Question

An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?

Answer 1

Using linear thermal expansion formula we get,

`\begin{gathered} \Delta L=\text{ }\alpha L_0L\Delta T \\ \Delta L=\text{ expansion in length;} \\ \alpha=\text{ coefficient of linear expansion;} \\ L_0=\text{ original length;} \\ L=\text{ expanded length;} \\ \Delta T=\text{ change in temperature ;} \end{gathered}`

For aluminum rod

`\begin{gathered} \Delta L_1=\text{ }\alpha_1*5* L_1*50;\begin{cases}L={5\text{ m}} \\ \Delta T={70-20=50\degree C}\end{cases} \\ L_1=\text{ }(\Delta L_1)/(250\alpha_1);\text{ -------\lparen1\rparen} \end{gathered}`

For nickel rod

`\begin{gathered} \Delta L_2=\text{ }\alpha_2*5* L_2*50; \\ L_2=\text{ }(\Delta L_2)/(250\alpha_2)\text{ -------\lparen2\rparen} \end{gathered}`

Now equation( 1 ) - (2)

`L_1-L_2=(\Delta L_1\alpha_2-\Delta L_2\alpha_1)/(250\alpha_1\alpha_2)`

Above relation gives the final answer