PhysicsCBSE Class 11 · Thermal Properties of Matter

Calorimetry, Specific Heat and Latent Heat Class 11

Updated 17 October 2026 · Fact-checked

Heat Q = mcΔT changes the temperature of a substance of specific heat c; during a change of state, heat Q = mL changes the state at constant temperature (L is the latent heat). Calorimetry rests on the principle of mixtures: heat lost by the hot bodies equals heat gained by the cold ones in an isolated system. Energy from a heater of power P in time t is Pt.

Sloped parts use Q = mcΔT as temperature rises; the flat orange parts use Q = mL, where heat changes state and temperature stays constant.

Illustration drawn for QuestionBuddy.

Understand Calorimetry, Specific Heat and Latent Heat

Heat and temperature. Heat is energy in transit because of a temperature difference (SI unit joule); temperature measures the degree of hotness. The 1 calorie is about 4.186 J.

Specific heat capacity. The heat needed to change the temperature of mass $m$ by $Δ T$ is
$ $Q = m c Δ T = n C Δ T,$ $
where $c$ is the specific heat (J kg⁻¹ K⁻¹) and $C$ the molar heat capacity. The heat capacity of a body is $m c$ . Water has a large specific heat (about 4200 J kg⁻¹ K⁻¹), which is why it is a good coolant and why coastal climates are mild.

Change of state. At the melting or boiling point, heat produces a change of state with no change in temperature: $Q = m L$ . $L_{f}$ is the latent heat of fusion and $L_{v}$ the latent heat of vaporisation. For water (round figures used here): $L_{f} \approx 3.35 \times 1 0^{5}$ J/kg, $L_{v} \approx 2.26 \times 1 0^{6}$ J/kg, $c_{i ce} \approx 2100$ J kg⁻¹ K⁻¹. The latent heat is the energy that changes the molecular arrangement, not the molecular kinetic energy.

Calorimetry (principle of mixtures). In an isolated system, heat lost $=$ heat gained. If a hot body, cold water and a calorimeter exchange heat, include the calorimeter's heat capacity. For a mixture involving a change of state, first check by energy accounting whether all of the ice melts (or all of the steam condenses) before writing the final temperature.

Heating curve. A body heated at constant power $P$ shows stages: warming the solid ( $m c_{s} Δ T$ ), melting at constant temperature ( $m L_{f}$ ), warming the liquid, boiling at constant temperature ( $m L_{v}$ ) and warming the vapour. Each stage takes $t = Q / P$ .

Heaters. With efficiency $η$ , the useful energy is $η P t$ .

Calorimetry, Specific Heat and Latent Heat: formulas and key facts

Heat for temperature change: $Q = m c Δ T$; $c$ in J kg⁻¹ K⁻¹.
Heat for change of state: $Q = m L$; Constant temperature.
Principle of mixtures: $Q_{l os t} = Q_{g ain e d}$; Isolated system.
Heat capacity: $C = m c$; J/K.
Heater energy: $E = η P t$; Efficiency $η$ .
Final temperature (no state change): $T = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}$; Weighted average.

How to solve Calorimetry, Specific Heat and Latent Heat questions

Write down every heat exchange with its sign, check which phase changes are complete, then equate heat lost to heat gained.

1List every body with m, c, and the starting temperature.
2Compute the heat needed to bring the cold body to the melting or boiling point, and to complete the phase change.
3Compare with the heat the hot body can give up; decide whether the change of state is complete.
4If it is, write Q_lost = Q_gained with the final temperature as the unknown.
5If it is not, the final temperature is the melting or boiling point, and the unmelted mass follows from the leftover heat.
6Include the calorimeter (m c) if given.

Calorimetry, Specific Heat 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.

Cooling a drink with ice

Suppose 200 g of a drink (treat as water, c = 4200 J/kg K) at 25 °C is cooled by adding ice at 0 °C until the drink reaches 5 °C; treat the melted ice as warming to 5 °C (L_f = 3.35 × 10⁵ J/kg).

Find the mass of ice needed.

Try it first, then show the solutionSolution

Heat given by the drink = 0.2 × 4200 × 20 = 16800 J.
Per kg of ice: L_f + c(5 − 0) = 3.35 × 10⁵ + 4200 × 5 = 3.56 × 10⁵ J.
m = 16800/3.56 × 10⁵ = 0.047 kg.

Answer: About 47 g.

Memory hook: Melting needs far more heat than warming.

Bathwater from a geyser

Suppose a geyser heats water from 20 °C to 45 °C using a 3 kW element at 100% efficiency.

How long does it take for 30 kg of water (c = 4200 J/kg K)?

Try it first, then show the solutionSolution

Q = 30 × 4200 × 25 = 3.15 × 10⁶ J.
t = Q/P = 3.15 × 10⁶/3000 = 1050 s = 17.5 min.

Answer: 17.5 minutes.

Memory hook: Energy equals power times time.

Tricks to remember Calorimetry, Specific Heat and Latent Heat

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

Check before you balance
See if all the ice melts before writing the final T.
Use it when: Ice-water mixtures.
Plateaus
Temperature stays constant while the state changes.
Use it when: Heating curves.
mc is the heat capacity
Calorimeter counts as an extra mc.
Use it when: Mixing problems.
Steam burns more
Steam gives up L_v on condensing.
Use it when: Latent heat reasoning.

Calorimetry, Specific Heat and Latent Heat: solved examples

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

Example 1: Heating oil · 2 marks · Very short answerBoard level

How much heat is needed to raise the temperature of 0.75 kg of oil (c = 2000 J/kg K) from 25 °C to 55 °C? How long would a 500 W immersion heater take, if all its energy goes into the oil?

Try it first, then show the solutionSolution

Q = mcΔT = 0.75 × 2000 × 30 = 4.5 × 10⁴ J. (1)
t = Q/P = 45000/500 = 90 s. (1)

Answer: 4.5 × 10⁴ J; 90 s.

Example 2: Hot metal in water · 3 marks · Short answerHarder board level

A 0.30 kg copper block (c = 390 J/kg K) at 100 °C is dropped into 0.50 kg of water at 20 °C in an insulated container of negligible heat capacity. Find the final temperature (c of water = 4200 J/kg K).

Try it first, then show the solutionSolution

Heat lost = heat gained: 0.3 × 390 × (100 − T) = 0.5 × 4200 × (T − 20). (1)
117(100 − T) = 2100(T − 20), so 11700 + 42000 = 2217T. (1)
T = 53700/2217 ≈ 24.2 °C. (1)

Answer: About 24.2 °C.

Example 3: Ice at −10 °C in warm water · 3 marks · Short answerHarder board level

100 g of ice at −10 °C is added to 300 g of water at 40 °C. Find the final temperature (c_ice = 2100, c_water = 4200 J/kg K, L_f = 3.35 × 10⁵ J/kg).

Try it first, then show the solutionSolution

Heat to bring the ice to 0 °C and melt it: 0.1 × 2100 × 10 + 0.1 × 3.35 × 10⁵ = 2100 + 33500 = 35600 J. (1)
Heat the water can give up in cooling to 0 °C: 0.3 × 4200 × 40 = 50400 J > 35600 J, so all the ice melts. (1)
Remaining 14800 J warms 0.4 kg of water: ΔT = 14800/(0.4 × 4200) = 8.8 °C. (1)

Answer: About 8.8 °C.

Example 4: Kettle with boil-off · 4 marks · Long answerHarder board level

A 1 kW kettle is 80% efficient. It heats 1.5 kg of water from 25 °C to 100 °C and then boils away 0.1 kg of it (c = 4200 J/kg K, L_v = 2.26 × 10⁶ J/kg). Find the energy needed and the time taken.

Try it first, then show the solutionSolution

Heating: 1.5 × 4200 × 75 = 4.725 × 10⁵ J. (1)
Boiling: 0.1 × 2.26 × 10⁶ = 2.26 × 10⁵ J. (1)
Total = 6.985 × 10⁵ J; useful power = 0.8 × 1000 = 800 W. (1)
t = 6.985 × 10⁵/800 ≈ 873 s ≈ 14.6 min. (1)

Answer: 6.99 × 10⁵ J; about 14.6 min.

Example 5: Heating curve of ice · 4 marks · Long answerHOTS / competency

0.5 kg of ice at −20 °C is heated at a steady rate of 1000 W with no losses (c_ice = 2100, c_water = 4200 J/kg K, L_f = 3.35 × 10⁵ J/kg). (a) Find the time at which melting starts and ends. (b) Describe the contents at t = 100 s. (c) Find the temperature at t = 300 s.

Try it first, then show the solutionSolution

(a) Warm the ice: 0.5 × 2100 × 20 = 21000 J, 21 s. Melting: 0.5 × 3.35 × 10⁵ = 167500 J, 167.5 s; melting ends at 188.5 s. (1)
(b) At 100 s, 79 s of melting has taken place: mass melted = 79000/3.35 × 10⁵ = 0.236 kg. (1)
The beaker holds 0.236 kg water and 0.264 kg ice at 0 °C. (1)
(c) At 300 s, 111.5 s of warming water: ΔT = 111500/(0.5 × 4200) = 53 °C, so the temperature is 53 °C. (1)

Answer: (a) 21 s to 188.5 s; (b) 0.236 kg water + 0.264 kg ice at 0 °C; (c) about 53 °C.

Where students lose marks in Calorimetry, Specific Heat and Latent Heat

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

Mistake: Forgetting the calorimeter's heat capacity.
Treating only water as the cold body.
Do this: Add the calorimeter's mc to the cold side.
Usually costs: 1 to 2 marks.
Mistake: Writing Q = mcΔT during a phase change.
Not noticing the constant temperature.
Do this: Use Q = mL during the change of state.
Usually costs: Method marks.
Mistake: Assuming all ice melts without checking.
Writing the equation immediately.
Do this: Compare the available heat with the heat to melt first.
Usually costs: The answer mark.
Mistake: Using grams with J/kg K.
Unit mix.
Do this: Convert mass to kg first.
Usually costs: The answer mark.

Exam tips

List all heat terms with signs before solving.
State "no heat is lost to the surroundings" as an assumption.
For ice questions show the check on whether the ice melts completely.
Write the final temperature with units (°C).
Use consistent units for mass (kg) and c.

Practice questions on Calorimetry, Specific Heat 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
The heat required to change 1 kg of ice at 0 °C into water at 0 °C is called the
(a) specific heat of ice (b) latent heat of fusion (c) latent heat of vaporisation (d) heat capacity
Hint
Same temperature, change of state.
Answer
(b).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Steam at 100 °C causes a more severe burn than boiling water at 100 °C.
Reason (R): Steam releases its latent heat of vaporisation when it condenses on the skin.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Extra energy per kg.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
4.2 kJ of heat raises the temperature of 100 g of a substance by 20 K. Find its specific heat.
Hint
c = Q/(mΔT).
Answer
c = 4200/(0.1 × 20) = 2100 J kg⁻¹ K⁻¹.
Q4 · 3 marks · Short answerHarder board level
50 g of ice at 0 °C is added to 200 g of water at 30 °C. Find the final temperature (L_f = 3.35 × 10⁵ J/kg, c_water = 4200 J/kg K).
Hint
Check if all the ice melts.
Answer
Water can give 0.2 × 4200 × 30 = 25200 J; melting needs 0.05 × 3.35 × 10⁵ = 16750 J, so all melts. Remaining 8450 J warms 0.25 kg: ΔT = 8450/1050 ≈ 8.0 °C.
Q5 · 4 marks · Long answerHOTS / competency
A 0.5 kg block of unknown metal at 100 °C is dropped into 0.4 kg of water at 20 °C held in a 0.2 kg aluminium calorimeter (c_Al = 900 J/kg K). The final temperature is 30 °C. (a) Find the specific heat of the metal. (b) If 5% of the heat lost by the block was lost to the surroundings, what is the corrected value? (c) Why is the uncorrected value too low?
Hint
Include the calorimeter.
Answer
(a) Gained = 0.4 × 4200 × 10 + 0.2 × 900 × 10 = 18600 J = 0.5 × c × 70, c ≈ 531 J/kg K. (b) 0.95 × 0.5 × c × 70 = 18600, c ≈ 559 J/kg K. (c) Some heat lost by the block never reached the water, so the heat gained underestimates the heat lost.

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.Heat for temperature change?Show answer
mcΔT.
2.Heat for change of state?Show answer
mL.
3.Latent heat of fusion of ice (round)?Show answer
3.35 × 10⁵ J/kg.
4.Principle of calorimetry?Show answer
Heat lost = heat gained.
5.Why is water a good coolant?Show answer
Large specific heat.
6.Temperature during melting?Show answer
Constant.

Calorimetry, Specific Heat and Latent Heat: frequently asked questions

Why does temperature stay constant during melting?

The heat supplied goes into increasing the potential energy of the molecules (breaking bonds), not their kinetic energy.

Is specific heat the same for all temperatures?

Approximately; in this chapter it is taken as constant.

What is a calorimeter?

A well-insulated container used to measure heat exchanged in a mixture.

Does the principle of mixtures need the final temperature to lie between the initial ones?

Yes, if no change of state occurs.

Why is the latent heat of vaporisation larger than that of fusion?

Boiling separates the molecules completely against their attractions and also does work against atmospheric pressure.