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)(T2 – 30oC)
0.2 = (1920 x 10-6)(T2 – 30)
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
[wpdm_package id=’698′]
- 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
