Table of contents |
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What is Thermal Expansion? |
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Types of Expansion |
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Water's Unique Behavior |
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Practical Applications of Expansion |
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Solved Examples |
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Most substances expand with heat and contract with cold. This change in dimensions with temperature variation is known as thermal expansion.
Example of Thermal Expansion
There are 3 types of expansion which occur in solids-
Here, αA represents the coefficient of area expansion of the given solid.
Here, αV is another characteristic of the substance, but it varies with temperature.
The graph depicts the Coefficient of volume expansion of copper concerning temperature:
It shows how the Coefficient of volume expansion of copper changes with temperature.
Example 1: A steel tape 1 m long is correctly calibrated for a temperature of 27°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27°C? Coefficient of linear expansion of steel = 1.20 × 10⁻⁶ °C⁻¹?
Sol: Length of rod at 45.0°C = 63 cm
t₁ = 27.0°C, t₂ = 45.0°C, α = 1.20 × 10⁻⁶ °C⁻¹
Length of steel rod at 45°C is given by
L₂ = L₁ [1 + α(t₂ - t₁)]
= 63 [1 + 1.2 × 10⁻⁶ × (45 - 27)]
= 63 [1 + 1.2 × 10⁻⁶ × 18]
= 63.0136 cm
The change in length = 63.0136 - 63 = 0.0136 cm.
As the data is correct up to three significant figures, hence the actual length of rod at 27°C as measured by tape = 63.0 cm.
Example 2: A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 m? Coefficient of linear expansion of brass = 2.0 × 10⁻⁵ °C⁻¹; Young’s modulus of brass = 0.91 × 10¹¹ Pa.
Solution:When the wire is cooled from 27°C to -39°C, the change in temperature Δt = 27 - (-39) = 66°C. The corresponding decrease in length Δl = αlΔt.
This will produce the contraction strain in the wire given by:
Thus, the stress produced = Y × strain = Y × α × Δt.
Now the force or tension in the wire = stress × area of cross-section
Substitute the values:
=
So, the tension developed in the wire is approximately 370 N.
Example 3: Determine the pressure needed to maintain the original length of a steel wire when its temperature is increased by 100°C. Given: ΔT = 100°C, Y = 2 x 1011 Nm−2 , α = 1.1 x 10−5 K−1 .
Sol: Given:
Thermal Strain (ε) = αΔT
ε = (1.1 x 10−5 K−1 ) × (100 K)
ε = 1.1 x 10−3
The thermal strain represents the fractional change in length due to the temperature increase. To maintain the original length of the wire, this strain must be countered by an applied stress (σ).
σ = Y × ε
⇒ F = σ × A
Where: A = Cross-sectional area of the wire
Substituting the values:
F = (Y × ε) × A
F = (2 x 1011 Nm−2 ) × (1.1 x 10−3 ) × A
F = 2.2 x 108 N/m2 × A
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1. What is thermal expansion and how does it occur? | ![]() |
2. What are the different types of thermal expansion? | ![]() |
3. Why is water's behavior during thermal expansion unique? | ![]() |
4. What are some practical applications of thermal expansion? | ![]() |
5. Can thermal expansion lead to any problems in engineering or construction? | ![]() |