मराठी

Calorimetry

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Estimated time: 23 minutes
  • Introduction
  • Definition: Calorimetry
  • Definition: Calorimeter
  • Calorimeter
  • Law: 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).

OR

An experimental technique for the quantitative measurement of heat exchange is called calorimetry.

CBSE: Class 12
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

CBSE: Class 12

Law: Principle of Calorimetry

Statement: When different parts of an isolated system are at different temperatures, heat transfers from the part at higher temperature to the part at lower temperature. The heat lost by the hot object is equal to the heat gained by the cold object, provided no heat is allowed to escape to the surroundings.

Heat lost by hot body = Heat gained by cold body
m1c1Δθ1 = m2c2Δθ2

(For liquid in calorimeter: m1c1Δθ + mcccΔθ)

Key Points:

  • A system is said to be isolated if no exchange of heat occurs between the system and its surroundings.
  • Calorimetry literally means measurement of heat.
  • Energy supplied by heater = VIt (voltage × current × time).
  • This principle is based on the Law of Conservation of Energy.
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

  • A calorimeter is an insulated device used to measure heat transfer; measurement of specific heat of a substance is carried out using it.
  • Principle of Calorimetry: Heat lost by hot body = Heat gained by cold body, which represents the law of conservation of heat energy.
  • In the method of mixtures, a heated sample is placed in the calorimeter and the temperature change is measured to calculate specific heat using the formula Q = msΔt.
  • Specific heat of a substance depends on the nature of the substance; water is preferred in calorimetry due to its high specific heat, allowing it to absorb large amounts of heat with minimal temperature change.
  • For accurate results, the sample must be transferred quickly into the calorimeter and stirred well to ensure uniform heat distribution.

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