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. Abban 30 oC the volume of an aluminum sphere is 30 cm3. The coefficient of lineáris expanzió is 24 x 10-6 óraC-1. If the final volume is 30.5 cm3, what is the final temperature of the aluminum sphere?
Ismert:
A lineáris tágulási együttható (α) = 24 x 10-6 óraC-1
Az együttható kötetbővítés (β) = 3 α = 3 x 24 x 10-6 óraC-1 = 72 x 10-6 óraC-1
A kezdeti hőmérséklet (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
Keresett: A végső hőmérséklet (T2)
megoldás:
ΔV= β (V1)(AT)
ΔV= β (V1)(T2 - T1)
0.5 cm3 = (X 72 10-6 óraC-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
X 0.23 103 = T.2 - 30
0.23 x 1000 = T2 - 30
230=T2 - 30
230 + 30 = T2
T2 = 260oC
2. The coefficient of linear expansion of a metal sphere is 9 x 10-6 óraC-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!
Ismert:
A lineáris tágulási együttható (α) = 9 x 10-6 óraC-1
The coefficient of volume expansion (β) = 3 α = 3 x 9 x 10-6 óraC-1 = 27 x 10-6 óraC-1
A kezdeti hőmérséklet (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
Keresett: A végső hőmérséklet (T2)
megoldás:
ΔV= β (V1)(AT)
5.91 cm3 = (X 27 10-6 óraC-1)(5.57 cm3)(T2 - 20oC)
5.91 = (150.39 x 10-6)(T2 - 20)
5.91/150.39 x 10-6 = T.2 - 20
X 0.039 106 = T.2 - 20
X 39 103 = T.2 - 20
39,000=T2 - 20
39,000 + 20 = T2
T2 = 39,020 oC
3. A 2000-cm3 alumínium tartály, 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.
Ismert:
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
A víz térfogat-tágulási együtthatója (γ) = 6.3 x 10-4 (oC)-1
Keresett: The volume of spilled water
megoldás:
A térfogat-tágulás egyenlete:
V = Vo + γ Vo AT
V – Vo = γ Vo AT
ΔV = γ Vo AT
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.X 3 10-4)(2 x 103)(90) = 1134 x 10-1 = 113.4 cm3
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 cm3 - 12.960 cm3 = 100.44 cm3
[wpdm_csomag azonosítója='702']
- Hőmérsékleti skálák átváltása
- Lineáris tágulás
- Területbővítés
- Hangerő bővítés
- Hőség
- A hő mechanikai megfelelője
- Fajhő és hőkapacitás
- Latent heat, the heat of fusion, the heat of vaporization
- Energiamegmaradás a hőátadásban
