder), and include the calorimeter (and stirrer) through its water equivalent.

Change of state. During melting or boiling the temperature stays constant while heat is absorbed to overcome the forces between molecules. The latent heat L is the heat per unit mass for the change at constant temperature: Q = mL. Typical values used in questions: latent heat of fusion of ice 3.36 × 10⁵ J kg⁻¹, latent heat of vaporisation of water 2.26 × 10⁶ J kg⁻¹, specific heat of ice 2100 J kg⁻¹ K⁻¹ (quoted in the problem when needed).

Heating curve. On a graph of temperature against heat supplied, sloping parts represent heating of a single phase (slope inversely proportional to mc) and flat parts represent a change of state (length proportional to mL).

Method. First check what the final state could be: does all the ice melt? Compare the heat available from cooling the warm part to 0 °C with the heat needed to melt the ice. Then write the energy balance.

Calorimetry and Latent Heat: formulas and key facts

Heat exchange: $Q = m c Δ T ∣ Q = m L$; Q = mL is at constant temperature.
Heat capacity: $C = m c ∣ w a t er e q u i v a l e n t = m c / c_{w} a t er$; C in J K⁻¹.
Calorimetry: $H e a tl os t = H e a t g ain e d$; For an insulated system.
Power and time: $Q = P \times t$; For an electric heater of power P.

Calorimetry and Latent Heat 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 into a warm drink

Suppose a steel flask of negligible heat capacity holds 200 g of water at 30 °C and a 50 g ice cube at 0 °C (L = 3.36 × 10⁵ J kg⁻¹, c = 4200 J kg⁻¹ K⁻¹) is dropped in.

Find the final temperature.

Try it first, then show the solutionSolution

Heat to melt ice = 0.05 × 3.36 × 10⁵ = 16 800 J. Heat the water can give up reaching 0 °C = 0.2 × 4200 × 30 = 25 200 J, which is enough.
Balance: 0.2 × 4200 (30 − T) = 16 800 + 0.05 × 4200 × T.
25 200 − 840T = 16 800 + 210T → 8400 = 1050T → T = 8.0 °C.

Answer: 8.0 °C

Memory hook: Check first that the ice melts completely.

Tricks to remember Calorimetry and Latent Heat

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

Flat means latent
On a heating curve, a horizontal part is a change of state, using mL.
Use it when: Reading temperature-time graphs.
Check the final state first
Compare available heat with the heat to melt all the ice before assuming a final temperature above 0 °C.
Use it when: Ice and water mixture questions.

Calorimetry and Latent Heat: solved examples

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

Example 1: Unit of latent heat · 1 mark · MCQBoard level

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

Try it first, then show the solutionSolution

L = Q/m, so the unit is J kg⁻¹.

Answer: (c) J kg⁻¹

Example 2: Heating water · 3 marks · Short answerBoard level

How much heat raises 2 kg of water from 20 °C to 80 °C? How long does a 2 kW heater (all heat absorbed) take? (c = 4200 J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

Q = mcΔT = 2 × 4200 × 60. (1 mark)
Q = 5.04 × 10⁵ J. (1 mark)
t = Q/P = 5.04 × 10⁵/2000 = 252 s. (1 mark)

Answer: Q = 5.04 × 10⁵ J; t = 252 s.

Example 3: Mixing hot and cold water · 3 marks · Short answerHarder board level

200 g of water at 80 °C is mixed with 300 g of water at 20 °C in an insulated container of negligible heat capacity. Find the final temperature.

Try it first, then show the solutionSolution

Heat lost = 0.2 × 4200 (80 − T); heat gained = 0.3 × 4200 (T − 20). (1 mark)
0.2(80 − T) = 0.3(T − 20) → 16 − 0.2T = 0.3T − 6. (1 mark)
22 = 0.5T → T = 44 °C. (1 mark)

Answer: 44 °C

Example 4: Ice to steam · 4 marks · Short answerHarder board level

How much heat turns 250 g of ice at −20 °C completely into steam at 100 °C? (c_ice = 2100, c_water = 4200 J kg⁻¹ K⁻¹; L_fusion = 3.36 × 10⁵ J kg⁻¹; L_vap = 2.26 × 10⁶ J kg⁻¹)

Try it first, then show the solutionSolution

Ice −20 → 0 °C: 0.25 × 2100 × 20 = 10 500 J. (1 mark)
Melting: 0.25 × 3.36 × 10⁵ = 84 000 J. Water 0 → 100 °C: 0.25 × 4200 × 100 = 105 000 J. (1 mark)
Vaporisation: 0.25 × 2.26 × 10⁶ = 565 000 J. (1 mark)
Total = 10 500 + 84 000 + 105 000 + 565 000 = 7.65 × 10⁵ J. (1 mark)

Answer: 7.65 × 10⁵ J (764 500 J).

Example 5: Calorimeter with ice · 5 marks · Long answerHOTS / competency

A copper calorimeter of mass 100 g (c = 400 J kg⁻¹ K⁻¹) contains 200 g of water at 30 °C. A 50 g piece of ice at 0 °C is added. Find the final temperature (L = 3.36 × 10⁵ J kg⁻¹).

Try it first, then show the solutionSolution

Heat capacity of water + calorimeter = 0.2 × 4200 + 0.1 × 400 = 880 J K⁻¹. (1 mark)
Heat available in cooling to 0 °C = 880 × 30 = 26 400 J > 16 800 J needed to melt the ice, so all the ice melts. (1 mark)
880(30 − T) = 16 800 + 0.05 × 4200 × T = 16 800 + 210T. (1 mark)
26 400 − 16 800 = 1090T. (1 mark)
T = 9600/1090 = 8.8 °C. (1 mark)

Answer: T ≈ 8.8 °C

Where students lose marks in Calorimetry and Latent Heat

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

Mistake: Forgetting the calorimeter
Leaving out the heat taken by the vessel.
Do this: Add the calorimeter through its water equivalent or m c.
Usually costs: 1 to 2 marks.
Mistake: Using mcΔT for a change of state
Computing melting heat from a temperature change.
Do this: Use Q = mL at constant temperature.
Usually costs: 2 marks.
Mistake: Assuming all the ice melts
Taking the final temperature above 0 °C without checking.
Do this: Compare heat available with mL first.
Usually costs: 1 to 2 marks.

Exam tips

Draw a timeline of the stages: heating, melting, heating, boiling.
Quote the values of L and c used in the question.
Write the energy balance in words first: heat lost = heat gained.
Give temperatures in °C and heat in J (or kJ).

Practice questions on Calorimetry and Latent Heat

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
Melting of ice at 0 °C involves:
(a) a rise of temperature
(b) absorption of heat at constant temperature
(c) release of heat
(d) a fall of temperature
Hint
Latent heat is absorbed during the change of state.
Answer
(b) absorption of heat at constant temperature
Q2 · 2 marks · Short answerBoard level
A 1 kW heater is used to heat 3 kg of water by 10 °C (c = 4200 J kg⁻¹ K⁻¹). Find the time taken, assuming no loss.
Hint
Q = mcΔT = Pt.
Answer
t = 3 × 4200 × 10/1000 = 126 s.
Q3 · 4 marks · Short answerHarder board level
20 g of steam at 100 °C is passed into 400 g of water at 20 °C in a calorimeter of negligible heat capacity. Find the final temperature (L_vap = 2.26 × 10⁶ J kg⁻¹, c = 4200 J kg⁻¹ K⁻¹).
Hint
Steam condenses, then the condensed water cools to T.
Answer
0.02[2.26 × 10⁶ + 4200(100 − T)] = 0.4 × 4200 (T − 20) → 45 200 + 8400 − 84T = 1680T − 33 600 → 87 200 = 1764T → T = 49.4 °C.
Q4 · 3 marks · Short answerHarder board level
How much ice at 0 °C can 1 kg of water at 10 °C just melt (L = 3.36 × 10⁵ J kg⁻¹)? Assume the final temperature is 0 °C.
Hint
Heat given by water cooling to 0 °C melts the ice.
Answer
Q = 1 × 4200 × 10 = 42 000 J; m = 42 000/3.36 × 10⁵ = 0.125 kg.
Q5 · 4 marks · Short answerHarder board level
A 5 kW source heats 0.5 kg of ice at −20 °C. Find the time to bring it to water at 100 °C, giving the time for each stage (c_ice = 2100, c_water = 4200 J kg⁻¹ K⁻¹; L = 3.36 × 10⁵ J kg⁻¹).
Hint
Time = Q/P for each stage.
Answer
Ice to 0 °C: 21 000 J, 4.2 s. Melting: 168 000 J, 33.6 s. Water to 100 °C: 210 000 J, 42 s. Total 79.8 s.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student drops a 150 g metal block at 100 °C into 200 g of water at 25 °C in a calorimeter of water equivalent 20 g; the final temperature is 30 °C (c_water = 4200 J kg⁻¹ K⁻¹).
(i) Find the heat gained by the water and calorimeter. (1 mark)
(ii) Find the specific heat of the metal. (1 mark)
(iii) If 5% of the heat lost by the block escaped to the room, find the true specific heat. (1 mark)
(iv) By what percentage was the measured value too low? (1 mark)
Hint
Gain = (0.2 + 0.02) × 4200 × 5.
Answer
(i) 0.22 × 4200 × 5 = 4620 J. (ii) c = 4620/(0.15 × 70) = 440 J kg⁻¹ K⁻¹. (iii) Heat lost by the block = 4620/0.95 = 4863 J, so c = 4863/10.5 = 463 J kg⁻¹ K⁻¹. (iv) (463 − 440)/463 = 5.0%.

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.Define specific heat capacity.Show answer
Heat needed to raise 1 kg of a substance by 1 K.
2.What is water equivalent?Show answer
The mass of water having the same heat capacity as the body.
3.What happens to the temperature during a change of state?Show answer
It stays constant.
4.State the principle of calorimetry.Show answer
Heat lost by hot bodies equals heat gained by cold bodies.

Calorimetry and Latent Heat: frequently asked questions

Where does the heat go while ice at 0 °C is melting?

The heat supplied goes into overcoming the bonds between molecules (increasing potential energy), not into raising their average kinetic energy.

Why is steam at 100 °C more dangerous than water at 100 °C?

Steam also carries the latent heat of vaporisation, which it releases on condensing on the skin.

Can the specific heat of a substance change with temperature?

Slightly. In ISC questions it is taken as constant over the range.