all materials / substances (solid, liquid, and gas) generally expand when heated and contract when cooled. expansion in solids and liquids ( in general) can
be explained
Length Expansion
on the long expansion talks, We will only pay attention to expansion in the long direction only. Expansion in the other direction we ignore (this may be done if the rod is very long and small has a cross section)
a long rod is initially lo. rod was then heated so that its temperature is changed as much as ΔT. Experiments show that the length of the stem is proportional to the lo and comparable with ΔT.
Δl α loΔT
to make this into the equation, right-hand sides must be multiplied by a constant comparison α (coefficient of expansion length):
Δl = lo α ΔT. . . . . . . . . . . . . . . . . . . . . . (6)
lo = length at first (m)
ΔT = T - To = Changes in temperature in ℃ or K
Δl = l - lo = change in length in meters
l = length at temperature T in meters
from formula (6) we can write the formula length expansion coefficient.
α = Δl / lo ΔT
According to the above formula length expansion coefficient can be defined as length changes of an object per unit length per degrees Celsius (or Kelvin). Α is a unit 1/derajat or (℃)-1 or K-1. For example, the value of α = 24 x 10-6 (℃)-1 means the length of an object will increase 24 milion part of its original length when the temperature is raised 1 degrees centigrade.
expansion coefficient of the length of magnitude depending on the temperature, but due to changes in the coefficients due to temperature changes is very small, length expansion coefficient can be considered constant.
an example
a. steel bar of length 2 meters the temperature is raised to 30 ℃.
What is the length of steel that now ? (α = 11 x 10-6 (℃)-1.)
completion:
the length of rod can be calculated by the formula (6). Stem length is now calculated by adding the length of the rod mul-onset and the length of this rod. use the data provided:
α = 11 x 10-6 (℃)-1 u = 2 m, ΔT = 30 ℃
the result:
Δl = lo α ΔT = 2 x (11 x 10-6) x 30
= 6.6 x 10-4 m = 0066 cm
l = l o + Δl = 2 + 6.6 x 10-4 = 2.00066 m
Extensive Expansion
when a thin rectangular or heated thin disc, then there is extensive expansion. The amount of large objects due to the temperature increment is increased by ΔT is given by the formula:
ΔA = A o β ΔT. . . . . . . . . . . . . . . . . (7).
β = coefficient of expansion in the area (℃)-1
A o = original area in m2
ΔT = T - To = Changes in temperature in ℃ or K
ΔA = A - A o = change in area in m2
A = area in m2
at a temperature T extensive expansion coefficient β associated with the long expansion coefficient α via the formula:
β = 2 α. . . . . . . . . . . . . . . . . . (8)
This formula can be proved as follows;
consider a rectangular Lo Po and width. Temperature is raised by ΔT. increase the length and width of the rectangle:
ΔP = α Po ΔT
ΔL = α Lo ΔT
broad temperature is raised is:
A = (Po + ΔP) (Lo + ΔL),
so the extent of the increase are:
Δ A = A - A o
= (Po + ΔP) (L o + ΔL) - Po L o
Po L o (1 + 2 + α ΔT (α ΔT)2) - Po L o
Po Lo (2 + α ΔT (α ΔT)2 )
A o (2 + α ΔT (α ΔT) 2)
Since α is very small then the second term in the equation above can be ignored so that the added area of ​​the rectangle to be:
A o (2 α ΔT).
compare this with equation (7) we obtain that β = 2 α.
Volume expansion.
when the beams or ball-shaped object is heated, the volume of these objects can be calculated by the formula:
ΔV = V o γ ΔT. . . . . . . . . . . . . . . . . . (9)
γ = coefficient of volume expansion (℃)-1
V o = initial volume in m3
ΔT = T - To = Changes in temperature in ℃ or K
ΔV = V - V o = change in volume in m3
V = Volume in m3 at a temperature T
γ coefficient of volume expansion associated with the long expansion coefficient α via the formula:
γ = 3 α. . . . . . . . . . . . . . . . . (10)
This formula can be proved in the same way with the vast expansion coefficient. we just derive a formula for the volume expansion of solids. Fluid volume expansion formula as formula (9).
Expansion coefficient of the liquid volume is relatively larger than the coefficient of volume expansion of solids (almost 10 x). Which is why when the thermometer heated volume of liquid (alcohol or mercury) is much larger rise in comparison with the increase in the volume of solids.
Note: because it is small, usually increases the volume of solids to be neglected for the increased volume of liquid.
Questions are Resolved.
1. A steel beam in a bridge extends 25 m across a small stream. What is it change in leght from the winter, Pls its temperature is - 20 ℃, to the summer, it is 38 ℃ Pls?
solution.
the change the temperature of the bridge is
ΔT = 38 ℃ - (- 20 ℃) = 58 ℃
using the equation (6),
We find the bridge That's change in length is
Δl = lo α ΔT = (12 x 10-6 (℃) -1) (25 m) (58 ℃) = 0017 m = 1.7 cm .
2. The aluminum lid of a jar of dill pickles is stuck to the jar. To Loosen the lid so it can be opened, hot water is poured over the lid, Causing it to expand. If the increase of temperature of the lid and glass is 40 ℃, calculate the change in circumference of the lid and of the glass on the which it is screwed. The diameter of the lid before heating is 22 cm.
solution.
The glass lid expand uniformly in all direction. Thus, as each warm, its diameter d increase by amount Δ d equal to
Δ d = α d ΔT
where α is the coefficient of the linear expansion of the aluminum lid (25 x 10 (℃)-1 ) or glass rim (10 x 10-6 (℃)-1 ) and ΔT is the temperature change. the circumference change from πd to π (d + Δ d) or by amount π Δ π α d = d ΔT. for the aluminum lid, the circumference change is from
π d ΔT = π (25 x 10-6 (℃)-1 ) (22 cm) (40 ℃)
= 0069 cm = 0.69 mm
The circumference of the glass change by αglass
d π ΔT = π (10 x 10-6 (℃)-1 ) (22 cm) (40 ℃)
= 0028 cm = 0.28 mm.
since the aluminum expands more than the glass, We Should now be the Able to open the jar of pickles. Because a material usually expands uniformly in all direction as it Becomes warmer, its volume increases. The change the volume ΔV of a substance depends on its change in temperature, its original volume V, and the materials of the which it is made ​​equation (9).
3. During a summer night the temperature is 20 ℃, your house contains 453 m3 of water. What volume of water leaves the house through an open window if the warm water to 40 ℃ on a very hot summer day? That assume the dimension of the house experience negligible change And that all other condition are constant.
solution.
as the warm water, increase of its volume by an amount ΔV = V o ΔT where γ, the coefficient of volume expansion of water is γ = 3670 x 10-6 (℃)-1 .
Thus, ΔV = (3670 x 10-6 (℃)-1 ) (453 m3) (40 ℃ - 20 ℃) = 33 m3 since the house stay about the same size, the water must find the extra space outside. Thus, about 33 m3 of water Will leave the house, approximately the volume of water in a typical bedroom.
Thermal Expansion Exercise.
1. If one steel beam extended the entire height of the sears tower in a Chicago (434 m), by how much would the beam contract from its summer high temperature of 37 ℃ to its winter low of - 20 ℃?
2. The volume of an iron ball of radius r is given by 4 / 3 π r3. A particular ball has a diameter of 10.00 cm and 0.01 cm is too large to fit through a hole in a metal plate. What temperature change of the ball Will allow it to fit through the hole?
3. An interstate highway is made ​​of concrete slabs 25 m long placed end to end.
(A) what the expansion gap must be left the between the slabs to Prevent buckling Pls the temperature changes from 20 ℃ to 50 ℃?
(B) starting whit the gap calculate in part (a), the gap Pls find the temperature decreases to - 20 ℃.
sources :
all book physics references
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