# Thermal expansion – problems and solutions

1. A sheet of steel at 20^{o}C 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^{o}C.

__Known :__

Length of steel = 40 cm

Width of steel = 20 cm

The initial of steel’s area (A_{o}) = (40)(20) = 800 cm^{2}

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^{o}C – 20^{o}C = 40^{o}C

__Wanted :__

^{o}C

__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^{o}C has shown in figure below. If the temperature is raised to 100^{o}C 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 m^{2}

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^{o}C – 20^{o}C = 80^{o}C

__Wanted ____:__ Area of iron at 100^{o}C

__Solution :__

The increase in length :

ΔA = β A_{o }ΔT

ΔA = (2.2 x 10^{-7})(4)(80) = 704 x 10^{-7 }= 0,0000704 m^{2}

__Area of iron :__

Area of iron = initial area + the increase in area

Area of iron = 4 m^{2 }+ 0.0000704 m^{2}

Area of iron = 4.0000704 m^{2}

3. A bronze plate with the coefficient of linear expansion α = 18.10^{-6} ^{o}C^{-1 }at 0^{o}C has size as shown in figure below. If the plate heated at 80 ^{o}C, 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 m^{2}

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^{o}C – 0^{o}C = 80^{o}C

__Wanted :__ The increase of area for bronze at 80^{o}C

__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} m^{2}

4. A glass container with volume of 4 liters filled with water, then heated until the increase in temperature is 20^{o}C. 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^{o}C

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^{o}C to 40^{o}C, 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^{o}C

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 s**pills.*

The volume of acetone spilled :

0.360 liters – 0.00720 liters = 0.3528 liters = 0.35 liters