PhysicsICSE Class 10 · Calorimetry

Specific Heat Capacity and Heat Capacity: C′ = Q/ΔT, c = Q/(mΔT) and Q = mcΔT (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

Heat capacity C′ is the heat needed to raise the temperature of a whole body by 1 K: C′ = Q/ΔT, in J K⁻¹. Specific heat capacity c is the heat needed to raise unit mass of a substance by 1 K: c = Q/(mΔT), in J kg⁻¹ K⁻¹. They are related by C′ = mc, so Q = mcΔT.

Cutting a block in half halves its heat capacity C′ (it depends on mass), but specific heat c stays the same because the material is the same.

Illustration drawn for QuestionBuddy.

Understand Specific Heat Capacity and Heat Capacity

Heat and temperature.

Heat is energy that flows from a hotter body to a colder one because of the difference in temperature. SI unit: joule (J). Older unit: the calorie (cal), the heat needed to raise the temperature of 1 g of water by 1 °C; $1 cal \approx 4.2$ J.
Temperature tells how hot or cold a body is. Units: degree Celsius (°C) and kelvin (K), with $T (K) = t (° C) + 273$ . A change of 1 °C is the same as a change of 1 K, so $Δ T$ has the same value in both.

**Heat capacity (thermal capacity), $C^{'}$ . The heat needed to raise the temperature of a body** by 1 K (or 1 °C):

$C^{'} = \frac{Q}{Δ T}$ , SI unit J K⁻¹.

It depends on both the material and the mass: a large pot of water has a far larger heat capacity than a cup of water. On a graph of heat supplied ( $y$ -axis) against rise in temperature ( $x$ -axis), the slope is the heat capacity.

**Specific heat capacity, $c$ . The heat needed to raise the temperature of unit mass (1 kg) of a substance** by 1 K:

$c = \frac{Q}{m Δ T}$ , SI unit J kg⁻¹ K⁻¹ (in problems often J g⁻¹ °C⁻¹; $1$ J g⁻¹ °C⁻¹ $= 1000$ J kg⁻¹ K⁻¹).

It depends only on the material and its state, not on how much there is. Cutting a copper block in half halves its heat capacity but leaves its specific heat unchanged.

The relation. $C^{'} = m c$ , and so the heat for a temperature change is

$Q = m c Δ T = C^{'} Δ T .$

	Heat capacity $C^{'}$	Specific heat capacity $c$
Refers to	A particular body	A substance (per kg)
Depends on	Material and mass	Material (and state) only
SI unit	J K⁻¹	J kg⁻¹ K⁻¹
Formula	$\frac{Q}{Δ T}$	$\frac{Q}{m Δ T}$

Some values (approximate):

Substance	$c$ (J kg⁻¹ K⁻¹)
Water	4200
Ice	2100
Copper	about 390 to 400

Water and ice are the same substance in different states, yet their specific heats differ: specific heat changes when the state changes.

The inverse rule. For the same heat given to the same mass, $Δ T = \frac{Q}{m c}$ , so $Δ T \propto \frac{1}{c}$ . The substance with the lower specific heat warms up (and cools down) more. A larger temperature rise means a smaller $c$ , never a larger one.

Why it matters. A calorimeter is made of copper partly because copper's low specific heat means it takes up little of the heat being measured. Cooking pots have metal bodies that warm quickly; water, with its very high $c$ , is a good store and carrier of heat.

Specific Heat Capacity and Heat Capacity: formulas and key facts

Heat capacity: $C^{'} = \frac{Q}{Δ T}$; J K⁻¹; slope of a Q–ΔT graph.
Specific heat capacity: $c = \frac{Q}{m Δ T}$; J kg⁻¹ K⁻¹; material and state only.
Relation and heat equation: $C^{'} = m c, Q = m c Δ T$; ΔT in K or °C (same value).
Units and values: $1 c a l \approx 4.2 J ∣ T (K) = t (° C) + 273∣ W a t er 4200 J k g^{- 1} K^{- 1} ∣ I ce 2100 J k g^{- 1} K^{- 1} ∣ C o pp er \approx 400 J k g^{- 1} K^{- 1}$; 4.2 J g⁻¹ °C⁻¹ = 4200 J kg⁻¹ K⁻¹.

How to solve Specific Heat Capacity and Heat Capacity questions

Decide whether the question is about a body (heat capacity) or a substance (specific heat), then keep the units consistent.

1Write $Q = m c Δ T$ (or $Q = C^{'} Δ T$ ) and list the known values.
2Choose one unit system: kg with J kg⁻¹ K⁻¹, or g with J g⁻¹ °C⁻¹. Never mix them.
3Find $Δ T$ as final minus initial temperature; it is the same in °C and K.
4If heat comes from an electric heater, $Q = P t$ with $t$ in seconds (assuming no loss).
5For comparisons with the same heat and mass, use $Δ T \propto \frac{1}{c}$ .
6Remember that $c$ does not change with mass, but $C^{'}$ does.

Specific Heat Capacity and Heat Capacity 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.

How long a geyser takes

Suppose your family's geyser holds 150 L (150 kg) of water at 15 °C and has a 2 kW heating element. You want the water at 40 °C.

Estimate the time taken, assuming no heat is lost.

Try it first, then show the solutionSolution

$Q = m c Δ T = 150 \times 4200 \times 25 = 1.575 \times 1 0^{7}$ J.
$t = \frac{Q}{P} = \frac{1.575 \times 1 0 ^{7}}{2000} = 7875$ s, about 2.2 hours.

Answer: About 7875 s, a little over 2 hours.

Memory hook: Big mass, big c: big wait.

Spoon versus tea

Imagine you leave a steel spoon in a cup of hot tea. In a minute the spoon is hot, but the tea has hardly cooled.

Explain using heat capacity.

Try it first, then show the solutionSolution

The spoon has a small mass and a low specific heat, so its heat capacity is small: a little heat raises its temperature a lot.
The tea is mostly water, with a larger mass and a high specific heat, so losing that little heat lowers its temperature only slightly.

Answer: The spoon's heat capacity is much smaller than the tea's.

Memory hook: Small C′, big change; big C′, small change.

Tricks to remember Specific Heat Capacity and Heat Capacity

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

'Specific' means per kilogram
Specific heat = heat capacity ÷ mass, just as density = mass ÷ volume.
Use it when: Telling C′ from c.
MCAT
Q = m c ΔT: 'em-see-delta-tee'.
Use it when: Writing the heat equation.
Low c, quick change
Metals have low c: they heat fast and cool fast. Water has high c: slow both ways.
Use it when: Comparison questions.
Slope is capacity
Q on y, ΔT on x: slope = Q/ΔT = heat capacity.
Use it when: Graph questions.

Specific Heat Capacity and Heat Capacity: solved examples

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

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

The SI unit of heat capacity is:
(a) J kg⁻¹ K⁻¹
(b) J K⁻¹
(c) J kg⁻¹
(d) cal g⁻¹

Try it first, then show the solutionSolution

$C^{'} = \frac{Q}{Δ T}$ : joule per kelvin. (J kg⁻¹ K⁻¹ is the unit of specific heat capacity.)

Answer: (b) J K⁻¹

Example 2: Warming water · 2 marks · Short answerBoard level

A bottle holds 750 g of water at 18 °C. How much heat must it absorb to reach 58 °C? ( $c = 4200$ J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

$m = 0.75$ kg; $Δ T = 58 - 18 = 40$ K. (½)
$Q = m c Δ T = 0.75 \times 4200 \times 40 = 126000$ J. (1½)

Answer: 126 000 J (126 kJ)

Example 3: Heating an iron block · 3 marks · Short answerHarder board level

A 500 W heater is embedded in a 2 kg iron block and switched on for 3 minutes. Assuming no heat is lost, find the rise in temperature of the block. ( $c_{i r o n} = 450$ J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

$Q = P t = 500 \times 180 = 90000$ J. (1)
$Δ T = \frac{Q}{m c} = \frac{90 000}{2 \times 450}$ (1)
$= 100$ K (or 100 °C). (1)

Answer: 100 K

Example 4: From one block to another · 3 marks · Short answerHarder board level

A 400 g metal block needs 1520 J to warm by 10 K. Find (a) its heat capacity, (b) the specific heat capacity of the metal. (c) Find the heat capacity and specific heat capacity of a 1.2 kg block of the same metal.

Try it first, then show the solutionSolution

(a) $C^{'} = \frac{1520}{10} = 152$ J K⁻¹. (1)
(b) $c = \frac{C ^{'}}{m} = \frac{152}{0.4} = 380$ J kg⁻¹ K⁻¹. (1)
(c) $C^{'} = 1.2 \times 380 = 456$ J K⁻¹ (three times the mass, three times $C^{'}$ ); $c$ stays 380 J kg⁻¹ K⁻¹. (1)

Answer: (a) 152 J K⁻¹ (b) 380 J kg⁻¹ K⁻¹ (c) 456 J K⁻¹; 380 J kg⁻¹ K⁻¹

Example 5: Which has the higher c? · 2 marks · Short answerHarder board level

The same quantity of heat warms 1 g of substance P by 8 °C and 1 g of substance Q by 24 °C. (a) Which has the higher specific heat capacity, and by what factor? (b) Which would be better for making a calorimeter?

Try it first, then show the solutionSolution

(a) $Δ T \propto \frac{1}{c}$ : P warms less, so P has the higher $c$ ; $c_{P} : c_{Q} = 24 : 8 = 3 : 1$ . (1)
(b) Q, because its lower specific heat means the calorimeter absorbs less of the heat being measured. (1)

Answer: (a) P, three times (b) Q

Example 6: Reading two graphs · 3 marks · Short answerHOTS / competency

Bodies A (mass 0.5 kg) and B (mass 0.25 kg) are heated. A straight-line graph of heat supplied against temperature rise passes through (5 K, 3000 J) for A and (15 K, 3000 J) for B. Find (a) the heat capacity of each, (b) the specific heat capacity of each, (c) which line is steeper.

Try it first, then show the solutionSolution

(a) $C_{A}^{'} = \frac{3000}{5} = 600$ J K⁻¹; $C_{B}^{'} = \frac{3000}{15} = 200$ J K⁻¹. (1)
(b) $c_{A} = \frac{600}{0.5} = 1200$ J kg⁻¹ K⁻¹; $c_{B} = \frac{200}{0.25} = 800$ J kg⁻¹ K⁻¹. (1)
(c) A's line is steeper: its slope (heat capacity) is larger. (1)

Answer: (a) 600 and 200 J K⁻¹ (b) 1200 and 800 J kg⁻¹ K⁻¹ (c) A

Example 7: Case study: ice and water are different · 4 marks · Case-basedHOTS / competency

Take 200 g of ice at −10 °C and, separately, 200 g of water at 0 °C. ( $c_{i ce} = 2100$ J kg⁻¹ K⁻¹, $c_{w a t er} = 4200$ J kg⁻¹ K⁻¹)
(a) Find the heat needed to warm the ice to 0 °C (without melting it).
(b) Find the heat needed to warm the water from 0 °C to 10 °C.
(c) Express the answer to (a) in calories (1 cal = 4.2 J).
(d) A student says, 'Specific heat stays the same when a substance changes state.' Is this true? Use your answers.

Try it first, then show the solutionSolution

(a) $Q = 0.2 \times 2100 \times 10 = 4200$ J. (1)
(b) $Q = 0.2 \times 4200 \times 10 = 8400$ J. (1)
(c) $\frac{4200}{4.2} = 1000$ cal. (1)
(d) False: the same mass and the same 10 K rise need different heat, because ice and water have different specific heats. (1)

Answer: (a) 4200 J (b) 8400 J (c) 1000 cal (d) False

Where students lose marks in Specific Heat Capacity and Heat Capacity

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

Mistake: Heat capacity treated as specific heat
Specific heat does not change with mass; heat capacity does.
Wrong working
Mass tripled, so 'specific heat becomes three times'.
Right working
c = C′/m stays the same; only C′ triples.
Do this: Ask: is it about the body (C′) or the substance (c)?
Usually costs: Reported by CISCE examiners for 2025 Q1(x).
Mistake: Wrong quantity from the slope
Slope of heat against temperature change is Q/ΔT, the heat capacity.
Wrong working
'The slope gives specific heat capacity.'
Right working
'The slope gives heat capacity.'
Do this: Write slope = Δy/Δx with the axis quantities.
Usually costs: Reported by CISCE examiners for 2024 Q1(xiv).
Mistake: Bigger temperature rise taken as bigger c
For the same heat and mass, ΔT ∝ 1/c.
Wrong working
'Y warms by 40 °C, so Y has the higher specific heat.'
Right working
'The one that warms less has the higher specific heat.'
Do this: Write ΔT = Q/(mc) and read it.
Usually costs: Reported by CISCE examiners for 2025 Q9(ii).
Mistake: Thinking c is the same in every state
Ice and water have different specific heats.
Wrong working
'Specific heat stays the same when a solid melts: True.'
Right working
'False; for example, ice ≈ 2100 and water 4200 J kg⁻¹ K⁻¹.'
Do this: Learn the ice and water values together.
Usually costs: Flagged in the CISCE examiners' suggestions for 2025 Q9(ii), which asked whether specific heat stays the same from solid to liquid.

Exam tips

Quote the unit with every heat capacity (J K⁻¹) and specific heat (J kg⁻¹ K⁻¹ or J g⁻¹ °C⁻¹).
Convert grams to kilograms when c is in J kg⁻¹ K⁻¹; or keep grams with J g⁻¹ °C⁻¹.
In 'which heats faster' MCQs, pick the lowest specific heat.
State 'assuming no heat loss' when using an electric heater's energy.

Practice questions on Specific Heat Capacity and Heat Capacity

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 specific heat capacity of a substance depends on:
(a) its mass
(b) its material
(c) its volume
(d) the heat supplied
Hint
Per unit mass.
Answer
(b) its material
Q2 · 1 mark · Very short answerBoard level
Express 30 °C in kelvin.
Hint
Add 273.
Answer
303 K.
Q3 · 2 marks · Short answerBoard level
Define heat capacity and state its SI unit.
Hint
The whole body, 1 K.
Answer
The heat capacity of a body is the heat needed to raise its temperature by 1 K (1 °C). SI unit: J K⁻¹.
Q4 · 2 marks · Short answerHarder board level
Find the heat needed to warm a 750 g copper block from 25 °C to 85 °C. ( $c = 400$ J kg⁻¹ K⁻¹)
Hint
$Q = m c Δ T$ .
Answer
$Q = 0.75 \times 400 \times 60 = 18000$ J.
Q5 · 1 mark · Very short answerHarder board level
Convert 5000 cal into joules (1 cal = 4.2 J).
Hint
Multiply.
Answer
$5000 \times 4.2 = 21000$ J.
Q6 · 2 marks · Short answerHarder board level
4 kg of a liquid absorbs 168 kJ of heat and its temperature rises by 20 °C. Find its specific heat capacity.
Hint
$c = \frac{Q}{m Δ T}$ .
Answer
$c = \frac{168 000}{4 \times 20} = 2100$ J kg⁻¹ K⁻¹.
Q7 · 2 marks · Short answerHarder board level
Equal masses of copper ( $c = 400$ J kg⁻¹ K⁻¹) and water ( $c = 4200$ J kg⁻¹ K⁻¹) receive the same heat. Find the ratio of their temperature rises.
Hint
$Δ T \propto \frac{1}{c}$ .
Answer
$Δ T_{C u} : Δ T_{w a t er} = 4200 : 400 = 10.5 : 1$ .
Q8 · 1 mark · Very short answerHarder board level
Find the heat capacity of 2.5 kg of water.
Hint
$C^{'} = m c$ .
Answer
$C^{'} = 2.5 \times 4200 = 10500$ J K⁻¹.
Q9 · 3 marks · Short answerHOTS / competency
A 1.5 kW immersion rod heats 6 kg of water in an open bucket from 20 °C to 45 °C. (a) Find the heat absorbed by the water. (b) Find the time this would take with no heat loss. (c) It actually takes 840 s. What percentage of the electrical energy heats the water?
Hint
$Q = m c Δ T$ , then $t = \frac{Q}{P}$ .
Answer
(a) $Q = 6 \times 4200 \times 25 = 630000$ J. (b) $t = \frac{630 000}{1500} = 420$ s. (c) $\frac{630 000}{1500 \times 840} \times 100 = 50%$ ; the rest is lost from the open bucket to the air and the bucket itself.
Q10 · 2 marks · Short answerHOTS / competency
Block A (1 kg, $c = 900$ J kg⁻¹ K⁻¹) and block B (3 kg, $c = 300$ J kg⁻¹ K⁻¹) receive the same heat. Which warms more? Explain.
Hint
Compare heat capacities, not specific heats.
Answer
Neither: $C_{A}^{'} = 1 \times 900 = 900$ J K⁻¹ and $C_{B}^{'} = 3 \times 300 = 900$ J K⁻¹. Equal heat capacities give equal temperature rises for the same heat.

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.SI unit of specific heat capacity?Show answer
J kg⁻¹ K⁻¹.
2.Relation between heat capacity and specific heat?Show answer
C′ = mc.
3.Specific heat of water?Show answer
4200 J kg⁻¹ K⁻¹.
4.Does c change with mass?Show answer
No.
5.1 cal in joules?Show answer
About 4.2 J.

Specific Heat Capacity and Heat Capacity: frequently asked questions

What is specific heat capacity?

Specific heat capacity is the heat needed to raise the temperature of 1 kg of a substance by 1 K: c = Q/(mΔT). Its SI unit is J kg⁻¹ K⁻¹; for water it is 4200 J kg⁻¹ K⁻¹.

What is the difference between heat capacity and specific heat capacity?

Heat capacity (C′ = Q/ΔT, in J K⁻¹) belongs to a body and depends on its mass and material. Specific heat capacity (c = Q/(mΔT), in J kg⁻¹ K⁻¹) belongs to a substance and depends only on its material and state. C′ = mc.

Which heats up faster: a substance with high or low specific heat?

For the same mass and heat, the substance with the lower specific heat heats up (and cools down) faster, because ΔT = Q/(mc).