Chapter 13.4, Problem 40PP

To determine

Length of the steel at $\text{1221°C}$ .

Answer to Problem 40PP

  $\text{11}\text{.66}\text{cm}$

Explanation of Solution

Given information:

A piece of steel, ${\text{L}}_{\text{1}}\text{ = 11}\text{.5}c\text{m}=0.115m$

Initial temperature, ${\text{T}}_{\text{1}}\text{ = 22°C}$

Secondary temperature, ${\text{T}}_{\text{2}}\text{ = 1221°C}$

Coefficient of linear expansion, ${\text{α = 12×10}}^{\text{-6}}\text{°}{\text{C}}^{\text{-1}}$ and Final length of piece of steel, $L_2=?$

Formula used:

The length, L, of a solid at temperature T can be found with the following equation:

  $L_2=L_1+\alpha L_1(T_2-T_1)$

Where, $L_1$ is the length at temperature $T_1$

  $L_2$ is the length at temperature $T_2$

  $\alpha$ is coefficient of linear expansion

Calculation:

Apply the formula, $L_2=\alpha L_1(T_2-T_1)$

  ${\text{L}}_{\text{2}}\text{=}\left(\text{0}\text{.115}\text{m}\right)\text{+}\left({\text{12×10}}^{\text{6}}{\text{°C}}^{\text{-1}}\right)\left(\text{0}\text{.115}\text{m}\right)\left(\text{1221°C-22°C}\right)$

  ${\text{L}}_{\text{2}}\text{= 11}\text{.66}\text{cm}$

Conclusion: The length of steel will be $\text{11}\text{.66}\text{cm}$ at $\text{1221°C}$ .