Hi Alan,
I don't know about this stuff so I am just using your formula and a bit of logic (most likely incorrect)
"If α is the linear coefficient of thermal expansion then
α = change in length/(original length*temperature change)"
$$\\11\times 10^{-6}=\frac{\triangle l_s}{23.89\times 195}\\\\
\triangle l_s = 11\times 10^{-6}\times23.89\times 195 \\\\$$
$${\mathtt{11}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{6}}\right)}{\mathtt{\,\times\,}}{\mathtt{23.89}}{\mathtt{\,\times\,}}{\mathtt{195}} = {\mathtt{0.051\: \!244\: \!05}}$$
$$\\\alpha_R=\frac{\triangle l_s+(24.01-23.89)}{23.89*195}\\\\
\alpha_R=\frac{0.05124405+0.12}{4658.55}\\\\$$
$${\frac{\left({\mathtt{0.051\: \!244\: \!05}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.12}}\right)}{{\mathtt{4\,658.55}}}} = {\mathtt{0.000\: \!036\: \!759\: \!088\: \!128\: \!3}}$$
$$\\\alpha_R=3.7\times 10^{-5} $ per degree celcius$ \qquad \\
$(correct to 2 significant figures)$$$
Is this correct?
+118733 
