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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

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Ebook PDF volume expansion sample problems with solutions 81.98 KB

  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, the heat of fusion, the heat of vaporization
  9. Energy conservation for heat transfer

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