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Calorimetry

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Estimated time: 22 minutes
  • Introduction
  • Definition: Calorimetry
  • Definition: Calorimeter
  • Calorimeter
  • The Principle of Calorimetry
  • Experiment: Method of Mixtures
  • Derivation of the Calorimetry Equation
  • Example
  • Key Points: Calorimetry
Maharashtra State Board: Class 11

Introduction

A thermometer tells you how hot something is, but it doesn't tell you how much heat energy an object contains or exchanges. To answer that question, scientists developed a technique called calorimetry and a device called a calorimeter.

Maharashtra State Board: Class 8, 11

Definition: Calorimetry

Calorimetry is the science of measuring heat exchange during physical or chemical processes. The word comes from the Latin calor (heat) + Greek metron (measure).

CISCE: Class 10

Definition: Calorimeter

A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained (or lost) by a body when it is mixed with another body or substance.

Maharashtra State Board: Class 8, 11

Calorimeter

A calorimeter is an insulated device used to measure the heat gained or lost by a substance during a thermal process. It works on a simple idea: trap all the heat inside the device so nothing escapes, and then measure the temperature change.

Labelled cross-section diagram of a calorimeter showing outer jacket, inner copper vessel, water, stirrer, thermometer and lid

Fig.: Cross-section of a simple water calorimeter showing all key components

Part Material Function
Inner Vessel Copper or Aluminium Holds water; copper is an excellent thermal conductor for quick heat exchange
Outer Jacket Insulated material Prevents heat loss to the surroundings
Insulating Air Gap Air / Padding Creates thermal isolation between inner and outer vessels
Stirrer Metal (copper) Ensures uniform temperature distribution throughout the water
Thermometer Glass / Digital Measures the temperature of the water accurately
Lid Insulated Prevents heat escape from the top

Analogy showing a thermos flask compared to a calorimeter — both use thermal insulation

Fig.: A calorimeter works just like a thermos flask — both trap heat inside using insulation

The Principle of Calorimetry

The entire working of a calorimeter is based on one fundamental law of physics — the Law of Conservation of Energy. Since the calorimeter is thermally insulated, no heat enters or leaves the system. This means:

Principle of Calorimetry: In an isolated system, the total heat lost by the hot body is exactly equal to the total heat gained by the cold body.
Heat Lost = Heat Gained

Diagram showing heat exchange: hot object loses heat, cold water plus calorimeter gains heat, reaching thermal equilibrium

Fig.: Heat flows from the hot object to the cold water and calorimeter until thermal equilibrium is reached

Maharashtra State Board: Class 11

Experiment: Method of Mixtures

The method of mixtures is the standard experimental technique for determining the specific heat capacity of an unknown substance using a calorimeter.

Procedure:

  1. Fill the Calorimeter: Pour a known mass of water (m₃) into the inner copper vessel. Note the initial temperature (T₂) of both water and calorimeter.
  2. Heat the Sample: Heat the sample (substance whose specific heat you want to find) to a high temperature (T₁). For example, place it in boiling water at 100°C.
  3. Quick Transfer: Quickly transfer the hot sample into the calorimeter water. Speed matters — minimise heat loss during the transfer!
  4. Stir Continuously: Use the stirrer to ensure uniform temperature distribution throughout the water.
  5. Record Final Temperature: Wait until the thermometer shows a stable reading. This is the final equilibrium temperature (T).
  6. Apply the Formula: Use the calorimetry equation: Heat lost by sample = Heat gained by water + Heat gained by calorimeter.
Maharashtra State Board: Class 11

Derivation of Calorimetry Equation

Let's build the master formula step by step. First, define all our variables clearly.

Variable Definitions

Symbol Meaning
m₁ Mass of the hot sample
m₂ Mass of the calorimeter (+ stirrer)
m₃ Mass of water in the calorimeter
s₁ Specific heat capacity of the sample (unknown)
s₂ Specific heat capacity of calorimeter material (e.g., copper = 387 J kg⁻¹ K⁻¹)
s₃ Specific heat capacity of water = 4186 J kg⁻¹ K⁻¹
T₁ Initial temperature of the hot sample
T₂ Initial temperature of calorimeter + water
T Final equilibrium temperature of the entire system

Step-by-Step Derivation

Step 1: Heat lost by the hot sample
Qlost = m₁ × s₁ × (T₁ − T)
The sample cools from T₁ down to T, so ΔT = T₁ − T
Step 2: Heat gained by the calorimeter
Qcal = m₂ × s₂ × (T − T₂)
The calorimeter heats up from T₂ to T, so ΔT = T − T₂
Step 3: Heat gained by the water
Qwater = m₃ × s₃ × (T − T₂)
Water also heats from T₂ to T (same ΔT as the calorimeter)
Step 4: Apply the Principle of Calorimetry
Qlost = Qcal + Qwater
m₁ s₁ (T₁ − T) = m₂ s₂ (T − T₂) + m₃ s₃ (T − T₂)   ... Eq. (1)
Step 5: Solve for the unknown specific heat s₁
s₁ = [ (m₂ s₂ + m₃ s₃) × (T − T₂) ] / [ m₁ × (T₁ − T) ]   ... Eq. (2)

This is the master formula. If you know all masses, temperatures, and specific heats of water and copper, you can calculate s₁ for any unknown material.
Maharashtra State Board: Class 11

Example

Problem: A 0.06 kg aluminium sphere is heated to 100°C and placed in a 0.12 kg copper calorimeter containing 0.30 kg of water at 25°C. The mixture reaches a steady temperature of 28°C. Find the specific heat capacity of aluminium.

Solution:

Given:
Mass of aluminium sphere (m₁) = 0.06 kg
Mass of copper calorimeter (m₂) = 0.12 kg
Mass of water (m₃) = 0.30 kg
Specific heat of copper (s₂) = 387 J kg⁻¹ K⁻¹
Specific heat of water (s₃) = 4180 J kg⁻¹ K⁻¹
Initial temp of sphere (T₁) = 100°C
Initial temp of water + calorimeter (T₂) = 25°C
Final equilibrium temp (T) = 28°C

Step 1: Calculate temperature changes

  • Temperature drop of sphere: T₁ − T = 100 − 28 = 72°C
  • Temperature rise of water & calorimeter: T − T₂ = 28 − 25 = 3°C

Step 2: Apply Eq.

s₁ = [(m₂ × s₂) + (m₃ × s₃)] × (T − T₂) / [m₁ × (T₁ − T)]

Step 3: Substitute values

s₁ = [(0.12 × 387) + (0.30 × 4180)] × 3 / [0.06 × 72]

Step 4: Compute the numerator

  • 0.12 × 387 = 46.44
  • 0.30 × 4180 = 1254
  • Sum = 46.44 + 1254 = 1300.44
  • Numerator = 1300.44 × 3 = 3901.32

Step 5: Compute the denominator

  • 0.06 × 72 = 4.32

Step 6: Final answer

s₁ = 3901.32 / 4.32 = 903.08 J kg⁻¹ K⁻¹
Specific heat capacity of aluminium ≈ 903 J kg⁻¹ K⁻¹
Maharashtra State Board: Class 8, 11

Key Points: Calorimetry

  • Calorimeter: An insulated device used to measure heat transfer during a process.
  • Principle of Calorimetry: Heat lost = Heat gained (energy is conserved).
  • Method of Mixtures: Heat a sample, place it in the calorimeter, and measure the temperature change to calculate the heat.
  • Formula: Used to calculate the specific heat of a substance from measured temperatures and masses.
  • Good Practice: Transfer the sample quickly and stir well for accurate results.
  • Why Water is Used: Water has high specific heat, so it absorbs a lot of heat without a large temperature change.

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