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

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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

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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

  1. Converting temperature scales
  2. Linear expansion
  3. Area expansion
  4. Volume expansion
  5. Heat
  6. Mechanical equivalent of heat
  7. Specific heat and heat capacity
  8. Latent heat, heat of fusion, heat of vaporization
  9. Energy conservation for heat transfer