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 o}C^-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 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 24 x 10^{-6 o}C^-1 = 72 x 10^{-6 o}C^-1

The initial temperature (T₁) = 30^oC

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 o}C^-1)(30 cm³)(T₂ – 30^oC)

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 = 260^oC

2. The coefficient of linear expansion of an metal sphere is 9 x 10^{-6 o}C^-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 o}C^-1

The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10^{-6 o}C^-1 = 27 x 10^{-6 o}C^-1

The initial temperature (T₁) = 20^oC

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 o}C^-1)(5.57 cm³)(T₂ – 20^oC)

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 0^oC. And then heated to 90^oC. 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 (V_o) = 2000 cm³ = 2 x 10³ cm³

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

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

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 o}C^-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 = V_o + γ V_o ΔT

V – V_o = γ V_o ΔT

ΔV = γ V_o ΔT

V = final volume, V_o = 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 V_o 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 V_o 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

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