[tex]\theta_0=5\ºC\\ \theta=50\ºC\\l_0=4\ cm=4\times10^{-2}\ m\\ \Delta l=0.6\ cm=6\times10^{-3}\ m\\\\ \boxed{\Delta l=l_0\alpha\Delta\theta}\ \therefore\ 6\times10^{-3}=4\times10^{-2}\alpha(50-5)\ \therefore[/tex]
[tex]0.6=180\alpha\ \therefore\ \alpha=\dfrac{1}{300}\ \therefore\ \boxed{\alpha\approx3.333\times10^{-3}\ \ºC^{-1}}[/tex]
