The linear expansivity \(\alpha\) of a substance is defined as the increase in length per unit length of a
piece of that substance per degree rise in temperature.
$$\alpha = \frac{\text{increase in length}}{\text{original length × Temp. rise}} $$
$$ \alpha = \frac{∆l}{{l_1}{∆\theta}} $$
$$ \alpha = \frac{{l_2}-{l_1}}{{l_1}({\theta_2} - {\theta_1})} $$
$$ \text{where } \alpha = \text{linear expansivity in }{K^{-1}} $$
$$ {l_1} = \text{length before expansion} $$
$$ {l_2} = \text{length after expansion} $$
$$ {\theta_1} = \text{Temperature before expansion} $$
$$ {\theta_2} = \text{temperature after expansion} $$
$$∆l = \text{change in length} $$
$$ ∆\theta = \text{change in temperature} $$
Calculations
Example 1: A brass is 2m long at a certain temperature. Calculate the linear expansion of the rod for a
temperature change of 100K. [ Take the linear expansivity of brass as 1.8 × 10-5K-1
(WAEC)
Solution
$$ {l_1} = 2m $$
$$ \alpha = 1.8 × 10^-5K^-1 $$
$$ ∆\theta = 100K $$
$$ ∆l = ? $$
$$ \alpha = \frac{∆l}{{l_1}{∆\theta}} $$
$$ ∆l = \alpha{{l_1}{∆\theta}} $$
$$ ∆l = 1.8 × 10^{-5} × 2 × 100 $$
$$ ∆l = 1.8 × 10^{-5} × 200 $$
$$ ∆l = 3.6 × 10^{-3}m $$
Example 2: On a fairly cool rainy day when the temperature is 20°C, the length of a steel rail road track
is 20m. What will be its length on a hot dry day when the temperature is 40°c? [coefficient of linear expansivity
of steel = 11 × \(10^{-6}\)\(K^{-1}\)] (JAMB)
Solution
$$ {l_1} = 20m $$
$$ {l_2} = ? $$
$$ {\theta_1} = 20°C $$
$$ {\theta_2} = 40°C $$
$$ \alpha = 11 × 10^-6K^-6 $$
$$ ∆\theta = {\theta_2} - {\theta_1} $$
$$ ∆\theta = 40°C - 20°C = 20°C $$
$$ ∆l = \alpha{{l_1}{∆\theta}}$$
$$ ∆l = 11 × 10^-6 × 20 × 20 $$
$$ ∆l = 0.0044m $$
$$ ∆l = {l_2}-{l_1} $$
$$ {l_2} = ∆l + {l_1} $$
$$ {l_2} = 0.0044 + 20 $$
$$ \text{length after expansion} = 20.0044m $$
Example 3: The ratio of the linear expansivity of copper to that of iron is approximately 1.5. A specimen
of iron and a specimen of copper expand by the same amount per unit rise in temperature. The ratio of their length
is? (JAMB)
Solution
$$\text{Since all other parameters are constant} $$
$$ \text{For copper} $$
$$ {\alpha_{Cu}} = \frac{1}{{l_1}_{Cu}} $$
$$ \text{For Iron (Fe)} $$
$$ {\alpha_{Fe}} = \frac{1}{{l_1}_{Fe}} $$
$$\therefore \text{Equating both} $$
$$ {\alpha_{(Cu)}}{l_1}_{(Cu)} = {\alpha_{(Fe)}}{l_1}_{(Fe)}$$
$$ \frac{{\alpha_{(Cu)}}}{{\alpha_{(Fe)}}} = \frac{3}{2} $$
$$ \frac{{l_1}_{(Cu)}}{{l_1}_{(Fe)}} = \frac{2}{3} = 0.67 $$