Question

The coefficient of linear expansion for some common materials are listed in the table. Material Suppose Mary would like to design an emergency overheating alarm using one of these materials. In her design, at room temperature (23.0 "C), a 6.01 cm long cylinder of the material will sit with one end against a wall and the other just barely touching a button. Once the heat causes the cylinder to lengthen by 0.0238 cm, the button will be activated, turning on the alarm. aluminum brass copper steel cast iron titanium glass ycarbonate702x 10-6 22.2x 10-6 18.7x 106 16.6x 106 130x 10-6 104x 106 8.60 x 10-6 5.92 x 106 At what temperature will the alarm be triggered if the cylinder is made out of aluminum? temperature: At what temperature will the alarm be triggered if the cylinder is made out of steel? "С emperature: netbeans-8.2-java..e ie Dropshipping-Ten..rtf ^ ki netbeans-s2-javailneree.Pa ︿ exe
Two part question

Answer 1

We Knar, in eene a expansion d is linear expansum cretAaen Giver L 4レ20.0238 cm To 23.0。 OC 0238 ジ (11-23.0)。 う !T, 201.4℃

O 23& 4,01 x13시

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Answer 2

The linear expansion of a typical solid is governed by the relation

$$\mathrm{\Delta }L=\alpha L\mathrm{\Delta }T$$

Here, ?$L$ is the length of the material at the initial temperature, ?$\alpha$ is the coefficient of (linear) thermal expansion, Δ?$\mathrm{\Delta }T$ is the change in temperature, and Δ?$\mathrm{\Delta }L$ is the change in length. As the temperature increases, so does the length of an object.

Since the alarm is designed with the material initially at room temperature (23.0 ∘C)$(23.0{}^{\circ }\text{C})$ , Δ?$\mathrm{\Delta }T$ can be written as

$$\begin{array}{rl}\mathrm{\Delta }T& =T_{\text{f}}-T_{\text{i}}\\ & =T_{\text{f}}-\left(23.0{}^{\circ }\text{C}\right)\end{array}$$

Combining this with the linear expansion equation and solving for the final temperature gives

$$T_{\text{f}}=\left(23.0{}^{\circ }\text{C}\right)+\frac{\mathrm{\Delta }L}{\alpha L}$$

Converting Δ?$\mathrm{\Delta }L$ from millimeters to centimeters and taking the appropriate expansion coefficient from the table, the alarm temperature for a copper cylinder is

$$\begin{array}{rl}{T}_{\text{f}}& =\left(23.0{}^{\circ }\text{C}\right)+\frac{0.0233\text{ cm}}{(16.6\times {10}^{-6}1{/}^{\circ }\text{C})\left(5.26\text{ cm}\right)}\\ & =2.90\times {10}^2{}^{\circ }\text{C}\end{array}$$

For the titanium cylinder, the alarm temperature would be

?f=(23.0 ∘C)+0.0233 cm(8.60×10−6 1/∘C)(5.26 cm)=538 ∘C