Physics Lesson 13.2.3 - Area Thermal Expansion and Contraction

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Welcome to our Physics lesson on Area Thermal Expansion and Contraction, this is the third lesson of our suite of physics lessons covering the topic of Thermal Expansion, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Area Thermal Expansion and Contraction

When an object is foil-like, it has two relevant dimensions, length and width. Height is very small to take into consideration. Therefore, another coefficient, known as area thermal expansion coefficient, β is introduced. Mathematically, we ca write β = 2α for the same material, because each dimension experiences the same degree of thermal expansion or contraction. For example, since the linear expansion coefficient of iron is 11.8 × 10^-6 K-1 (as discussed in the previous example), the area thermal expansion coefficient of iron is 2 × 11.8 × 10^-6 K-1 = 23.6 × 10^-6 K-1. The formula is similar to that used for linear thermal expansion, i.e.

A = A₀ × (1 + β × ∆T)

where A is the final area and A₀ is the original area, and ΔT = T - T₀ is the change in temperature.

Example 2

A tin sheet must cover the upper part of a rectangular storage room of floor dimensions 5 m × 4 m as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Thermal Expansion

If the linear expansion coefficient of tin (in 200 C is 20 × 10^-6 K - 1, calculate the minimum area of the tin cover, so that the materials inside the storage room be protected from weather conditions. Take the minimum temperature in winter in the specific region equal to -30°C. Suppose that the setup is made at normal weather conditions, i.e. at 20°C.

Solution 2

First, we calculate the area coefficient of thermal expansion, β. Thus, given that the linear coefficient of thermal expansion for tin is α = 20 × 10^-6 K-1, we obtain for the corresponding area coefficient of thermal expansion β:

β = 2 × α
= 2 × 20 × 10^-6 K^-1
= 40 × 10^-6 K^-1

The area to be covered represents the original area A₀ in the formula. Thus, we have

A₀ = 5m × 4m
= 20 m²

Also, the change in temperature is calculated by taking the difference between final and initial temperature, i.e.

∆t = t - t₀
= -30°C - 20°C
= -50°C

Therefore, the area of the tin sheet in winter after experiencing thermal contraction is:

A = A₀ × (1 + β × ∆T)
= 20 × [1 + 40 × 10^-6 × (-50)]
= 20 × (1 - 2000 × 10^-6 )
= 20 × (1 - 2 × 10^-3 )
= 20 × (1 - 0.002)
= 20 × 0.998 = 19.96 m²

This means at least other 0.04 m₂ of tin are needed at the setup moment to make possible the entire area coverage of the room. Therefore, at least 20 m₂ + 0.04 m₂ = 20.04 m₂ of tin are needed to cover the entire room in order to prevent issues in winter.

You have reached the end of Physics lesson 13.2.3 Area Thermal Expansion and Contraction. There are 7 lessons in this physics tutorial covering Thermal Expansion, you can access all the lessons from this tutorial below.

More Thermal Expansion Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.2	Thermal Expansion
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.2.1	What is Thermal Expansion and Thermal Contraction?
13.2.2	Linear Thermal Expansion and Contraction
13.2.3	Area Thermal Expansion and Contraction
13.2.4	Volume Thermal Expansion and Contraction
13.2.5	Thermal Expansion and Contraction in Daily Life
13.2.6	Bimetallic Strip. Thermostat
13.2.7	Types of Thermometers

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