Question

The length of a rod increases from 50 cm to 50.12 cm, when its temperature is increased from 12 °C to 212 °C. Calculate its coefficient of liner expansion.

Answer 1

Initial length of rod, $\sf{L_1=50\ cm}$

Final length of rod, $\sf{L_2=50.12\ cm}$

Change in length of rod,

$\sf{\longrightarrow \Delta L=L_2-L_1}$

$\sf{\longrightarrow \Delta L=50.12\ cm-50\ cm}$

$\sf{\longrightarrow \Delta L=0.12\ cm}$

Initial temperature, $\sf{T_1=12^oC}$

Final temperature, $\sf{T_2=212^oC}$

Change in temperature,

$\sf{\longrightarrow \Delta T=T_2-T_1}$

$\sf{\longrightarrow \Delta T=212^oC-12^oC}$

$\sf{\longrightarrow \Delta T=200^oC}$

or,

$\sf{\longrightarrow \Delta T=200\ K}$

(°C and K are considered same in case of temperature difference.)

We know change in length is given by,

$\sf{\longrightarrow \Delta L=L_1\alpha\,\Delta T}$

Then coefficient of linear expansion is,

$\sf{\longrightarrow \alpha=\dfrac{\Delta L}{L_1\ \Delta T}}$

$\sf{\longrightarrow \alpha=\dfrac{0.12}{50\times200}}$

$\sf{\longrightarrow\underline{\underline{\alpha=1.2\times10^{-5}\ K^{-1}}}}$