Question

A metal rod .50 m is heated from 15C to 95C. The length of the rod increases by 0.96
mm. What is the coefficient of expansion for the rod?

Answer 1

$\Large\boxed{\sf{\quad\alpha=2.4\times10^{-5}\ K^{-1}\quad}}$

Length of the metal rod,

• $\sf{L=0.5\ m}$

Initial temperature,

• $\sf{T_1=15^oC}$

And final temperature,

• $\sf{T_2=95^oC}$

So the difference in temperature is,

• $\sf{\Delta T=T_2-T_1}$

• $\sf{\Delta T=95^oC-15^oC}$

• $\sf{\Delta T=80^oC}$

Or,

• $\sf{\Delta T=80\ K}$

[Note: The units $\sf{\ ^oC,\ K}$ are same in case of temperature difference.]

The extension formed in the rod,

• $\sf{\Delta L=0.96\ mm}$

• $\sf{\Delta L=0.96\times10^{-3}\ m}$

We have,

$\longrightarrow\sf{\Delta L=L\cdot\alpha\cdot\Delta T}$

Then the coefficient of linear expansion,

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

$\longrightarrow\sf{\alpha=\dfrac{0.96\times10^{-3}}{0.5\times80}}$

$\longrightarrow\sf{\underline{\underline{\alpha=2.4\times10^{-5}\ K^{-1}}}}$