As a copper wire is heated, its length increases by 0.200%. What is the change of the temperature of the wire? ($\alpha$_Cu= 16.6 *10 ^-⁶/C $\degree$) A) 120 $\degree$C B) 60.2 $\degree$C C) 30.1 $\degree$C D) 6.0 $\degree$C

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The change in length is given by:$$\Delta L = \alpha L_0 \Delta T$$where:* $\Delta L$ is the change in length* $\alpha$ is the coefficient of linear expansion* $L_0$ is the original length* $\Delta T$ is the change in temperatureWe are given that $\Delta L = 0.200%$ of $L_0$, so we can write:$$\frac{\Delta L}{L_0} = 0.00200 = \alpha \Delta T$$Solving for $\Delta T$, we get:$$\Delta T = \frac{\Delta L}{\alpha L_0} = \frac{0.00200}{16.6 	imes 10^{-6} 	ext{/C}} = 120 	ext{C}$$Therefore, the change in temperature of the wire is 120C.

As a Copper Wire Is Heated, Its Length Increases by 0.200