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: Vo = Initial volume, V = Final volume, ΔV = V – Vo = The change in volume, To = Initial temperature, T = Final temperature, ΔT = T – To = The change in temperature, β = the coefficient of volume expansion. Unit of β = (Co) -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. У 30 oC the volume of an aluminum sphere is 30 cm3. The coefficient of лінійне розширення is 24 x 10-6°CC-1. If the final volume is 30.5 cm3, what is the final temperature of the aluminum sphere?
Відомо:
Коефіцієнт лінійного розширення (α) = 24 x 10-6°CC-1
Коефіцієнт розширення обсягу (β) = 3 α = 3 x 24 x 10-6°CC-1 = 72 х 10-6°CC-1
Початкова температура (T1) = 30oC
The initial volume (V1) = 30см3
The final volume (V2) = 30.5см3
The change in volume (ΔV) = 30.5 cm3 - 30 см3 = 0.5 см3
Розшукується: Кінцева температура (T2)
Рішення:
ΔV = β (V1)(ΔT)
ΔV = β (V1)(Т2 - T1)
0.5 см3 = (72 х 10-6°CC-1)(30 см3)(Т2 - 30oC)
0.5 = (2160 х 10-6)(T2 - 30)
0.5 = (2.160 х 10-3)(T2 - 30)
0.5 = (2.160 х 10-3)(T2 - 30)
0.5 / (2.160 х 10-3) = Т2 - 30
0.23 х 103 = Т2 - 30
0.23 x 1000 = T2 - 30
230 = Т2 - 30
230 + 30 = Т2
T2 = 260oC
2. The coefficient of linear expansion of a metal sphere is 9 x 10-6°CC-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!
Відомо:
Коефіцієнт лінійного розширення (α) = 9 x 10-6°CC-1
The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10-6°CC-1 = 27 х 10-6°CC-1
Початкова температура (T1) = 20oC
The initial diameter (D1) = 2.2см
The final diameter (D2) = 2.8см
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) = 5.57см3
The final volume (V2) = 4/3 π r23 = (4/3)(3.14)(1.4 cm)3 = (4/3)(3.14)(2.744 cm3) = 11.48см3
The change in volume (ΔV) = 11.48 cm3 - 5.57 см3 = 5.91 см3
Розшукується: Кінцева температура (T2)
Рішення:
ΔV = β (V1)(ΔT)
5.91 см3 = (27 х 10-6°CC-1)(5.57 см3)(Т2 - 20oC)
5.91 = (150.39 х 10-6)(T2 - 20)
5.91/150.39 х 10-6 = Т2 - 20
0.039 х 106 = Т2 - 20
39 х 103 = Т2 - 20
39,000 = Т2 - 20
39,000 + 20 = Т2
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.
Відомо:
The initial volume of the aluminum container and water (Vo) = 2000см3 = 2 х 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
Розшукується: The volume of spilled water
Рішення:
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 х 10-6)(2 x 103)(90) = 12960 × 10-3 = 12.960 см3
Calculate the change in volume of the water :
ΔV = γ Vo ΔT = (6.3 х 10-4)(2 x 103)(90) = 1134 × 10-1 = 113.4 см3
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 см3 - 12.960 см3 = 100.44 см3
[ідентифікатор wpdm_package='702′]
- Перетворення температурних шкал
- Лінійне розширення
- Розширення території
- Розширення обсягу
- Тепло
- Механічний еквівалент тепла
- Питома теплоємність та теплоємність
- Latent heat, the heat of fusion, the heat of vaporization
- Збереження енергії для теплопередачі
