PhysicsICSE Class 10 · Calorimetry

Principle of Method of Mixtures: Heat Lost = Heat Gained, Calorimeter and Numericals (ICSE Class 10)

Updated 16 October 2026 · Fact-checked

The principle of method of mixtures states that when a hot and a cold body are mixed with no heat lost to the surroundings, heat lost by the hot body equals heat gained by the cold body: m₁c₁(t₁ − t) = m₂c₂(t − t₂), where t is the final common temperature. It follows from conservation of energy.

With no heat escaping, heat lost by the hot water equals heat gained by the cold water, and the final t lies between t₂ and t₁.

Illustration drawn for QuestionBuddy.

Understand Principle of Method of Mixtures

The principle. When two bodies at different temperatures are put in contact (or mixed), heat flows from the hotter to the colder until both reach the same final temperature $t$ . If no heat is lost to or gained from the surroundings:

Heat lost by the hot body = heat gained by the cold body.

This is simply the conservation of energy applied to heat.

Mathematical statement. A hot body (mass $m_{1}$ , specific heat $c_{1}$ , temperature $t_{1}$ ) is mixed with a cold body ( $m_{2}$ , $c_{2}$ , $t_{2}$ ):

$m_{1} c_{1} (t_{1} - t) = m_{2} c_{2} (t - t_{2}) .$

Write each temperature change as higher minus lower, so both sides are positive.

Final temperature. Solving for $t$ :

$t = \frac{m _{1} c _{1} t _{1} + m _{2} c _{2} t _{2}}{m _{1} c _{1} + m _{2} c _{2}} .$

For two masses of the same liquid ( $c_{1} = c_{2}$ ), this becomes a mass-weighted average: $t = \frac{m _{1} t _{1} + m _{2} t _{2}}{m _{1} + m _{2}}$ . Equal masses give the simple average. The final temperature always lies between $t_{1}$ and $t_{2}$ , nearer to the temperature of the body with the larger heat capacity $m c$ .

Including the calorimeter. The vessel also gains (or loses) heat. If the calorimeter (mass $m_{c a l}$ , specific heat $c_{c a l}$ ) starts at the same temperature as the cold water, add it to the 'gained' side:

$m_{1} c_{1} (t_{1} - t) = (m_{w} c_{w} + m_{c a l} c_{c a l}) (t - t_{2}) .$

The calorimeter. A cylindrical copper vessel with a copper stirrer, placed inside an insulated outer jacket with a lid through which a thermometer passes.

Copper is used because it is a good conductor (it quickly reaches the temperature of its contents) and has a low specific heat (it takes up little of the heat).
The insulating jacket and lid reduce heat lost by conduction, convection and evaporation.
Stirring spreads the heat evenly so the thermometer reads the true common temperature.

Finding the specific heat of a solid.
1. Weigh the calorimeter with stirrer, then with water; note the water's temperature $t_{2}$ .
2. Heat the solid (for example in boiling water) to a known temperature $t_{1}$ .
3. Transfer it quickly into the calorimeter, stir, and note the highest steady temperature $t$ .
4. Apply: $m_{s} c_{s} (t_{1} - t) = (m_{w} c_{w} + m_{c a l} c_{c a l}) (t - t_{2})$ and solve for $c_{s}$ .

Sources of error. Some heat is always lost: while the hot solid is being transferred, and from the calorimeter to the air. The water then gains less heat than the solid lost, so the calculated specific heat comes out too low. Ignoring the calorimeter's own heat gain also gives too low a value. Minimise losses by transferring quickly, using a lid and jacket, and stirring.

(Mixtures that include melting ice also need latent heat; they are treated in the latent heat chapter.)

Principle of Method of Mixtures: formulas and key facts

Principle of mixtures: $m_{1} c_{1} (t_{1} - t) = m_{2} c_{2} (t - t_{2})$; Heat lost = heat gained, no loss to surroundings.
Final temperature: $t = \frac{m _{1} c _{1} t _{1} + m _{2} c _{2} t _{2}}{m _{1} c _{1} + m _{2} c _{2}}$; Same liquid: mass-weighted average.
With a calorimeter: $m_{s} c_{s} (t_{1} - t) = (m_{w} c_{w} + m_{c a l} c_{c a l}) (t - t_{2})$; Calorimeter starts with the cold water.
Useful values: $W a t er : 4.2 J g^{- 1} ° C^{- 1} ∣ C o pp er : ab o u t 0.4 J g^{- 1} ° C^{- 1}$; Keep g with J g⁻¹ °C⁻¹.

How to solve Principle of Method of Mixtures questions

Every mixtures problem is one equation: write 'heat lost = heat gained' in words first, then fill it in.

1Identify which bodies lose heat (start hotter than $t$ ) and which gain (start colder), including the calorimeter.
2Write: heat lost by hot body = heat gained by water + heat gained by calorimeter.
3Express each as $m c Δ T$ with $Δ T$ = higher − lower temperature.
4Use one unit system throughout (g and J g⁻¹ °C⁻¹, or kg and J kg⁻¹ K⁻¹).
5Solve for the unknown ( $t$ , $m$ or $c$ ) and check that $t$ lies between the starting temperatures.

Principle of Method of Mixtures 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.

Milk in hot tea

Suppose you pour 50 g of cold milk at 10 °C into 150 g of tea at 90 °C. Treat both as water and ignore heat lost to the cup.

Find the temperature of your tea now.

Try it first, then show the solutionSolution

Heat lost by tea = heat gained by milk: $150 \times 4.2 \times (90 - t) = 50 \times 4.2 \times (t - 10)$ .
$3 (90 - t) = t - 10$ , so $280 = 4 t$ and $t = 70$ °C.

Answer: About 70 °C.

Memory hook: Three parts hot, one part cold: three-quarters of the way to hot.

Quenching a horseshoe

Imagine a blacksmith plunges a red-hot iron horseshoe into a large bucket of water. The horseshoe cools by hundreds of degrees, but the water warms by only a few degrees.

Explain using the method of mixtures.

Try it first, then show the solutionSolution

Heat lost by the iron = heat gained by the water.
The water's heat capacity ( $m c$ ) is far larger: a large mass of a substance with a high specific heat.
So the same quantity of heat gives the water only a small temperature change.

Answer: Equal heat, very different heat capacities, so very different temperature changes.

Memory hook: Same heat, big mc, small change.

Tricks to remember Principle of Method of Mixtures

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

LOST = GAINED
Write the words before the numbers; then fill in each side with mcΔT.
Use it when: Setting up the equation.
Higher minus lower
Both ΔTs are positive: hot side (t₁ − t), cold side (t − t₂).
Use it when: Avoiding sign errors.
The see-saw
The final temperature sits nearer the body with the bigger mc, like a see-saw balancing nearer the heavier child.
Use it when: Checking answers.
Copper calorimeter: C and C
Copper: Conducts well, small Capacity for heat.
Use it when: Why copper is used.

Principle of Method of Mixtures: solved examples

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

Example 1: What the principle rests on · 1 mark · MCQBoard level

The principle of method of mixtures is a consequence of:
(a) conservation of mass
(b) conservation of energy
(c) Newton's law of cooling
(d) conservation of momentum

Try it first, then show the solutionSolution

Heat is energy; with no loss, the energy lost by one body is gained by the other.

Answer: (b) conservation of energy

Example 2: Mixing two lots of water · 3 marks · Short answerBoard level

150 g of water at 64 °C is mixed with 450 g of water at 24 °C. Find the final temperature, assuming no heat loss.

Try it first, then show the solutionSolution

Heat lost = heat gained: $150 \times 4.2 \times (64 - t) = 450 \times 4.2 \times (t - 24)$ . (1)
$64 - t = 3 (t - 24) \Rightarrow 64 - t = 3 t - 72$ . (1)
$4 t = 136$ , so $t = 34$ °C. (1)

Answer: 34 °C

Example 3: Specific heat of a metal · 3 marks · Short answerHarder board level

A 0.4 kg metal block at 100 °C is dropped into 0.6 kg of water at 20 °C. The final temperature is 25 °C. Neglecting the container and heat losses, find the specific heat capacity of the metal. ( $c_{w} = 4200$ J kg⁻¹ K⁻¹)

Try it first, then show the solutionSolution

Heat gained by water $= 0.6 \times 4200 \times (25 - 20) = 12600$ J. (1)
Heat lost by metal $= 0.4 \times c \times (100 - 25) = 30 c$ . (1)
$30 c = 12600 \Rightarrow c = 420$ J kg⁻¹ K⁻¹. (1)

Answer: 420 J kg⁻¹ K⁻¹

Example 4: Hot water into a calorimeter · 3 marks · Short answerHarder board level

A copper calorimeter of mass 210 g ( $c = 0.4$ J g⁻¹ °C⁻¹) contains 180 g of water at 15 °C. 100 g of water at 75 °C is poured in. Find the final temperature, assuming no heat loss.

Try it first, then show the solutionSolution

Heat lost by hot water $= 100 \times 4.2 \times (75 - t) = 420 (75 - t)$ . (1)
Heat gained $= (180 \times 4.2 + 210 \times 0.4) (t - 15) = (756 + 84) (t - 15) = 840 (t - 15)$ . (1)
$420 (75 - t) = 840 (t - 15) \Rightarrow 75 - t = 2 t - 30 \Rightarrow t = 35$ °C. (1)

Answer: 35 °C

Example 5: Design of the calorimeter · 3 marks · Short answerHarder board level

Give a reason for each: (a) a calorimeter is made of copper, (b) it is kept in an insulated jacket with a lid, (c) the contents are stirred before reading the temperature.

Try it first, then show the solutionSolution

(a) Copper is a good conductor and has a low specific heat, so it quickly reaches the mixture's temperature and absorbs little heat. (1)
(b) To reduce heat lost to (or gained from) the surroundings, so that heat lost = heat gained holds. (1)
(c) To spread the heat evenly so the whole mixture reaches one common temperature. (1)

Answer: (a) Good conductor, low c (b) Reduce heat exchange with surroundings (c) Uniform temperature

Example 6: Running a bath · 3 marks · Short answerHOTS / competency

A bath has 28 kg of cold water at 20 °C. How much hot water at 70 °C must be added to make the bath 35 °C? Ignore heat losses.

Try it first, then show the solutionSolution

Heat lost by hot water = heat gained by cold water: $m \times c \times (70 - 35) = 28 \times c \times (35 - 20)$ . (1)
$c$ cancels (both water): $35 m = 420$ . (1)
$m = 12$ kg. (Check: $\frac{12 \times 70 + 28 \times 20}{40} = 35$ °C.) (1)

Answer: 12 kg

Example 7: Case study: a student's experiment · 4 marks · Case-basedHOTS / competency

A student drops a 250 g metal block at 84 °C into 140 g of water at 18 °C held in a copper calorimeter of mass 105 g, also at 18 °C. The final temperature is 28 °C. ( $c_{w} = 4.2$ , $c_{C u} = 0.4$ J g⁻¹ °C⁻¹)
(a) Write the heat-balance equation in words.
(b) Find the specific heat capacity of the metal.
(c) Another student forgets the calorimeter. What value does she get?
(d) In practice some heat escapes while the block is transferred. Will the student's value be higher or lower than the true value? Explain.

Try it first, then show the solutionSolution

(a) Heat lost by the block = heat gained by the water + heat gained by the calorimeter. (1)
(b) Gained $= (140 \times 4.2 + 105 \times 0.4) \times 10 = 630 \times 10 = 6300$ J; lost $= 250 \times c \times 56 = 14000 c$ ; $c = 0.45$ J g⁻¹ °C⁻¹. (1)
(c) $\frac{588 \times 10}{14 000} = 0.42$ J g⁻¹ °C⁻¹, too low. (1)
(d) Lower: the water and calorimeter receive less heat than the block really lost, so the calculated heat lost (and hence $c$ ) is too small. (1)

Answer: (a) Lost by block = gained by water + calorimeter (b) 0.45 J g⁻¹ °C⁻¹ (c) 0.42 J g⁻¹ °C⁻¹ (d) Lower

Where students lose marks in Principle of Method of Mixtures

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

Mistake: Wrong heat-balance equation
Each side must be the heat change of the bodies losing or gaining heat, written as mcΔT.
Wrong working
'mcΔt = ml' for a problem with no change of state, or equating temperatures.
Right working
'Heat lost by hot body = heat gained by cold body (+ calorimeter)', each as mcΔT.
Do this: Write the equation in words before substituting.
Usually costs: Reported by CISCE examiners for 2025 Q9(i).
Mistake: Mixed units
c in J kg⁻¹ K⁻¹ with mass in grams gives an answer 1000 times wrong.
Wrong working
200 g × 4200 J kg⁻¹ K⁻¹ × 10.
Right working
200 g × 4.2 J g⁻¹ °C⁻¹ × 10, or 0.2 kg × 4200 × 10.
Do this: Choose one system and convert everything at the start.
Usually costs: Reported by CISCE examiners for 2025 Q9(i).
Mistake: Forgetting the calorimeter
When its mass and specific heat are given, it also gains heat.
Wrong working
Heat lost by block = heat gained by water only.
Right working
Heat lost by block = heat gained by water + calorimeter.
Do this: List every body in the problem and decide whether it loses or gains.
Usually costs: Usually costs the method mark.
Mistake: Negative temperature differences
Writing (t − t₁) on the hot side gives a negative heat and a wrong answer.
Wrong working
m₁c₁(t − t₁) = m₂c₂(t − t₂).
Right working
m₁c₁(t₁ − t) = m₂c₂(t − t₂).
Do this: Always higher minus lower.
Usually costs: Usually costs the answer mark.

Exam tips

State the principle with its condition: 'provided no heat is lost to the surroundings'.
Draw and label the calorimeter: copper vessel, stirrer, thermometer, insulated jacket, lid.
Check that your final temperature lies between the initial temperatures.
Give c with units J g⁻¹ °C⁻¹ or J kg⁻¹ K⁻¹ to match your working.

Practice questions on Principle of Method of Mixtures

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
In the method of mixtures, assuming no heat is lost to the surroundings:
(a) heat lost by the hot body is more than heat gained by the cold body
(b) heat lost by the hot body equals heat gained by the cold body
(c) the final temperature equals the higher initial temperature
(d) temperature lost equals temperature gained
Hint
Energy is conserved; temperatures are not.
Answer
(b) heat lost by the hot body equals heat gained by the cold body
Q2 · 2 marks · Short answerBoard level
Write the mathematical statement of the principle of method of mixtures for a hot solid dropped into cold water, explaining each symbol.
Hint
Two $m c Δ T$ terms.
Answer
$m_{s} c_{s} (t_{1} - t) = m_{w} c_{w} (t - t_{2})$ , where $m_{s}$ , $c_{s}$ , $t_{1}$ are the mass, specific heat and initial temperature of the solid; $m_{w}$ , $c_{w}$ , $t_{2}$ those of the water; and $t$ the final common temperature.
Q3 · 2 marks · Short answerBoard level
Give two reasons why a calorimeter is made of copper.
Hint
Conduction and heat capacity.
Answer
(i) Copper is a good conductor of heat, so the calorimeter quickly reaches the temperature of its contents. (ii) It has a low specific heat, so it absorbs very little heat.
Q4 · 2 marks · Short answerHarder board level
1 kg of water at 90 °C is mixed with 2 kg of water at 30 °C. Find the final temperature.
Hint
Same liquid: weighted average.
Answer
$t = \frac{1 \times 90 + 2 \times 30}{3} = 50$ °C.
Q5 · 3 marks · Short answerHarder board level
A 1.05 kg copper block ( $c = 400$ J kg⁻¹ K⁻¹) at 140 °C is placed in 0.5 kg of water at 20 °C. Find the final temperature, ignoring heat losses.
Hint
Heat capacities 420 and 2100 J K⁻¹.
Answer
$420 (140 - t) = 2100 (t - 20) \Rightarrow 140 - t = 5 t - 100 \Rightarrow t = 40$ °C.
Q6 · 2 marks · Short answerHarder board level
200 g of a liquid at 25 °C is mixed with 100 g of water at 85 °C, and the final temperature is 55 °C. Find the specific heat capacity of the liquid.
Hint
Water loses, liquid gains.
Answer
$100 \times 4.2 \times 30 = 200 \times c \times 30 \Rightarrow c = 2.1$ J g⁻¹ °C⁻¹.
Q7 · 2 marks · Short answerHarder board level
State two precautions that reduce heat loss when finding the specific heat of a solid by the method of mixtures.
Hint
Transfer and insulation.
Answer
Any two: transfer the hot solid quickly; keep the calorimeter in an insulated jacket with a lid; stir and read the highest temperature promptly; avoid splashing water.
Q8 · 2 marks · Short answerHarder board level
A small hot metal nail is dropped into a large beaker of cold water. Is the final temperature closer to the nail's or the water's initial temperature? Explain.
Hint
Compare $m c$ .
Answer
Closer to the water's: the water has a much larger heat capacity (large mass, high specific heat), so it changes temperature only slightly for the heat the nail gives out.
Q9 · 2 marks · Short answerHOTS / competency
Equal masses of water at 10 °C, 30 °C and 80 °C are mixed. Find the final temperature.
Hint
Same liquid, equal masses.
Answer
Heat lost = heat gained gives $3 t = 10 + 30 + 80$ , so $t = 40$ °C (the 80 °C water loses heat; the other two gain).
Q10 · 2 marks · Short answerHOTS / competency
A shower mixer receives 6 kg of hot water per minute at 60 °C. How much cold water at 15 °C must it add each minute to deliver water at 40 °C?
Hint
Balance per minute.
Answer
$6 \times (60 - 40) = m \times (40 - 15) \Rightarrow m = \frac{120}{25} = 4.8$ kg per minute.

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.Principle of method of mixtures?Show answer
Heat lost by hot body = heat gained by cold body (no loss).
2.Law it is based on?Show answer
Conservation of energy.
3.Material of a calorimeter?Show answer
Copper.
4.Final temperature of equal masses of water at 20 °C and 60 °C?Show answer
40 °C.
5.Effect of heat loss on the measured c of a solid?Show answer
It comes out too low.

Principle of Method of Mixtures: frequently asked questions

What is the principle of method of mixtures?

When a hot and a cold body are mixed and no heat is lost to the surroundings, the heat lost by the hot body equals the heat gained by the cold body: m₁c₁(t₁ − t) = m₂c₂(t − t₂).

How do you find the final temperature of a mixture?

Set heat lost equal to heat gained and solve for t: t = (m₁c₁t₁ + m₂c₂t₂)/(m₁c₁ + m₂c₂). For two masses of the same liquid it is the mass-weighted average temperature.

Why is a calorimeter made of copper?

Copper is a good conductor, so the calorimeter quickly reaches the temperature of its contents, and it has a low specific heat, so it absorbs very little of the heat.