# Area expansion – problems and solutions

1. At 20 ^{o}C, the length of a sheet of steel is 50 cm and the width is 30 cm. If the coefficient of linear expansion for steel is 10^{-5} ^{o}C^{-1}, determine the change in area and the final area at 60 ^{o}C.

__Known :__

The initial temperature (T_{1}) = 20^{o}C

The final temperature (T_{2}) = 60^{o}C

The change in temperature (ΔT) = 60^{o}C – 20^{o}C = 40^{o}C

The initial area (A_{1}) = length x width = 50 cm x 30 cm = 1500 cm^{2}

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

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

__Wanted :__ The change in area (ΔA)

__Solution :__

The change in area (ΔA) :

ΔA = β A_{1} ΔT

ΔA = (2 x 10^{-5} ^{o}C^{-1})(1500 cm^{2})(40^{o}C)

ΔA = (80 x 10^{-5})(1500 cm^{2})

ΔA = 120,000 x 10^{-5 }cm^{2}

ΔA = 1.2 x 10^{5} x 10^{-5 }cm^{2}

ΔA = 1.2 cm^{2}

The final area (A_{2}) :

A_{2 }= A_{1 }+ ΔA

A_{2 }= 1500 cm^{2} + 1.2 cm^{2}

A_{2} = 1501.2 cm^{2}

2. At 30 ^{o}C, the area of a sheet of aluminum is 40 cm^{2 }and the coefficient of linear expansion is 24 x 10^{-6} /^{o}C. Determine the final temperature if the final area is 40.2 cm^{2}.

__Known :__

The initial temperature (T_{1}) = 30^{o}C

The coefficient of linear expansion (α) = 24 x 10^{-6} ^{o}C^{-1}

The coefficient of area expansion (β) = 2a = 2 x 24 x 10^{-6} ^{o}C^{-1} = 48 x 10^{-6} ^{o}C^{-1}

The initial area (A_{1}) = 40 cm^{2}

The final area (A_{2}) = 40.2 cm^{2}

The change in area (ΔA) = 40.2 cm^{2 }– 40 cm^{2} = 0.2 cm^{2}

__Wanted : __ Determine the final temperature (T_{2})

__Solution :__

Formula of the change in area (ΔA) :

ΔA = β A_{1 }ΔT

The final temperature (T_{2}) :

ΔA = β A_{1 }(T_{2} – T_{1})

0.2 cm^{2} = (48 x 10^{-6} ^{o}C^{-1})(40 cm^{2})(T_{2} – 30^{o}C)

0.2 = (1920 x 10^{-6})(T_{2} – 30)

0.2 = (1.920 x 10^{-3})(T_{2} – 30)

0.2 = (2 x 10^{-3})(T_{2} – 30)

0.2 / (2 x 10^{-3}) = T_{2} – 30

0.1 x 10^{3} = T_{2} – 30

1 x 10^{2} = T_{2} – 30

100 = T_{2 }– 30

100 + 30 = T_{2}

T_{2} = 130

The final temperature = 130^{o}C

3. The radius of a ring at 20 ^{o}C is 20 cm. If the final radius at 100 ^{o}C is 20.5 cm, determine the coefficient of area expansion and the coefficient of linear expansion…

__Known :__

The initial temperature (T_{1}) = 30^{o}C

The final temperature (T_{2}) = 100^{o}C

The change in temperature (ΔT) = 100^{o}C – 30^{o}C = 70^{o}C

The initial radius (r_{1}) = 20 cm

The final radius (r_{2}) = 20.5 cm

__Wanted :__ The coefficient of area expansion (β)

__Solution :__

The initial area (A_{1}) = π r_{1}^{2} = (3.14)(20 cm)^{2} = (3.14)(400 cm^{2}) = 1256 cm^{2 }

The final area (A_{2}) = π r_{2}^{2} = (3.14)(20.5 cm)^{2} = (3.14)(420.25 cm^{2}) = 1319.585 cm^{2 }

The change in area (ΔA) = 1319.585 cm^{2 }– 1256 cm^{2} = 63.585 cm^{2 }

Formula of the change in area (ΔA) :

ΔA = β A_{1 }ΔT

The coefficient of area expansion :

ΔA = β A_{1 }ΔT

63.585 cm^{2 }= b (1256 cm^{2})(70 ^{o}C)

63.585 = b (87,920 ^{o}C)

β = 63.585 / 87,920 ^{o}C

β = 0.00072 /^{o}C

β = 7.2 x 10^{-4} /^{o}C

β = 7.2 x 10^{-4} ^{o}C^{-1}

The coefficient of linear expansion (α) :

β = 2 α

α = β / 2

α = (7.2 x 10^{-4}) / 2

α = 3.6 x 10^{-4 }^{o}C^{-1}

**Ebook PDF area expansion sample problems with solutions 82.52 KB**

- Converting temperature scales
- Linear expansion
- Area expansion
- Volume expansion
- Heat
- Mechanical equivalent of heat
- Specific heat and heat capacity
- Latent heat, heat of fusion, heat of vaporization
- Energy conservation for heat transfer