Answer
(a) $b_{container}=49.7\times10^{-6} \frac{1}{^{\circ}C}$
(b) Most likely copper.
Work Step by Step
(a) The expansion of the container causes the water inside the container to increase as if it were made of the same material as the container. Thereforce, the volume of water lost is $$\Delta V=\Delta V_{water} - \Delta V_{container}= b_{water}V_{0}\Delta T - b_{container}V_{0}\Delta T=V_{0}(b_{water}-b_{container})\Delta T$$ $$(0.35g)(\frac{1 mL}{0.98324 g})=(55.50mL)(210\times10^{-6}-b_{container})(60^{\circ}C-20^{\circ}C)$$ $$b_{container}=49.7\times10^{-6} \frac{1}{^{\circ}C}$$
(b) Based on the Table 13-1, the material of the container is most likely copper, which has a coefficient of volume expansion $50\times10^{-6}\frac{1}{^{\circ}C}$.
