Example 20.5. Expansion of an Aluminum and a Copper Rod with Rise in Temeperature.
An aluminum and a copper bar each of length \(1.0\text{ m}\) at \(27^{\circ}\text{C}\) are put in boiling water at \(99^{\circ}\text{C}\text{.}\) What is the difference in their lengths at \(99^{\circ}\text{C}\text{?}\)
Data: \(\alpha_{\text{aluminum}} = 23.1\times 10^{-6}\ \text{per }^{\circ}\text{C}\) and \(\alpha_{\text{copper}} = 16.5\times 10^{-6}\ \text{per }^{\circ}\text{C}\text{.}\)
Answer.
\(0.47\text{ mm}. \)
Solution.
Using \(\Delta L = \alpha L_0\Delta T,\) we find the change in the lengths of the two bars when their temperatures are raised to \(99^{\circ}\text{C}\text{.}\)
\begin{align*}
\amp \Delta L_{\text{aluminum}} = \frac{23.1\times 10^{-6}}{\ ^{\circ}\text{C}}\times 1\ \text{m} \times (99-27)^{\circ}\text{C} = 1.66\ \text{mm}.\\
\amp \Delta L_{\text{copper}} = \frac{16.5\times 10^{-6}}{\ ^{\circ}\text{C}}\times 1\ \text{m} \times (99-27)^{\circ}\text{C} = 1.19\ \text{mm}.
\end{align*}
Therefore, the difference in the lengths of the the two bars at \(99^{\circ}\text{C}\) will be \(1.66\text{ mm} - 1.19\text{ mm} = 0.47\text{ mm}\text{.}\)
