Answer
a) $y^2 = .25(L_0^2-d^2)+\frac{1}{2}L_0^2 \alpha \Delta T$
b) $\alpha = 2.35 \times 10^{-6} ; d=.8 \ meters $
c) Aluminum
Work Step by Step
a) Looking for a linear relationship between some function of y and T, we find:
$y^2 = .25(L_0^2-d^2)+\frac{1}{2}L_0^2 \alpha \Delta T$
Note, this is linear because the function of y, in this case $y^2$, is equal to a constant, $.25(L_0^2-d^2)$, plus T raised to the first power multiplied by a constant, $\frac{1}{2}L_0^2 \alpha \Delta T$.
b) We plot the data in Microsoft Excel, and we plot the best fit line. Doing this, we find that: $\alpha = 2.35 \times 10^{-6}$, and $ d=.8 \ meters $
c) Using the tables for the values of $\alpha$, we find that the substance is Aluminum.
