Linear expansion is experienced only by solid objects; volume expansion is experienced by all objects, both solid, liquid, and gas. The equation of volume expansion is similar to the equation of linear expansion.

*Description: V*_{o}* = Initial volume, V = Final volume, ΔV = V – V*_{o }*= The change in volume, T*_{o}* = Initial temperature, T = Final temperature, ΔT = T – T*_{o }*= The change in temperature, β = the coefficient of volume expansion. Unit of β = (C*^{o}*) *^{-1}

The above volume expansion equation applies only when the changes in the volume of the objects (both solid, liquid, and gas) are smaller than the original volume of the object. If the change in volume of an object is greater than the initial volume of the object, the equation of the volume expansion does not give the right results. Usually, the changes in volume experienced by solid objects are not too large. Conversely, the coefficient of volume expansion of the liquid and gas is large. The coefficient of volume expansion for gaseous substances is also easy to change if the temperature changes. Therefore the formula above is used only for the expansion of solid objects.

1. At 30 ^{o}C the volume of an aluminum sphere is 30 cm^{3}. The coefficient of linear expansion is 24 x 10^{-6 o}C^{-1}. If the final volume is 30.5 cm^{3}, 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 (T_{1}) = 30^{o}C

The initial volume (V_{1}) = 30 cm^{3}

The final volume (V_{2}) = 30.5 cm^{3}

The change in volume (ΔV) = 30.5 cm^{3} – 30 cm^{3 }= 0.5 cm^{3}

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

__Solution :__

ΔV = β (V_{1})(ΔT)

ΔV = β (V_{1})(T_{2} – T_{1})

0.5 cm^{3 }= (72 x 10^{-6 o}C^{-1})(30 cm^{3})(T_{2} – 30^{o}C)

0.5 = (2160 x 10^{-6})(T_{2} – 30)

0.5 = (2.160 x 10^{-3})(T2 – 30)

0.5 = (2.160 x 10^{-3})(T_{2} – 30)

0.5 / (2.160 x 10^{-3}) = T_{2} – 30

0.23 x 10^{3} = T_{2} – 30

0.23 x 1000 = T_{2} – 30

230 = T_{2} – 30

230 + 30 = T_{2}

T_{2} = 260^{o}C

2. The coefficient of linear expansion of a metal sphere is 9 x 10^{-6 o}C^{-1}. The internal diameter of the metal sphere at 20 ^{o}C 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 (T_{1}) = 20^{o}C

The initial diameter (D_{1}) = 2.2 cm

The final diameter (D_{2}) = 2.8 cm

The initial radius (r_{1}) = D_{1} / 2 = 2.2 cm^{3 }/ 2 = 1.1 cm^{3}

The final radius (r_{2}) = D_{2} / 2 = 2.8 cm^{3 }/ 2 = 1.4 cm^{3}

The initial volume (V_{1}) = 4/3 π r_{1}^{3} = (4/3)(3.14)(1.1 cm)^{3} = (4/3)(3.14)(1.331 cm^{3}) = 5.57 cm^{3}

The final volume (V_{2}) = 4/3 π r_{2}^{3} = (4/3)(3.14)(1.4 cm)^{3} = (4/3)(3.14)(2.744 cm^{3}) = 11.48 cm^{3}

The change in volume (ΔV) = 11.48 cm^{3 }– 5.57 cm^{3 }= 5.91 cm^{3 }

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

__Solution :__

ΔV = β (V_{1})(ΔT)

5.91 cm^{3 }= (27 x 10^{-6 o}C^{-1})(5.57 cm^{3})(T_{2} – 20^{o}C)

5.91 = (150.39 x 10^{-6})(T_{2} – 20)

5.91 / 150.39 x 10^{-6 }= T_{2} – 20

0.039 x 10^{6} = T_{2} – 20

39 x 10^{3} = T_{2} – 20

39,000 = T_{2} – 20

39,000 + 20 = T_{2}

T_{2 }= 39,020 ^{o}C

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

__Known :__

The initial volume of the aluminum container and water (V_{o}) = 2000 cm^{3} = 2 x 10^{3 }cm^{3}

The initial temperature of the aluminum container and water (T_{1}) = 0^{o}C

The final temperature of the aluminum container and water (T_{2}) = 90^{o}C

The coefficient of linear expansion for aluminum (α) = 24 x 10^{-6 }(^{o}C)^{-1 }

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

The coefficient of volume expansion for water (γ) = 6.3 x 10^{-4} (^{o}C)^{-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 10^{3})(90) = 12960 x 10^{-3} = 12.960 cm^{3 }

Calculate the change in volume of the water :

ΔV = γ V_{o} ΔT = (6.3 x 10^{-4})(2 x 10^{3})(90) = 1134 x 10^{-1} = 113.4 cm^{3}

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

Calculate the volume of spilled water :

113.4 cm^{3} – 12.960 cm^{3 }= 100.44 cm^{3}

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- Converting temperature scales
- Linear expansion
- Area expansion
- Volume expansion
- Heat
- Mechanical equivalent of heat
- Specific heat and heat capacity
- Latent heat, the heat of fusion, the heat of vaporization
- Energy conservation for heat transfer