Answer
a.
overflows
b.
$24\mathrm{c}\mathrm{m}^{3}$
Work Step by Step
We use $\qquad \Delta V=\beta V\Delta T \qquad$ 16-6.
$\beta_{w}=0.21\times 10^{-3}(\mathrm{C}^{\mathrm{o}})^{-1}=21\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}$
$\beta_{\mathrm{A}\mathrm{l}}=3\alpha_{\mathrm{A}\mathrm{l}}=3\cdot 2.4\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}=7.2\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}$
a.
Water has a larger $\beta$, so its volume increases more than the saucepan's.
It overflows.
b.
Initial volumes:
$V_{0}=\displaystyle \pi r^{2}h=\pi(\frac{23\mathrm{c}\mathrm{m}}{2})^{2}(6.0 \mathrm{c}\mathrm{m} )=2493\mathrm{c}\mathrm{m}^{3}$
$V_{overflow}=\Delta V_{\mathrm{w}}-\Delta V_{\mathrm{A}1}$
$=\beta_{\mathrm{w}}V_{0}\Delta T-3\alpha_{\mathrm{A}1}V_{0}\Delta T$
$=(\beta_{\mathrm{w}}-3\alpha_{\mathrm{A}1})V_{0}\Delta T$
$=(21-7.2)\times 10^{-5}(\mathrm{C}^{\mathrm{o}})^{-1}(2493\mathrm{c}\mathrm{m}^{3})[(88-19)^{\mathrm{o}}\mathrm{C}]$
$=24\mathrm{c}\mathrm{m}^{3}$
