Linear expansion – problems and solutions

1. A steel is 40 cm long at 20 ^oC. The coefficient of linear expansion for steel is 12 x 10^-6 (C^o)^-1. The increase in length and the final length when it is at 70 ^oC will be…

Known :

The change in temperature (ΔT) = 70^oC – 20^oC = 50^oC

The original length (L₁) = 40 cm

Coefficient of linear expansion for steel (α) = 12 x 10^-6 (C^o)^-1

Wanted : The change in length (ΔL) and the final length (L₂)

Solution :

a) The change in length (ΔL)

ΔL = α L₁ ΔT

ΔL = (12×10^{-6 o}C^-1)(40 cm)(50^oC)

ΔL = (10^-6)(24 x 10³) cm

ΔL = 24 x 10^-3 cm

ΔL = 24 / 10³ cm

ΔL = 24 / 1000 cm

ΔL = 0.024 cm

b) The final length (L₂)

L₂ = L₁ + ΔL

L₂ = 40 cm + 0.024 cm

L₂ = 40.024 cm

2. An iron rod heated from 30 ^oC to 80 ^oC. The final length of iron is 115 cm and the coefficient of linear expansion is 3×10^{-3 o}C^-1. What is the original length and the change in length of the iron ?

Solution :

The change in temperature (ΔT) = 80 ^oC – 30 ^oC = 50 ^oC

The final length (L₂) = 115 cm

The coefficient of linear expansion (α) = 3×10^{-3 o}C^-1

Wanted : the original length (L₁) and the change in length (ΔL)

Solution :

a) The original length (L₁)

Formula of the change in length for the linear expansion :

ΔL = α L₁ ΔT

Formula of the final length :

L₂ = L₁ + ΔL

L₂ = L₁ + α L₁ ΔT

L₂ = L₁ (1 + α ΔT)

115 cm = L₁ (1 + (3.10^{-3 o}C^-1)(50^oC)

115 cm = L₁ (1 + 150.10^-3)

115 cm = L₁ (1 + 0.15)

115 cm = L₁ (1.15)

L1 = 115 cm / 1.15 L1 = 100 cm b) the change in length (ΔL) ΔL = L2 – L1 ΔL = 115 cm – 100 cm ΔL = 15 cm

3. At 25 ^oC, the length of the glass is 50 cm. After heated, the final length of the glass is 50.9 cm. The coefficient of linear expansion is α = 9 x 10^-6 C^-1. Determine the final temperature of the glass…

Known :

The original length (L₁) = 50 cm

The final length (L₂) = 50.09 cm

The change in length (ΔL) = 50.2 cm – 50 cm = 0.09 cm

The coefficient of linear expansion (α) = 9 x 10^{-6 o}C^-1

The original temperature (T₁) = 25^oC

Wanted : The final temperature (T₂)

Solution :

ΔL = α L₁ ΔT

ΔL = α L₁ (T₂ – T₁)

0.09 cm = (9 x 10^{-6 o}C)(50 cm)(T₂ – 25 ^oC)

0.09 = (45 x 10^-5)(T₂ – 25)

0.09 / (45 x 10^-5) = T₂ – 25

0.002 x 10⁵ = T₂ – 25

2 x 10² = T₂ – 25

200 = T₂ – 25

T₂ = 200 + 25

T₂ = 225^oC

The final temperature is 225 ^oC.

4. The original length of metal is 1 meter and the final length is 1.02 m. The change in temperature is 50 Kelvin. Determine the coefficient of linear expansion!

Known :

The initial length (L₁) = 1 meter

The final length (L₂) = 1.02 meter

The change in length (ΔL) = L₂ – L₁ = 1.02 meter – 1 meter = 0.02 meter

The change in temperature (ΔT) = 50 Kelvin = 50^oC

Wanted : The coefficient of linear expansion

Solution :

ΔL = α L₁ ΔT

0.02 m = α (1 m)(50^oC)

0.02 = α (50^oC)

α = 0.02 / 50^oC

α = 0.0004 ^oC^-1

α = 4 x 10^{-4 o}C^-1

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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