# 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 (Co)-1. The increase in length and the final length when it is at 70 oC will be…

Known :

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

The original length (L1) = 40 cm

Coefficient of linear expansion for steel (α) = 12 x 10-6 (Co)-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 oC-1)(40 cm)(50oC)

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

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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 oC-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 oC-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 oC-1)(50oC)

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

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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 oC-1

The original temperature (T1) = 25oC

Wanted : The final temperature (T2)

Solution :

ΔL = α L1 ΔT

ΔL = α L1 (T2 – T1)

0.09 cm = (9 x 10-6 oC)(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 = 225oC

The final temperature is 225 oC.

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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 T) = 50 Kelvin = 50oC

Wanted : The coefficient of linear expansion

Solution :

ΔL = α L1 ΔT

0.02 m = α (1 m)(50oC)

0.02 = α (50oC)

α = 0.02 / 50oC

α = 0.0004 oC-1

α = 4 x 10-4 oC-1

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