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Welcome to our Physics lesson on Linear Thermal Expansion and Contraction, this is the second lesson of our suite of physics lessons covering the topic of Thermal Expansion, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Linear Thermal Expansion and Contraction
In reality, all objects expand or contract thermally in all dimensions (in 3 D). However, when one dimension is much greater than the other two (such as a long and thin bars for example), only the dimension corresponding to the greater value (usually the length) is considered. Given this, we can think about the factors affecting the amount of thermal expansion or contraction of a long bar of original length L0. They are:
- Type of material. Obviously, different types of materials experience different amounts of thermal expansion (contraction) for the same conditions. Usually denser materials expand or contract less than the other less dense materials as the particles of dense materials are more strongly bound to each other and it is more difficult to make them move away from their equilibrium positions. There is a quantity known as the coefficient of linear expansion, α (measured in Kelvin or Celsius degree at power -1), which represents numerically the type of material in the formula of thermal expansion (contraction).
- Length of material. Obviously, longer the material, more it expands for the same conditions. For example, a long iron bar expands more millimeters than an iron nail if they experience the same process of heating, despite they are made of the same material. Original length of material is represented in formula by the letter L0, as stated earlier.
- Change in temperature. When the change in any object's temperature is slight, it does not experience any considerable change in length. On the other hand, if there is a considerable change in object's temperature, its dimensions change a lot. We can use either Celsius or Kelvin scale but not Fahrenheit to represents the change in temperature of an expanding (contracting) object. Thus, for temperatures measured in Celsius scale, we can write Δt = t - t0 for the change in temperature where t is the final and t0 the initial temperature of the object, while for temperatures measured in Kelvin scale we can write ΔT = T - T0 for the same thing.
Putting all the above factors together, we obtain the formula of linear thermal expansion (contraction):
∆L = L0 × α × ∆T
where ΔL = L - L0 is the amount of linear extension (compression) a bar with original length experiences when the temperature changes by ΔT = T - T0 degrees, and L is the final length of the bar.
If we are interested to find the final length of the bar, the above formula becomes:
L - L0 = L0 × α × (T - T0)
or
L = L0 + L0 × α × (T - T0)
When factoring L0 we obtain for the final length L of the bar after experiencing thermal expansion or contraction:
L = L0 × (1 + α × ∆T)
Thus, when the bar heats up, ΔT is positive as T > T0. Therefore, the final length of the bar will be greater than its its initial length. On the other hand, when the bar cools down, ΔT is negative as T > T0. This means the bar will be shorter than before at the end of process.
Remark!
You can also use temperatures in Celsius scale in the formula as the change in temperature in both Kelvin and Celsius scale is the same. Or maybe you can switch from Kelvin to Celsius an vice-versa during the solution of a problem if needed.
Example 1
The linear thermal expansion coefficient of iron at 20°C is 11.8 × 10-6 K-1. What is the length of a 12 m long railway track during a hot summer day where the temperature of the track becomes 42°C?
Solution 1
Let's write the clues first. We have:
α = 11.8 × 10-6 K-1
t0 = 20°C
t = 42°C
L0 = 12 m
L = ?
Applying the linear thermal expansion formula
L = L0 [1 + α × (t - t0]
we obtain for the change in the bar's length after substituting the known values:
L=12 m × [1 + 11.8 × 10-6 K-1 × (42-20)°C]
= 12 m × [1 + 11.8 × 10-6 1/K × 22K]
= 12 m × [1 + 259.6 × 10-6 ]
= 12 m × [1 + 0.0002596]
= 12 m × 1.0002596
= 12.0031152 m
≈12.003 m
This means in this example, the track elongates by about 3 mm because of weather conditions.
You have reached the end of Physics lesson 13.2.2 Linear Thermal Expansion and Contraction. There are 7 lessons in this physics tutorial covering Thermal Expansion, you can access all the lessons from this tutorial below.
More Thermal Expansion Lessons and Learning Resources
Thermodynamics Learning Material| Tutorial ID | Physics Tutorial Title | Tutorial | Video
Tutorial | Revision
Notes | Revision
Questions |
|---|
| 13.2 | Thermal Expansion | | | | |
| Lesson ID | Physics Lesson Title | Lesson | Video
Lesson |
|---|
| 13.2.1 | What is Thermal Expansion and Thermal Contraction? | | |
| 13.2.2 | Linear Thermal Expansion and Contraction | | |
| 13.2.3 | Area Thermal Expansion and Contraction | | |
| 13.2.4 | Volume Thermal Expansion and Contraction | | |
| 13.2.5 | Thermal Expansion and Contraction in Daily Life | | |
| 13.2.6 | Bimetallic Strip. Thermostat | | |
| 13.2.7 | Types of Thermometers | | |
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