2. Thermal Expansion
Linear Expansion
The rate of expansion of solids, liquids, or gas differs from substance
to substance. For solids, the expansion depends on the change in temperature,
original length, and the type of material of which it is made. This is also known
as linear expansion, where solid expands linearly with the change in
temperature.
3.
4.
5. The amount of expansion may be
very small, However, if you are talking
about long pieces like railroad tracks,
expansion could pose a lot of problems.
This is why railroad tracks have gaps in
between pieces, called “expansion joints”.
6. These quantities are related mathematically as:
∆𝑳 = 𝜶𝑳𝒊𝜟𝑻
(𝛥𝑇 = Tf – Ti)
Tf = final temperature, Ti = initial temperature
where 𝛂 is the coefficient of thermal expansion of the object expressed in
per Celsius degree (°C -1, see Table 5.1 of your book), ∆𝑳 is the change in
length in meters, 𝑳𝒊 is the original length of the object, and 𝜟𝑻 is the change in
temperature in C°. The new length of the material after expansion may be
calculated by : Lf = Li + ∆𝑳
7. Example 1.
A 1.0-m iron rod is heated from 25°C to 55°C. By how much will
it expand and what will be its final length? (𝛼𝑖𝑟𝑜𝑛 = 12 x 10 -6/C°)
Given:
Li = 1.0 m
Ti = 25°C
Tf = 55°C
𝛼 = 12 x 10 -6/C°
8. Solution:
The amount of expansion could be determined as:
∆𝐿 = 𝛼𝐿𝑖𝛥𝑇 = 𝛼𝐿𝑖 (Tf – Ti)
= (12 x 10 -6/C°) (1.0m)(55°C - 25°C)
= (12 x 10 -6 ) (1.0m)(30 °C)
C°
= 360 x 10 -6 m = 3.60 x 102 (x 10 -6 m)
= 3.6 x 10-4 m = (0.00036m)
The length of expansion and final length of the iron rod is 0.00036m and 1.00036m
(Lf = Li + ∆𝐿), respectively.
9. Example 2.
At 30°C, A 1.5-m copper stick is heated to 62°C. By how much it
will expand and what will be its final length? (𝛼𝑐𝑜𝑝𝑝𝑒𝑟 = 17 x 10 -6/C°)
Given:
Li = 1.5 m
Ti = 30°C
Tf = 62°C
𝛼 = 17 x 10 -6/C°
10. Solution:
∆𝐿 = 𝛼𝐿𝑖𝛥𝑇 = 𝛼𝐿𝑖 (Tf – Ti)
= (17 x 10 -6/C°) (1.5 m)(62°C - 30°C)
= (
17 x 10 −6
C°
) (1.5m)(32°C)
= 816 x 10 -6 m = 8.16 x 102 (x 10 -6 m)
= 8.16 x 10-4 m = (0.000816m)
The length of expansion and final length of the iron rod is 0.000816m and
1.500816m (Lf = Li + ∆𝐿), respectively.
11. Volume Expansion
As materials become warmer, they expand uniformly in all directions,
therefore their volume increases also. The change in volume
∆𝑉 is determined by:
∆𝑉 = 𝛽𝑉𝑖 ∆𝑇
Where Vi is the initial volume, ∆𝑇 is the change in temperature, and 𝛽 is the
coefficient of volume expansion, which is related to the coefficient of the linear
expansion 𝛼 by 𝛽 x 𝛼.
12.
13. Example 1.
What will be the volume of 80.0 ml ethyl alcohol at 10.0 °C when it is
heated to 60.0 °C?
Given:
Vi = 80.0 ml
Ti = 10.0 °C
Tf = 60.0 °C
𝛽 = 1,100 x 10-6 /C°
14. Solution:
The volume expanded:
∆𝑉 = 𝛽𝑉𝑖 ∆𝑇
= (1,100 x 10-6 /C° )(80.0 ml)(60.0 °C - 10.0 °C)
= (
1,100 x 10−6
C°
)(80.0 ml)(50.0 °C)
= 4,400,000 x 10-6 ml
= 4.4 ml
(Vf = Vi + ∆𝑉) = 80.0 ml + 4.4 ml = 80.4 ml
The volume of the ethyl alcohol when heated at 60°C is 84.4 ml (80.0 ml + 4.4 ml).
