Expansion Of Solid And Liquids – Coefficient Of Linear Expansion
Experimentally it is observed that the linear expansion of a metal rod on heating is directly proportional to
- the initial length of the rod and
- the rise in temperature of the rod.
Let l1 be the length of a rod at temperature t1 and the length becomes l2 at temperature t2, (where t2> t1).
∴ The increase in length = (l2 – l1) and the increase in temperature = (t2 – t1)
As an expansion of length is directly proportional to the initial length and the rise in temperature, (l2 — l1) ∝ l1, for the same temperature change and (l2– l1) ∝ (t2– t1), for the same initial length.
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∴ \(\left(l_2-l_1\right) \propto l_1\left(t_2-t_1\right)\) when both \(l_1 and \left(t_2-t_1\right) vary\)
or, \(\left(l_2-l_1\right)=\alpha l_1\left(t_2-t_1\right)\) ……….(1)
Here α is the constant of proportionality, whose value is dif-ferent for different materials and is called the coefficient of linear expansion of that material.
From (1), \(\alpha=\frac{l_2-l_1}{l_1\left(t_2-t_1\right)}\)
= \(\frac{\text { increase in length }}{\text { initial length } \times \text { rise in temperature }}\)……(2)
From (2), \(l_2=l_1\left\{1+\alpha\left(t_2-t_1\right)\right\}\)….(3)
If the initial temperature t1 = 0, and the corresponding length is ZQ, then the length at temperature t2 = t is, from equation (3),
∴ \(l_t=l_0(1+\alpha t)\)….(4)
Also from equation (1), if l1 – 1 and (t2 -t1) then α= (l2 – l1). From this, the coefficient of linear expansion can be defined.
Definition: The increase in length for unit rise in temperature for a unit length of a solid is called the coefficient of linear expansion of the material of the solid.
The coefficient of linear expansion, α, is not a constant. In the above discussion we assume that the value of α does not depend on the temperature of the body. But it is not exactly true.
The value of α depends on the temperature of the body. A solid of fixed length, for unit degree rise in temperature, expands a little differently in different temperature ranges.
In fact, the value of α, calculated using equation (2) is the average value of a between the temperatures t1 and t2.
For more precise calculations, initial length of the material is taken as its length at 0°C. Thus, the expression for α modifies as
⇒ \(\alpha=\frac{\text { increase in length for } 1^{\circ} \mathrm{C} \text { rise in temperature }}{\text { length at } 0^{\circ} \mathrm{C}}\)
Unit of coefficient of linear expansion: From relation (2),
unit of \(\alpha=\frac{\text { unit of length }}{\text { unit of length } \times \text { unit of temperature }}\)
= \(\frac{1}{\text { unit of temperature }}\)
This leads to two conclusions:
1. Value of a is independent of the unit of length.
In equation (2), since \(\frac{l_2-l_1}{l_1}\) is a ratio of two lengths, the coefficient of linear expansion does not depend on the unit of length.
For example, α for iron is 12 x 10-6 °C-1 means that an iron rod of length 1 cm or 1 ft or 1 m, when heated through 1°C, will expand by 12 x 10-6 cm or 12 x 10-6 ft or 12 x 10-6 m respectively.
2. Value of a depends on the unit of temperature.
Unit of a is per °C or per °F.
As the change in temperature by 1°F = 5/9 °C change in temperature,
⇒ \(\alpha_F=\frac{5}{9} \alpha_C\), where αF = value of α in Fahrenheit and αC = value of a in Celsius scale.
For example, the coefficient of linear expansion of iron
= 12 x 10-6 °C-1 = 5/9 x 12 x 10-6 °F-1
= 6.67 x 10-6 °F-1.
Coefficients of linear expansion of some solids at 20 °C
Expansion Of Solid And Liquids – Coefficient Of Linear Expansions Numerical Examples
Example 1. A steel rod is 1.5 m long at 20°C. What will be the increase in its length if it is heated up to 100°C? a for steel =11 x 10-6 °C-1.
Solution:
Here, I1 = 1.5 m = 150 cm, α = 11 x 10-6 °C-1, t1 = 20°C and t2 = 100°C
Increase in length = l2 – l1= l1 α (t2 – t1)
= 150x 11 x 10-6 x (100 – 20)
= 150x 11 x 10-6 x 80 = 0.132 cm.
Example 2. The length of a zinc rod, when heated from 20°C to 80°C, increases by 0.6 cm. If the coefficient of linear expansion of zinc is 27 x 10-6 °C-1, what is the initial length of the rod?
Solution:
⇒ \(a=\frac{l_2-l_1}{l_1\left(t_2-t_1\right)} \text { or, } l_1=\frac{l_2-l_1}{\alpha\left(t_2-t_1\right)}\)
Given, α = 27 x 10-6 °C-1, t1 = 20°C, t2 = 80°C, l2 – l1 = 0.6 cm
∴ \(l_1=\frac{0.6}{27 \times 10^{-6} \times(80-20)}\)
= \(\frac{0.6}{27 \times 10^{-6} \times 60}\)
Example 3. The lengths of a copper rod are 200.166 cm and 200.664 cm at 50°C and 200°C respectively. What is the coefficient of linear expansion of copper?
Solution:
α = \(\frac{l_2-l_1}{l_1\left(t_2-t_1\right)}\)
Given, l1 = 200.166 cm, l2 = 200.664 cm, t1 = 50°C and t2 = 200°C
∴ α = \(\frac{200.664-200.166}{200.166(200-50)}=\frac{0.498}{200.166 \times 150}\)
= 16.6 x 10 °C
Example 4. A brass rod has a length of 150 cm at 40°C. What will be its length at 100°C ? The coefficient of linear expansion of brass is 18 x 10-6 °C-1.
Solution:
⇒ \(l_2=l_1\left\{1+\alpha\left(t_2-t_1\right)\right\}\)
Given, \(l_1 =150 \mathrm{~cm}, t_1=40^{\circ} \mathrm{C}, t_2=100^{\circ} \mathrm{C}, \alpha=18 \times 10^{-6 \circ} \mathrm{C}^{-1}\)
∴ \(l_2 =150\left\{1+18 \times 10^{-6}(100-40)\right\}\)
= \(150\left\{1+108 \times 10^{-5}\right\}=150 \times 1.00108\)
= 150.162 cm.
Example 5. Coefficient of linear expansion of aluminium is 19 x 10-6 per degree Celsius. What will be the value of this coefficient in the Fahrenheit scale?
Solution:
∴ \(a_F=\frac{5}{9} a_C=19 \times 10^{-6} \times \frac{5}{9}=10.56 \times 10^{-6 \circ} \mathrm{F}^{-1}\)