THERMAL EXPANSION
Thermal Expansion
10.1 Examples and Applications of Thermal Expansion
Thermal expansion is the tendency of matter to increase in volume when heated and decrease when cooled. When a substance is heated, the kinetic energy of its molecules increases, causing them to vibrate more vigorously and occupy more space.
Daily Life Examples:
- Railway Tracks: Small gaps are left between rails to allow for expansion in summer
- Concrete Roads: Expansion joints are provided to prevent cracking
- Telephone Wires: Sag in summer due to expansion, become tight in winter
- Thermostats: Bimetallic strips bend when heated to make or break electrical contacts
- Glassware: Thick glass breaks when hot water is poured due to uneven expansion
- Bridge Construction: One end is fixed, other is on rollers to allow expansion
Simple Experiments:
- Ball and Ring Experiment: A metal ball that passes through a ring at room temperature gets stuck after heating, demonstrating expansion
- Bimetallic Strip: Two different metals bonded together bend when heated due to different expansion rates
- Thermometer: Liquid expands in capillary tube showing temperature change
10.2 Types of Thermal Expansion
1. Linear Expansion
When a solid is heated, its length increases. This is called linear expansion.
Δl = l₁αΔT
Where: l₁ = initial length, l₂ = final length, α = coefficient of linear expansion, ΔT = temperature change
Coefficient of Linear Expansion (α):
It is defined as the fractional increase in length per degree rise in temperature.
Unit: K⁻¹ or °C⁻¹
2. Superficial (Area) Expansion
When a solid is heated, its area increases. This is called superficial expansion.
ΔA = A₁βΔT
Where: β = coefficient of superficial expansion
3. Cubical (Volume) Expansion
When a solid is heated, its volume increases. This is called cubical expansion.
ΔV = V₁γΔT
Where: γ = coefficient of cubical expansion
10.3 Relation Between Coefficients of Thermal Expansion
Mathematical Derivation:
Consider a square plate of side 'l' at temperature T.
Initial area A₁ = l²
When heated, new length l₂ = l₁(1 + αΔT)
New area A₂ = [l(1 + αΔT)]² = l²(1 + αΔT)²
Expanding: A₂ = l²(1 + 2αΔT + α²(ΔT)²)
Since α is very small, α²(ΔT)² ≈ 0
Therefore: A₂ = l²(1 + 2αΔT) = A₁(1 + 2αΔT)
Comparing with A₂ = A₁(1 + βΔT): β = 2α
Similarly for volume expansion of a cube:
V₂ = [l(1 + αΔT)]³ = l³(1 + αΔT)³
V₂ = l³(1 + 3αΔT + 3α²(ΔT)² + α³(ΔT)³)
Neglecting higher order terms: V₂ = l³(1 + 3αΔT) = V₁(1 + 3αΔT)
Comparing with V₂ = V₁(1 + γΔT): γ = 3α
Therefore: β = 2α and γ = 3α
10.4 Pullinger's Method for Coefficient of Linear Expansion
Apparatus:
- Metal rod with steam chamber
- Two vernier scales (one fixed, one movable)
- Steam generator
- Thermometer
Procedure:
- Measure initial length of metal rod (l₁) at room temperature (T₁)
- Record initial vernier reading
- Pass steam through the chamber to heat the rod
- Record final temperature (T₂) and vernier reading
- Calculate change in length (Δl) from vernier readings
- Use formula: α = Δl/(l₁ × ΔT)
Where: Δl = change in length, l₁ = original length, ΔT = temperature change
Precautions:
- Ensure uniform heating of the rod
- Take readings when temperature is steady
- Account for thermal expansion of vernier scales
- Use long rod for better accuracy
10.5 Force Due to Expansion and Contraction
When a rod is heated and prevented from expanding, or cooled and prevented from contracting, thermal stress is developed.
Expression for Thermal Stress:
Let a rod of length l, area of cross-section A, and Young's modulus Y be heated through ΔT.
Free expansion = lαΔT
If expansion is prevented, strain = lαΔT/l = αΔT
Stress = Y × strain = YαΔT
Force = Stress × Area = YαΔT × A
Thermal Force = YαΔT × A
Where: Y = Young's modulus, α = coefficient of linear expansion, ΔT = temperature change, A = cross-sectional area
Applications:
- Railway tracks: Heavy rails can develop enormous forces if expansion is prevented
- Concrete roads: Reinforcement prevents cracking due to thermal stress
- Bridge construction: Expansion joints relieve thermal stress
10.6 Differential Expansion
Differential expansion occurs when different materials expand by different amounts when subjected to the same temperature change due to their different coefficients of expansion.
Applications:
- Bimetallic Strips: Used in thermostats, fire alarms, and temperature control devices
- Compensated Pendulum: Invar and brass used to maintain constant length
- Reinforced Concrete: Steel and concrete have similar expansion coefficients
- Thermostats: Bimetallic strip bends to make/break electrical contact
Where: α₁, α₂ = coefficients of expansion of two materials, l = length, ΔT = temperature change
Bimetallic Strip Working:
Two strips of different metals (e.g., brass and steel) are riveted together. When heated, brass (α = 19 × 10⁻⁶/°C) expands more than steel (α = 12 × 10⁻⁶/°C), causing the strip to bend with brass on the outer side.
10.7 Variation of Density with Temperature
When a substance is heated, its volume increases while mass remains constant, leading to decrease in density.
Mathematical Expression:
Density ρ = m/V
Initial: ρ₁ = m/V₁
Final: ρ₂ = m/V₂ = m/[V₁(1 + γΔT)]
Therefore: ρ₂ = ρ₁/(1 + γΔT)
For small γΔT: ρ₂ ≈ ρ₁(1 - γΔT)
Where: ρ₁ = initial density, ρ₂ = final density, γ = coefficient of cubical expansion
Applications:
- Hot Air Balloons: Heated air becomes less dense and rises
- Ocean Currents: Temperature variations cause density differences
- Thermometers: Mercury expands and contracts with temperature
10.8 Real and Apparent Expansion of Liquids
Real Expansion:
Actual expansion of liquid when heated, measured when container expansion is accounted for.
Apparent Expansion:
Expansion of liquid as observed, without accounting for container expansion.
Relation Between Real and Apparent Expansion:
When a liquid in a container is heated:
Apparent expansion of liquid = Real expansion of liquid - Expansion of container
VₐγₐΔT = VᵣγᵣΔT - Vₐγ₉ΔT
(Since volume of liquid = volume of container)
γₐ = γᵣ - γ₉
Therefore: γᵣ = γₐ + γ₉
Where: γᵣ = coefficient of real expansion, γₐ = coefficient of apparent expansion, γ₉ = coefficient of expansion of container
Experiment:
Liquid in a flask with a long narrow stem shows apparent expansion. Real expansion is calculated by accounting for glass expansion.
10.9 Dulong and Petit's Experiment
This experiment determines the absolute expansivity (real expansion) of a liquid using a weight thermometer.
Apparatus:
- Weight thermometer (glass bulb with capillary tube)
- Balance
- Water bath
- Liquid whose expansion is to be determined
Procedure:
- Weigh the empty weight thermometer (W₁)
- Fill with liquid at temperature T₁ and weigh (W₂)
- Heat to temperature T₂ and weigh again (W₃)
- Mass of liquid expelled = (W₂ - W₃)
- Mass of liquid remaining = (W₃ - W₁)
Calculation:
Volume of liquid expelled = Volume of liquid remaining × γ × ΔT
(W₂ - W₃)/ρ₂ = (W₃ - W₁)/ρ₂ × γ × ΔT
γ = (W₂ - W₃)/[(W₃ - W₁) × ΔT]
Since ρ₂ ≈ ρ₁ for small temperature changes
Where: W₁ = weight of empty thermometer, W₂ = weight at T₁, W₃ = weight at T₂, ΔT = T₂ - T₁
10.10 Numerical Problems
Problem 1:
A steel rod is 100 cm long at 20°C. Find its length at 80°C if coefficient of linear expansion of steel is 1.2 × 10⁻⁵/°C.
Solution:
Given: l₁ = 100 cm, T₁ = 20°C, T₂ = 80°C, α = 1.2 × 10⁻⁵/°C
ΔT = 80 - 20 = 60°C
l₂ = l₁(1 + αΔT) = 100[1 + 1.2 × 10⁻⁵ × 60]
l₂ = 100[1 + 7.2 × 10⁻⁴] = 100 × 1.00072 = 100.072 cm
Problem 2:
A rectangular sheet of metal measures 20 cm × 10 cm at 15°C. Find the change in area when heated to 65°C. Coefficient of linear expansion = 1.8 × 10⁻⁵/°C.
Solution:
Given: A₁ = 20 × 10 = 200 cm², ΔT = 65 - 15 = 50°C, α = 1.8 × 10⁻⁵/°C
Coefficient of superficial expansion β = 2α = 2 × 1.8 × 10⁻⁵ = 3.6 × 10⁻⁵/°C
ΔA = A₁βΔT = 200 × 3.6 × 10⁻⁵ × 50 = 0.36 cm²
Problem 3:
A copper cube of side 10 cm is heated from 20°C to 120°C. Calculate the increase in volume. α for copper = 1.7 × 10⁻⁵/°C.
Solution:
Given: l = 10 cm, V₁ = 10³ = 1000 cm³, ΔT = 100°C, α = 1.7 × 10⁻⁵/°C
γ = 3α = 3 × 1.7 × 10⁻⁵ = 5.1 × 10⁻⁵/°C
ΔV = V₁γΔT = 1000 × 5.1 × 10⁻⁵ × 100 = 0.51 cm³
Problem 4:
A steel wire 2 m long at 20°C is heated to 80°C while being fixed at both ends. Calculate the thermal stress developed. Y for steel = 2 × 10¹¹ N/m², α = 1.2 × 10⁻⁵/°C.
Solution:
Given: l = 2 m, ΔT = 60°C, Y = 2 × 10¹¹ N/m², α = 1.2 × 10⁻⁵/°C
Thermal stress = YαΔT = 2 × 10¹¹ × 1.2 × 10⁻⁵ × 60
Thermal stress = 1.44 × 10⁸ N/m²
Problem 5:
The coefficient of apparent expansion of a liquid is 1.2 × 10⁻³/°C and that of the container is 3 × 10⁻⁵/°C. Find the real coefficient of expansion of the liquid.
Solution:
Given: γₐ = 1.2 × 10⁻³/°C, γ₉ = 3 × 10⁻⁵/°C
γᵣ = γₐ + γ₉ = 1.2 × 10⁻³ + 3 × 10⁻⁵ = 1.23 × 10⁻³/°C