## 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 l_{1 }be the length of a rod at temperature t_{1} and the length becomes l_{2} at temperature t_{2}, (where t_{2}> t_{1}).

∴ The increase in length = (l_{2} – l_{1}) and the increase in temperature = (t_{2} – t_{1})

As an expansion of length is directly proportional to the initial length and the rise in temperature, (l_{2} — l_{1}) ∝ l_{1}, for the same temperature change and (l_{2}– l_{1}) ∝ (t_{2}– t_{1}), for the same initial length.

**Read and Learn More: Class 11 Physics Notes**

∴ \(\left(l_2-l_1\right) \propto l_1\left(t_2-t_1\right)\) 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 t_{1} = 0, and the corresponding length is ZQ, then the length at temperature t_{2} = t is, from equation (3),

∴ \(l_t=l_0(1+\alpha t)\)….(4)

Also from equation (1), if l_{1} – 1 and (t_{2 }-t_{1}) then α= (l_{2} – l_{1}). 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 t_{1} and t_{2}.

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 α_{F} = value of α in Fahrenheit and α_{C} = 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, I_{1} = 1.5 m = 150 cm, α = 11 x 10^{-6 °}C^{-1}, t_{1} = 20°C and t_{2} = 100°C

Increase in length = l_{2} – l_{1}= l_{1} α (t_{2} – t_{1})

= 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}, t_{1} = 20°C, t_{2} = 80°C, l_{2} – l_{1} = 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, l_{1} = 200.166 cm, l_{2} = 200.664 cm, t_{1} = 50°C and t_{2} = 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:**

∴ \(a_F=\frac{5}{9} a_C=19 \times 10^{-6} \times \frac{5}{9}=10.56 \times 10^{-6 \circ} \mathrm{F}^{-1}\)