# Area expansion – problems and solutions

1. At 20 oC, 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 oC-1, determine the change in area and the final area at 60 oC.

Known :

The initial temperature (T1) = 20oC

The final temperature (T2) = 60oC

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

The initial area (A1) = length x width = 50 cm x 30 cm = 1500 cm2

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

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

Wanted : The change in area (ΔA)

Solution :

The change in area (ΔA) :

ΔA = β A1 ΔT

ΔA = (2 x 10-5 oC-1)(1500 cm2)(40oC)

ΔA = (80 x 10-5)(1500 cm2)

ΔA = 120,000 x 10-5 cm2

ΔA = 1.2 x 105 x 10-5 cm2

ΔA = 1.2 cm2

The final area (A2) :

A2 = A1 + ΔA

A2 = 1500 cm2 + 1.2 cm2

A2 = 1501.2 cm2

2. At 30 oC, the area of a sheet of aluminum is 40 cm2 and the coefficient of linear expansion is 24 x 10-6 /oC. Determine the final temperature if the final area is 40.2 cm2.

Known :

The initial temperature (T1) = 30oC

The coefficient of linear expansion (α) = 24 x 10-6 oC-1

The coefficient of area expansion (β) = 2a = 2 x 24 x 10-6 oC-1 = 48 x 10-6 oC-1

The initial area (A1) = 40 cm2

The final area (A2) = 40.2 cm2

The change in area (ΔA) = 40.2 cm2 – 40 cm2 = 0.2 cm2

Wanted : Determine the final temperature (T2)

Solution :

Formula of the change in area (ΔA) :

ΔA = β A1 ΔT

The final temperature (T2) :

ΔA β A1 (T2 – T1)

0.2 cm2 = (48 x 10-6 oC-1)(40 cm2)(T230oC)

0.2 = (1920 x 10-6)(T230)

0.2 = (1.920 x 10-3)(T2 – 30)

0.2 = (2 x 10-3)(T2 – 30)

0.2 / (2 x 10-3) = T2 – 30

0.1 x 103 = T2 – 30

1 x 102 = T2 – 30

100 = T2 – 30

100 + 30 = T2

T2 = 130

The final temperature = 130oC

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

Known :

The initial temperature (T1) = 30oC

The final temperature (T2) = 100oC

The change in temperature (ΔT) = 100oC – 30oC = 70oC

The initial radius (r1) = 20 cm

The final radius (r2) = 20.5 cm

Wanted : The coefficient of area expansion (β)

Solution :

The initial area (A1) = π r12 = (3.14)(20 cm)2 = (3.14)(400 cm2) = 1256 cm2

The final area (A2) = π r22 = (3.14)(20.5 cm)2 = (3.14)(420.25 cm2) = 1319.585 cm2

The change in area (ΔA) = 1319.585 cm2 1256 cm2 = 63.585 cm2

Formula of the change in area (ΔA) :

ΔA = β A1 ΔT

The coefficient of area expansion :

ΔA = β A1 ΔT

63.585 cm2 = b (1256 cm2)(70 oC)

63.585 = b (87,920 oC)

β = 63.585 / 87,920 oC

β = 0.00072 /oC

β = 7.2 x 10-4 /oC

β = 7.2 x 10-4 oC-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 oC-1

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