Calorimetry ICSE Class 10: Heat Capacity, Specific Heat Capacity and the Method of Mixtures

What Calorimetry is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Calorimetry is in the 5. Heat (i) 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

Heat, temperature and their units

Heat is energy transferred from a hotter to a colder body because of a temperature difference. SI unit: joule (J). Older unit: calorie (cal), the heat needed to raise 1 g of water by 1 °C; $1\text{ cal}\approx 4.2$ J ($4.186$ J more exactly).

Temperature measures how hot a body is (the average kinetic energy of its molecules). Units: degree Celsius (°C) and kelvin (K): $T(\text{K})=t(°\text{C})+273$. A change of 1 °C equals a change of 1 K.

Worked example

Convert (a) 27 °C to kelvin, (b) 840 J to calories. (c) Is a rise of 10 °C equal to a rise of 10 K?

Try it first, then show the solutionSolution

1. (a) $27+273=300$ K.
2. (b) $840÷4.2=200$ cal.
3. (c) Yes: the size of one degree is the same on both scales.

Answer: (a) 300 K (b) 200 cal (c) yes

Heat capacity (thermal capacity)

The heat capacity $C^{\prime }$ of a body is the heat needed to raise its temperature by one degree:
$C^{\prime }=\frac{Q}{\Delta T}$, SI unit J K⁻¹ (also J °C⁻¹).

It depends on both the material and the mass of the body: a bucket of water has a much larger heat capacity than a cup of water. On a graph of heat supplied ($y$) against temperature rise ($x$), the slope is the heat capacity.

Worked example

A block absorbs 2000 J and its temperature rises by 4 K. Find its heat capacity.

Try it first, then show the solutionSolution

1. $C^{\prime }=\frac{2000}{4}$.
2. $C^{\prime }=500$ J K⁻¹.

Answer: 500 J K⁻¹

Specific heat capacity and Q = mcΔT

The specific heat capacity $c$ of a substance is the heat needed to raise the temperature of unit mass by one degree:
$c=\frac{Q}{m\Delta T}$, so $Q=mc\Delta T$.
SI unit: J kg⁻¹ K⁻¹ (often J g⁻¹ °C⁻¹ in problems).

Relation: $C^{\prime }=mc$. Specific heat is a property of the material (and its state), not of the mass.

Values: water 4200 J kg⁻¹ K⁻¹ (4.2 J g⁻¹ °C⁻¹); ice about 2100 J kg⁻¹ K⁻¹; copper about 390–400 J kg⁻¹ K⁻¹.

For the same heat and mass, the temperature rise is inversely proportional to $c$: a substance with low $c$ heats up more.

Worked example

How much heat is needed to raise the temperature of 0.5 kg of water from 20 °C to 80 °C? ($c=4200$ J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

1. $\Delta T=60$ K.
2. $Q=0.5\times 4200\times 60=126000$ J.

Answer: 1.26 × 10⁵ J

Principle of method of mixtures

When a hot body and a cold body are mixed and no heat is lost to the surroundings:
heat lost by the hot body = heat gained by the cold body.

Mathematically, with final common temperature $t$:
$m_1{c}_1(t_1-t)=m_2{c}_2(t-t_2)$.

If a calorimeter holds the cold liquid, it also gains heat: add $m_{\text{cal}}{c}_{\text{cal}}(t-t_2)$ on the 'gained' side.

Worked example

200 g of water at 80 °C is mixed with 300 g of water at 20 °C. Find the final temperature (no heat loss).

Try it first, then show the solutionSolution

1. $200\times 4.2\times (80-t)=300\times 4.2\times (t-20)$.
2. $16000-200t=300t-6000$, so $500t=22000$.
3. $t=44$ °C.

Answer: 44 °C

Finding specific heat with a calorimeter

A calorimeter is a copper vessel with a stirrer, kept in an insulated outer jacket. Copper is used because it is a good conductor (quickly reaches the mixture's temperature) and has a low specific heat (absorbs little heat itself).

To find the specific heat of a metal: heat a known mass of it to a known temperature, drop it into a known mass of water in the calorimeter, stir and note the final temperature, then apply the principle of mixtures.

Worked example

200 g of a metal at 100 °C is dropped into 150 g of water at 20 °C. The final temperature is 30 °C. Find the specific heat capacity of the metal (ignore the calorimeter).

Try it first, then show the solutionSolution

1. Heat gained by water $=150\times 4.2\times 10=6300$ J.
2. Heat lost by metal $=200\times c\times 70$.
3. $14000c=6300$, so $c=0.45$ J g⁻¹ °C⁻¹.

Answer: 0.45 J g⁻¹ °C⁻¹

High specific heat of water and its consequences

Water has a very high specific heat capacity (4200 J kg⁻¹ K⁻¹). So it absorbs or gives out a lot of heat for a small change in temperature.

Consequences and uses:

• Coolant in car radiators and power plants: it carries away much heat.
• Hot-water bottles: water gives out heat for a long time.
• Moderate climate near the sea: the sea warms and cools slowly, so coastal places have smaller day–night and summer–winter changes; this also drives sea and land breezes.
• Farmers flood fields on frosty nights: the water releases heat slowly and protects the crops.
• Our body, mostly water, keeps a steady temperature.

Worked example

Why is water used in the radiator of a car engine?

Try it first, then show the solutionSolution

1. Water's high specific heat lets it absorb a large amount of heat from the engine with only a small rise in temperature.
2. It then gives this heat to the air in the radiator.

Answer: Its high specific heat lets it carry away much heat.

Physics in real life: Calorimetry

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 Heat Capacity and Heat Capacity
• Principle of Method of Mixtures
• High Specific Heat of Water

Calorimetry formula sheet

Heat capacity

$C^{\prime }=\frac{Q}{\Delta T}$

J K⁻¹; depends on mass and material.

Specific heat capacity

$c=\frac{Q}{m\Delta T}$

J kg⁻¹ K⁻¹; material only.

Heat for a temperature change

$Q=mc\Delta T$

Keep units consistent (g with J g⁻¹ °C⁻¹).

Relation

$C^{\prime }=mc$

Heat capacity = mass × specific heat.

Method of mixtures

$m_1{c}_1(t_1-t)=m_2{c}_2(t-t_2)$

Add the calorimeter on the side it is.

Units

$1\text{cal}\approx 4.2\text{J},T(\text{K})=t(°\text{C})+273$

Water: c = 4.2 J g⁻¹ °C⁻¹.

Calorimetry: diagrams to practise

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

How the board asks Calorimetry

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

Tricks to remember Calorimetry

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

• MCAT

  Q = m c ΔT: 'em-see-delta-tee'.

  Use it when: Heat formula.

• Capacity has mass, specific has not

  Heat capacity grows with mass; specific heat is fixed by the material.

  Use it when: Heat capacity vs specific heat.

• Lost = gained

  In mixtures, hot loses exactly what cold gains (no losses).

  Use it when: Every mixture problem.

• Water is slow to change

  High c: water heats slowly and cools slowly.

  Use it when: Consequences of water's specific heat.

Calorimetry: 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 Calorimetry

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

• ✗Mistake: Heat capacity confused with specific heat

  CISCE examiners' comments, 2025: told a 5 g metal had heat capacity 2.5 J K⁻¹ and its mass was tripled, candidates multiplied by 3 for the specific heat instead of finding c = 2.5 ÷ 5 = 0.5 J g⁻¹ K⁻¹ (unchanged).

  ✓Do this: c = C′ ÷ m and does not depend on mass; C′ = mc grows with mass.

  Usually costs: 1 mark.

• ✗Mistake: Not reading the graph's slope

  CISCE examiners' comments, 2024: for a graph of heat supplied against temperature change, candidates chose specific heat or latent heat instead of heat capacity.

  ✓Do this: Slope = Q/ΔT = heat capacity; it would be specific heat only if the mass were 1 unit.

  Usually costs: 1 mark.

• ✗Mistake: Larger temperature rise means larger c

  CISCE examiners' comments, 2025: candidates thought the material warmed more had the higher specific heat.

  ✓Do this: For the same heat and mass, ΔT ∝ 1/c: a bigger rise means a smaller c.

  Usually costs: 1 mark.

• ✗Mistake: Wrong equation or mixed units in mixtures

  CISCE examiners' comments, 2025: candidates set up wrong equations and converted c to SI units while leaving mass in grams.

  ✓Do this: Write 'heat lost = heat gained' first; use g with J g⁻¹ °C⁻¹, or kg with J kg⁻¹ K⁻¹, never a mix.

  Usually costs: 1 to 2 marks.

Practice questions on Calorimetry

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 heat capacity is:
   (a) J K⁻¹
   (b) J kg⁻¹
   (c) J kg⁻¹ K⁻¹
   (d) J kg K

   Hint

   Per kg per K.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   Equal masses of water, kerosene, glass and mercury receive the same amount of heat. Their specific heats are 4.2, 2.1, 0.84 and 0.14 J g⁻¹ °C⁻¹. The one with the greatest rise in temperature is:
   (a) water
   (b) kerosene
   (c) glass
   (d) mercury

   Hint

   Lowest c.

   Answer

   (d)

3. Q3 · 1 mark · MCQHarder board level

   Two blocks of the same metal have masses 0.2 kg and 0.6 kg. The smaller block has a heat capacity of 80 J K⁻¹. The heat capacity and the specific heat capacity of the larger block are:
   (a) 80 J K⁻¹ and 400 J kg⁻¹ K⁻¹
   (b) 240 J K⁻¹ and 400 J kg⁻¹ K⁻¹
   (c) 240 J K⁻¹ and 1200 J kg⁻¹ K⁻¹
   (d) 80 J K⁻¹ and 1200 J kg⁻¹ K⁻¹

   Hint

   C′ = mc grows with mass; c does not.

   Answer

   (b): $c=\frac{80}{0.2}=400$ J kg⁻¹ K⁻¹ for both; $C^{\prime }=0.6\times 400=240$ J K⁻¹.

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Water is used as a coolant in car radiators.
   Reason (R): Water has a high specific heat capacity, so it absorbs a lot of heat for a small rise in temperature.
   (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

   High c.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   Find the heat needed to raise the temperature of 2 kg of water by 30 °C, in kJ.

   Hint

   Q = mcΔT.

   Answer

   $2\times 4200\times 30=252000$ J $=252$ kJ

6. Q6 · 2 marks · Short answerHarder board level

   Find the heat capacity of a 250 g aluminium block ($c=0.9$ J g⁻¹ °C⁻¹) and the heat needed to warm it by 40 °C.

   Hint

   C′ = mc.

   Answer

   $C^{\prime }=225$ J °C⁻¹; $Q=225\times 40=9000$ J.

7. Q7 · 3 marks · Long answerBoard level

   (a) Define specific heat capacity and state its SI unit. (b) State the principle of method of mixtures.

   Hint

   Unit mass, one degree.

   Answer

   (a) Heat needed to raise the temperature of unit mass of a substance by 1 °C (1 K); J kg⁻¹ K⁻¹. (b) When hot and cold bodies are mixed with no heat lost, heat lost by the hot body equals heat gained by the cold body.

8. Q8 · 3 marks · Long answerHarder board level

   250 g of water at 80 °C is mixed with 750 g of water at 20 °C. Find the final temperature, assuming no heat loss.

   Hint

   Weighted average for the same liquid.

   Answer

   $250\times 4.2\times (80-t)=750\times 4.2\times (t-20)$, so $t=\frac{250\times 80+750\times 20}{1000}=35$ °C

9. Q9 · 4 marks · Long answerHOTS / competency

   A copper calorimeter of mass 50 g ($c=0.4$ J g⁻¹ °C⁻¹) contains 100 g of water at 20 °C. How much heat must be supplied to raise both to 40 °C? If this is done by dropping in a metal ball of mass 50 g at 340 °C, find the ball's specific heat capacity (no losses).

   Hint

   Water and calorimeter both gain heat.

   Answer

   Heat $=(100\times 4.2+50\times 0.4)\times 20=8800$ J. Ball: $50\times c\times 300=8800$, so $c\approx 0.59$ J g⁻¹ °C⁻¹.

10. Q10 · 3 marks · Long answerHarder board level

    Give three consequences of the high specific heat capacity of water.

    Hint

    Coolant, climate, crops.

    Answer

    Any three: used as a coolant in radiators; moderate climate near the sea (and sea breezes); hot-water bottles stay warm long; flooding fields protects crops from frost.

11. Q11 · 1 mark · MCQHOTS / competency

    Two bodies A and B have equal masses. A needs twice as much heat as B for the same rise in temperature. The ratio of their specific heats $c_A:c_B$ is:
    (a) 1 : 2
    (b) 2 : 1
    (c) 1 : 4
    (d) 4 : 1

    Hint

    c ∝ Q for equal m and ΔT.

    Answer

    (b)

12. Q12 · 4 marks · Long answerHOTS / competency

    A student mixes 50 g of water at 90 °C with 50 g of water at 10 °C in a copper calorimeter of mass 40 g that was initially at 10 °C, and finds the final temperature is 48 °C. (a) What final temperature would be expected without the calorimeter? (b) Show that the measured value fits when the calorimeter's heat is included ($c_{\text{copper}}=0.4$ J g⁻¹ °C⁻¹). (c) Why is the result lower than (a)?

    Hint

    The calorimeter starts cold.

    Answer

    (a) 50 °C. (b) Heat lost by hot water $=50\times 4.2\times 42=8820$ J; gained by cold water $=50\times 4.2\times 38=7980$ J and by calorimeter $=40\times 0.4\times 38=608$ J; total $=8588$ J, close to 8820 J (the small difference is heat lost to the room). (c) The cold calorimeter also absorbs heat, and some heat escapes, so the final temperature is below 50 °C.

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 heat capacity.Show answer

   Heat needed to raise the temperature of a body by 1 °C (1 K); C′ = Q/ΔT.

2. 2.Define specific heat capacity.Show answer

   Heat needed to raise the temperature of unit mass by 1 °C (1 K).

3. 3.SI unit of specific heat capacity?Show answer

   J kg⁻¹ K⁻¹

4. 4.Relation between C′ and c?Show answer

   C′ = mc

5. 5.Specific heat of water?Show answer

   4200 J kg⁻¹ K⁻¹ (4.2 J g⁻¹ °C⁻¹).

6. 6.State the principle of method of mixtures.Show answer

   Heat lost by the hot body = heat gained by the cold body (no losses).

7. 7.Why is a calorimeter made of copper?Show answer

   Good conductor with low specific heat.

Quick revision checklist

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

• ☐Can you tell heat capacity from specific heat?
• ☐Can you use Q = mcΔT with consistent units?
• ☐Can you set up heat lost = heat gained?
• ☐Do you include the calorimeter when given?
• ☐Can you explain four consequences of water's high specific heat?

Calorimetry: frequently asked questions