Volume expansion – problems and solutions

1. At 30 oC the volume of an aluminum sphere is 30 cm3. The coefficient of linear expansion is 24 x 10-6 oC-1. If the final volume is 30.5 cm3, what is the final temperature of the aluminum sphere?

Known :

The coefficient of linear expansion (α) = 24 x 10-6 oC-1

The coefficient of volume expansion (β) = 3 α = 3 x 24 x 10-6 oC-1 = 72 x 10-6 oC-1

The initial temperature (T1) = 30oC

The initial volume (V1) = 30 cm3

The final volume (V2) = 30.5 cm3

The change in volume (ΔV) = 30.5 cm3 – 30 cm3 = 0.5 cm3

Wanted : The final temperature (T2)

Solution :

ΔV = β (V1)(ΔT)

ΔV = β (V1)(T2 – T1)

0.5 cm3 = (72 x 10-6 oC-1)(30 cm3)(T2 – 30oC)

0.5 = (2160 x 10-6)(T2 – 30)

0.5 = (2.160 x 10-3)(T2 – 30)

0.5 = (2.160 x 10-3)(T2 – 30)

0.5 / (2.160 x 10-3) = T2 – 30

0.23 x 103 = T2 – 30

0.23 x 1000 = T2 – 30

230 = T2 – 30

230 + 30 = T2

T2 = 260oC

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2. The coefficient of linear expansion of an metal sphere is 9 x 10-6 oC-1. The internal diameter of the metal sphere at 20 oC is 2.2 cm. If the final diameter is 2.8 cm, what is the final temperature!

Known :

The coefficient of linear expansion (α) = 9 x 10-6 oC-1

The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10-6 oC-1 = 27 x 10-6 oC-1

The initial temperature (T1) = 20oC

The initial diameter (D1) = 2.2 cm

The final diameter (D2) = 2.8 cm

The initial radius (r1) = D1 / 2 = 2.2 cm3 / 2 = 1.1 cm3

The final radius (r2) = D2 / 2 = 2.8 cm3 / 2 = 1.4 cm3

The initial volume (V1) = 4/3 π r13 = (4/3)(3.14)(1.1 cm)3 = (4/3)(3.14)(1.331 cm3) = 5.57 cm3

The final volume (V2) = 4/3 π r23 = (4/3)(3.14)(1.4 cm)3 = (4/3)(3.14)(2.744 cm3) = 11.48 cm3

The change in volume (ΔV) = 11.48 cm3 – 5.57 cm3 = 5.91 cm3

Wanted : The final temperature (T2)

Solution :

ΔV = β (V1)(ΔT)

5.91 cm3 = (27 x 10-6 oC-1)(5.57 cm3)(T2 – 20oC)

5.91 = (150.39 x 10-6)(T2 – 20)

5.91 / 150.39 x 10-6 = T2 – 20

0.039 x 106 = T2 – 20

39 x 103 = T2 – 20

39,000 = T2 – 20

39,000 + 20 = T2

T2 = 39,020 oC

3. A 2000-cm3 aluminum container, filled with water at 0oC. And then heated to 90oC. If the coefficient of linear expansion for aluminum is 24 x 10-6 (oC)-1 and the coefficient of volume expansion for water is 6.3 x 10-4 (oC)-1, determine the volume of spilled water.

Known :

The initial volume of the aluminum container and water (Vo) = 2000 cm3 = 2 x 103 cm3

The initial temperature of the aluminum container and water (T1) = 0oC

The final temperature of the aluminum container and water (T2) = 90oC

The coefficient of linear expansion for aluminum (α) = 24 x 10-6 (oC)-1

The coefficient of volume expansion for aluminum (γ) = 3α = 3 (24 x 10-6 (oC)-1 ) = 72 x 10-6 oC-1

The coefficient of volume expansion for water (γ) = 6.3 x 10-4 (oC)-1

Wanted : The volume of spilled water

Solution :

The equation of the volume expansion :

V = Vo + γ Vo ΔT

V – Vo = γ Vo ΔT

ΔV = γ Vo ΔT

V = final volume, Vo = initial volume, ΔV = the change in volume, γ = the coefficient of volume expansion, ΔT = the change in temperature

Calculate the change in volume of the aluminum container :

ΔV = γ Vo ΔT = (72 x 10-6)(2 x 103)(90) = 12960 x 10-3 = 12.960 cm3

Calculate the change in volume of the water :

ΔV = γ Vo ΔT = (6.3 x 10-4)(2 x 103)(90) = 1134 x 10-1 = 113.4 cm3

The change in volume of the water is more greater than the aluminum container so that some water spilled.

Calculate the volume of spilled water :

113.4 cm3 – 12.960 cm3 = 100.44 cm3