Thermal expansion – problems and solutions

Area expansion

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

Known :

Length of steel = 40 cm Thermal expansion - problems and solutions 1

Width of steel = 20 cm

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

The coefficient of linear expansion (α) = 10-5 oC-1

The coefficient of area expansion (β) = 2 x coefficient of linear expansion (2α) = 2 x 10-5 oC-1

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

Wanted : The change in area of steel at 60oC

Solution :

Equation of area expansion :

ΔA = β Ao ΔT

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

The increase in area of steel :

ΔA = β Ao ΔT

ΔA = (2 x 10-5)(800)(40) = 0.64 cm

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

Known :

Length of plate = 2 m Thermal expansion - problems and solutions 2

Width of plate = 2 m

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

The coefficient of linear expansion for iron (α) = 1.1 x 10-7 oC-1

The coefficient of area expansion for iron (β) = 2 x the coefficient of linear expansion for iron (2α) = 2.2 x 10-7 oC-1

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

Wanted : Area of iron at 100oC

Solution :

The increase in length :

ΔA = β Ao 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

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3. A bronze plate with the coefficient of linear expansion α = 18.10-6 oC-1 at 0oC 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 metersThermal expansion - problems and solutions 3

Width of bronze = 20 cm = 0.2 meters

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

The coefficient of linear expansion for bronze (α) = 18 x 10-6 oC-1

The coefficient of area expansion for bronze (β) = 2 x The coefficient of linear expansion (2α) = 36 x 10-6 oC-1

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

Wanted : The increase of area for bronze at 80oC

Solution :

The increase of area for bronze :

ΔA = β Ao ΔT

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

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Volume expansion

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

Known :

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

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

The coefficient of linear expansion for glass (α) = 9 x 10-6 oC-1

The coefficient of volume expansion for glass (γ) = 3α = 3 (9 x 10-6 oC-1) = 27 x 10-6 oC-1

The coefficient of volume expansion for water (γ) = 2.1 x 10-4 oC-1

Wanted : Volume of spilled water

Solution :

The equation of the volume expansion :

V = Vo + γ Vo ΔT

V – Vo = γ Vo ΔT

ΔV = γ Vo ΔT

V = final volume, Vo = 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 Vo 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 Vo 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

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

Known :

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

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

The coefficient of linear expansion for steel (α) = 10-5 oC-1

The coefficient of volume expansion for steel (γ) = 3α = 3 (10-5 oC-1) = 3 x 10-5 oC-1

The coefficient of volume expansion for acetone (γ) = 1.5 x 10-3 oC-1

Wanted : The volume of spilled acetone

Solution :

The equation of volume expansion :

ΔV = γ Vo ΔT

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

The change in volume of the steel container :

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

The change in volume of the acetone :

ΔV = γ Vo 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