# Linear expansion – problems and solutions

1. A steel is 40 cm long at 20 ^{o}C. 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 ^{o}C will be…

__Known :__

The change in temperature (ΔT) = 70^{o}C – 20^{o}C = 50^{o}C

The original length (L_{1}) = 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 (L_{2})

__Solution :__

**a) The change in length (****ΔL)**

ΔL = α L_{1} ΔT

ΔL = (12×10^{-6 o}C^{-1})(40 cm)(50^{o}C)

ΔL = (10^{-6})(24 x 10^{3}) cm

ΔL = 24 x 10^{-3} cm

ΔL = 24 / 10^{3} cm

ΔL = 24 / 1000 cm

ΔL = 0.024 cm

**b) The final length (L**_{2}**) **

L_{2} = L_{1} + ΔL

L_{2} = 40 cm + 0.024 cm

L_{2} = 40.024 cm

2. An iron rod heated from 30 ^{o}C to 80 ^{o}C. 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 ^{o}C – 30 ^{o}C = 50 ^{o}C

The final length (L_{2}) = 115 cm

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

__Wanted :__ the original length (L_{1}) and the change in length (ΔL)

__Solution :__

**a) ****The original length ****(L**_{1}**)**

Formula of the change in length for the linear expansion :

ΔL = α L_{1} ΔT

Formula of the final length :

L_{2} = L_{1 }+ ΔL

L_{2} = L_{1} + α L_{1} ΔT

L_{2 }= L_{1} (1 + α ΔT)

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

115 cm = L_{1 }(1 + 150.10^{-3})

115 cm = L_{1 }(1 + 0.15)

115 cm = L_{1 }(1.15)

L L
ΔL = L ΔL = 115 cm – 100 cm ΔL = 15 cm |

3. At 25 ^{o}C, 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 (L_{1}) = 50 cm

The final length (L_{2}) = 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 (T_{1}) = 25^{o}C

__Wanted :__ The final temperature (T_{2})

__Solution :__

ΔL = α L_{1} ΔT

ΔL = α L_{1} (T_{2 }– T_{1})

0.09 cm = (9 x 10^{-6 o}C)(50 cm)(T_{2 }– 25 ^{o}C)

0.09 = (45 x 10^{-5})(T_{2 }– 25)

0.09 / (45 x 10^{-5}) = T_{2 }– 25

0.002 x 10^{5 }= T_{2} – 25

2 x 10^{2 }= T_{2} – 25

200 = T_{2} – 25

T_{2 }= 200 + 25

T_{2 }= 225^{o}C

The final temperature is 225 ^{o}C.

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 (L_{1}) = 1 meter

The final length (L_{2}) = 1.02 meter

The change in length (ΔL) = L_{2 }– L_{1} = 1.02 meter – 1 meter = 0.02 meter

The change in temperature (ΔT) = 50 Kelvin = 50^{o}C

__Wanted : __The coefficient of linear expansion

__Solution :__

ΔL = α L_{1} ΔT

0.02 m = α (1 m)(50^{o}C)

0.02 = α (50^{o}C)

α = 0.02 / 50^{o}C

α = 0.0004 ^{o}C^{-1}

α = 4 x 10^{-4} ^{o}C^{-1}

- 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