A 0.4-m (15.7-in.) rod of a metal elongates \(0.48\) \(\mathrm{mm}\left(0.019\right.\) in.) on heating from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) \(\left(68^{\circ} \mathrm{F}\right.\) to \(\left.212^{\circ} \mathrm{F}\right)\). Determine the value of the linear coefficient of thermal expansion for this material.

Short Answer

Expert verified
Answer: The linear coefficient of thermal expansion for the material of the rod is 1.5 x 10^-5 °C^(-1).

Step by step solution

01

Write down the given quantities

We are given the following data: - Initial length of the rod: \(L_0=0.4 \thinspace m\) - Elongation of the rod: \(\Delta L=0.48 \thinspace mm = 0.00048 \thinspace m\) - Change in temperature: \(\Delta T=100^{\circ}C - 20^{\circ}C=80^{\circ}C\)
02

Write down the formula for linear expansion

The formula for linear expansion is given by: \(\Delta L = L_0 \alpha \Delta T\) where \(\Delta L\) is the elongation of the rod, \(L_0\) is the initial length of the rod, \(\alpha\) is the linear coefficient of thermal expansion, and \(\Delta T\) is the change in temperature.
03

Solve for the linear coefficient of thermal expansion

We want to find the value of \(\alpha\). We can rearrange the linear expansion formula to solve for \(\alpha\): \(\alpha = \frac{\Delta L}{L_0 \Delta T}\) Now, we can plug in the given values to find the value of \(\alpha\): \(\alpha = \frac{0.00048 \thinspace m}{0.4 \thinspace m \times 80^{\circ}C}= \frac{0.00048}{32} = 1.5 \times 10^{-5} \thinspace \frac{1}{^{\circ}C}\)
04

Write down the result

The linear coefficient of thermal expansion for the material of the rod is \(\alpha = 1.5 \times 10^{-5} \thinspace ^{\circ}C^{-1}\).

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