ICSE Class 10 Physics Calorimetry Heat Exchange Problems Solved

Ever wondered why a metal spoon gets hot when you leave it in a cup of tea? That’s calorimetry in action – the science of measuring heat flow.

💡 In Simple Words: Calorimetry tells us how much heat moves between objects when they touch. By knowing the mass, material type and temperature change, we can calculate the exact energy transferred.

What is Calorimetry?

Calorimetry is a method to find the amount of heat (energy) transferred during a temperature change. Think of it like a kitchen scale that measures weight, but instead it “weighs” heat.

Key terms you’ll meet

• Heat (Q): Energy that moves because of a temperature difference, measured in joules (J).
• Specific heat capacity (c): Amount of heat needed to raise 1 kg of a substance by 1 °C, like how much fuel a car needs to go one mile.
• Mass (m): How much material you have, in kilograms.
• Temperature change (ΔT): Final temperature minus initial temperature, written as ΔT = Tfinal – Tinitial.

Heat Exchange Formula

The backbone of every calorimetry problem is:

Q = m c ΔT

It works for any substance as long as you use the right specific heat capacity. If you have more than one material exchanging heat, the heat lost by the hotter body equals the heat gained by the cooler one (ignoring losses to the surroundings).

Step‑by‑step method to solve heat exchange problems

When you see a question, follow this simple recipe:

1. Read the problem carefully and note down all given values (m, c, initial and final temperatures).
2. Identify which object is losing heat and which is gaining heat.
3. Write the heat equation for each object: Qlost = mhot chot ΔThot, Qgained = mcold ccold ΔTcold.
4. Set the magnitude of heat lost equal to heat gained (|Qlost| = Qgained) and solve for the unknown.
5. Check units and sign – heat lost is negative, heat gained is positive.

Worked Example 1 – Mixing Water

A 200 g piece of iron at 150 °C is dropped into 500 g of water at 25 °C. Specific heat capacities: iron = 0.45 J g⁻¹ °C⁻¹, water = 4.18 J g⁻¹ °C⁻¹. Find the final temperature of the mixture (ignore heat loss to the surroundings).

Solution:

1. Let T be the final temperature.
2. Heat lost by iron: Qiron = miron ciron (T – 150). Since T 
3. Heat gained by water: Qwater = mwater cwater (T – 25).
4. Set |Qiron| = Qwater: 200 × 0.45 × (150 – T) = 500 × 4.18 × (T – 25).
5. Simplify: 90 × (150 – T) = 2090 × (T – 25).
   90 × 150 – 90T = 2090T – 52250.
   13500 – 90T = 2090T – 52250.
   13500 + 52250 = 2090T + 90T.
   65750 = 2180T.
   T ≈ 30.2 °C.
6. So the mixture settles at about 30 °C.

Worked Example 2 – Calorimeter with Metal

A calorimeter (heat capacity 150 J °C⁻¹) contains 250 g of water at 20 °C. A 50 g piece of copper (c = 0.385 J g⁻¹ °C⁻¹) at 100 °C is placed inside. Find the final temperature.

Solution:

1. Let T be the final temperature.
2. Heat lost by copper: QCu = 50 × 0.385 × (100 – T).
3. Heat gained by water: Qwater = 250 × 4.18 × (T – 20).
4. Heat gained by calorimeter: Qcal = 150 × (T – 20).
5. Energy balance: QCu = Qwater + Qcal.
   50 × 0.385 × (100 – T) = 250 × 4.18 × (T – 20) + 150 × (T – 20).
6. Calculate left side: 19.25 × (100 – T). Right side: (1045 + 150) × (T – 20) = 1195 × (T – 20).
7. 19.25 × 100 – 19.25T = 1195T – 23900.
   1925 – 19.25T = 1195T – 23900.
   1925 + 23900 = 1195T + 19.25T.
   25825 = 1214.25T.
   T ≈ 21.3 °C.
8. The final temperature is roughly 21 °C.

Quick Revision Table

📝 Likely Exam Questions

• What is the heat absorbed when 150 g of aluminium (c = 0.90 J g⁻¹ °C⁻¹) is heated from 25 °C to 75 °C?
  Answer: Q = 150 × 0.90 × (75‑25) = 150 × 0.90 × 50 = 6750 J.
• A 100 g piece of ice at 0 °C melts completely in 200 g of water initially at 30 °C. Given latent heat of fusion of ice = 334 J g⁻¹, find the final temperature of the mixture.
  Answer: Heat to melt ice = 100 × 334 = 33400 J. Heat lost by water = 200 × 4.18 × (30‑T). Set 33400 = 200 × 4.18 × (30‑T). Solve: T ≈ 13.0 °C.
• Explain why the heat lost by a hotter object equals the heat gained by a cooler one in a calorimetry experiment (ignore surroundings).
  Answer: Energy is conserved; heat transferred out of one body must appear as heat transferred into another if no energy escapes.