A glass flask is filled 'to the mark" with $50.00 \mathrm{~cm}^{3}$ of mercury at $18^{\circ} \mathrm{C}$. If the flask and its contents are heated to $38^{\circ} \mathrm{C}$, how much mercury will be above the mark? $\alpha_{\text {glass }}=9.0 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and $\beta_{\text {mercury }}=182 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}$
We shall take $\beta_{\text {elass }}=3 \alpha_{\text {plass }}$ as a good approximation. The flask interior will expand just as though it were a solid piece of glass. Thus,
Volume of mercury above mark $=(\Delta V$ for mercury) $-(\Delta V$ for glass)
$$
\begin{array}{l}
=\beta_{m} V_{0} \Delta T-\beta_{g} V_{0} \Delta T=\left(\beta_{m}-\beta_{g}\right) V_{0} \Delta T \\
=\left[(182-27) \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\right]\left(50.00 \mathrm{~cm}^{3}\right)\left[(38-18){ }^{\circ} \mathrm{C}\right] \\
=0.15 \mathrm{~cm}^{3}
\end{array}
$$