Thermal expansion – problems and solutions

Thermal expansion – problems and solutions

Area expansion

1. A sheet of steel at 20^oC has size as shown in the figure below. If the coefficient of linear expansion for steel is 10^{-5 o}C^-1 then what is the change in the area at 60^oC.

Known :

Length of steel = 40 cm

Width of steel = 20 cm

The initial of steel’s area (A_o) = (40)(20) = 800 cm²

The coefficient of linear expansion (α) = 10^{-5 o}C^-1

The coefficient of area expansion (β) = 2 x coefficient of linear expansion (2α) = 2 x 10^{-5 o}C^-1

The change in temperature (ΔT) = 60^oC – 20^oC = 40^oC

Wanted : The change in area of steel at 60^oC

Solution :

Equation of area expansion :

ΔA = β A_o ΔT

ΔA = the increase in area of steel, β = The coefficient of area expansion, A_o = initial area, ΔT = the change in temperature = final temperature – initial temperature

The increase in area of steel :

ΔA = β A_o ΔT

ΔA = (2 x 10^-5)(800)(40) = 0.64 cm

2. A plate of iron at 20^oC has shown in figure below. If the temperature is raised to 100^oC and the coefficient of linear expansion of iron is 1.1 x 10^{-7 o}C^-1, then what is the final area of plate.

Known :

Length of plate = 2 m

Width of plate = 2 m

The initial area of iron (A_o) = (2)(2) = 4 m²

The coefficient of linear expansion for iron (α) = 1.1 x 10^{-7 o}C^-1

The coefficient of area expansion for iron (β) = 2 x the coefficient of linear expansion for iron (2α) = 2.2 x 10^{-7 o}C^-1

The change in temperature (ΔT) = 100^oC – 20^oC = 80^oC

Wanted : Area of iron at 100^oC

Solution :

The increase in length :

ΔA = β A_o ΔT

ΔA = (2.2 x 10^-7)(4)(80) = 704 x 10^-7 = 0,0000704 m²

Area of iron :

Area of iron = initial area + the increase in area

Area of iron = 4 m² + 0.0000704 m²

Area of iron = 4.0000704 m²

3. A bronze plate with the coefficient of linear expansion α = 18.10^{-6 o}C^-1 at 0^oC has size as shown in figure below. If the plate heated at 80 ^oC, then what is the increase in area of plate.

Known :

The length of bronze = 40 cm = 0.4 meters

Width of bronze = 20 cm = 0.2 meters

Initial area of bronze (A_o) = (0.4)(0.2) = 0.08 m²

The coefficient of linear expansion for bronze (α) = 18 x 10^{-6 o}C^-1

The coefficient of area expansion for bronze (β) = 2 x The coefficient of linear expansion (2α) = 36 x 10^{-6 o}C^-1

The change in temperature (ΔT) = 80^oC – 0^oC = 80^oC

Wanted : The increase of area for bronze at 80^oC

Solution :

The increase of area for bronze :

ΔA = β A_o ΔT

ΔA = (36 x 10^-6)(0.08)(80) = 230.4 x 10^-6 = 2.304 x 10^-4 m²

Volume expansion

4. A glass container with volume of 4 liters filled with water, then heated until the increase in temperature is 20^oC. Some water spilled. The coefficient of linear expansion for glass = 9 x 10^{-6 o}C^-1; the coefficient of volume expansion for water = 2.1 x 10^{-4 o}C^-1. Determine the volume of spilled water.

Known :

The initial volume of the gas and water (V_o) = 4 liters

The increase in temperature of the glass and water (ΔT) = 20^oC

The coefficient of linear expansion for glass (α) = 9 x 10^{-6 o}C^-1

The coefficient of volume expansion for glass (γ) = 3α = 3 (9 x 10^{-6 o}C^-1) = 27 x 10^{-6 o}C^-1

The coefficient of volume expansion for water (γ) = 2.1 x 10^{-4 o}C^-1

Wanted : Volume of spilled water

Solution :

The equation of the volume expansion :

V = V_o + γ V_o ΔT

V – V_o = γ V_o ΔT

ΔV = γ V_o ΔT

V = final volume, V_o = initial volume, ΔV = the change in volume, γ = the coefficient of volume expansion, ΔT = the change in temperature.

The change in volume of the glass container :

ΔV = γ V_o ΔT = (27 x 10^-6)(4)(20) = 2160 x 10^-6 = 2.160 x 10^-3 = 0.002160 liters

The change in volume of the water :

ΔV = γ V_o ΔT = (2.1 x 10^-4)(4)(20) = 168 x 10^-4 = 0.0168 liters

The change in volume of the water is greater than the glass container, so some water spills.

The volume of spilled water :

0.0168 liters – 0.002160 liters = 0.01464 liters = 0.015 liters

5. A steel container (the coefficient of linear expansion = 10^{-5 o}C^-1) with volume of 6 liters filled with acetone (the coefficient of volume expansion = 1.5 x 10^{-3 o}C^-1). If the container and acetone are heated from 0^oC to 40^oC, what is the volume of spilled acetone?

Known :

The initial volume of the container and acetone (V_o) = 6 liters

The change in temperature of the container and acetone (ΔT) = 40^oC

The coefficient of linear expansion for steel (α) = 10^{-5 o}C^-1

The coefficient of volume expansion for steel (γ) = 3α = 3 (10^{-5 o}C^-1) = 3 x 10^{-5 o}C^-1

The coefficient of volume expansion for acetone (γ) = 1.5 x 10^{-3 o}C^-1

Wanted : The volume of spilled acetone

Solution :

The equation of volume expansion :

ΔV = γ V_o ΔT

ΔV = the change in volume, γ = the coefficient of volume expansion, V_o = initial volume, ΔT = the change in temperature.

The change in volume of the steel container :

ΔV = γ V_o ΔT = (3 x 10^-5)(6)(40) = 720 x 10^-5 = 0.00720 liters

The change in volume of the acetone :

ΔV = γ V_o ΔT = (1.5 x 10^-3)(6)(40) = 360 x 10^-3 = 0.360 liters

The change in volume of the acetone is greater than the steel container, so some acetone spills.

The volume of acetone spilled :

0.360 liters – 0.00720 liters = 0.3528 liters = 0.35 liters

1. What is thermal expansion?
   • Answer: Thermal expansion refers to the tendency of matter to change its volume in response to a change in temperature.
2. How does thermal expansion affect the density of a substance?
   • Answer: As a substance undergoes thermal expansion, its volume increases, which leads to a decrease in its density, assuming its mass remains constant.
3. Why do gaps exist between sections of bridges and railway tracks?
   • Answer: These gaps, often called expansion joints, are designed to accommodate the expansion and contraction of the material due to temperature changes, preventing potential deformation or damage.
4. What’s the difference between linear, volumetric, and area expansion?
   • Answer: Linear expansion pertains to the change in one dimension (like the length of a rod); area expansion refers to the change in two dimensions (like the surface area of a sheet); and volumetric expansion relates to the change in all three dimensions (like the volume of a liquid).
5. How is the coefficient of linear expansion defined?
   • Answer: It is defined as the fractional change in length per degree change in temperature at a constant pressure. For a substance, the change in length (∆L) due to a temperature change (∆T) is given by ∆L=αL₀​∆T, where L₀​ is the initial length and $\alpha$ is the coefficient of linear expansion.
6. Why do bimetallic strips bend when heated or cooled?
   • Answer: A bimetallic strip consists of two different metals bonded together. Since each metal has its own coefficient of thermal expansion, they expand or contract at different rates when the temperature changes. This differential expansion causes the strip to bend.
7. How does the phenomenon of thermal expansion relate to the rising sea levels?
   • Answer: A portion of the rise in sea levels can be attributed to the thermal expansion of seawater. As the Earth’s temperature rises, the oceans warm up, leading to the expansion of water and a consequent increase in sea levels.
8. What happens in “anomalous expansion of water”?
   • Answer: Unlike most substances, water expands as it cools from 4°C to 0°C, and then contracts when it freezes. This anomaly means that water has its maximum density at 4°C. It’s why ice (which is less dense than liquid water) floats on water.
9. Why is thermal expansion important in engineering and construction?
   • Answer: Materials expand or contract as temperatures change. Without accounting for thermal expansion, structures may experience undue stress, deformation, or failure. Engineers and architects incorporate design features to handle these changes safely.
10. Can thermal expansion be reversed?

• Answer: Yes, typically when a material is cooled, it will undergo thermal contraction, which is the opposite of thermal expansion. The degree and nature of this contraction will depend on the material and the conditions.