解：$(1)ρ_{冰}= 0.9×10^3\ \mathrm {kg/m}^3=0.9\ \mathrm {g/cm}^3，$冰的体积$V_{冰}=Sh=30\ \mathrm {cm}^2×10\ \mathrm {cm}=300\ \mathrm {cm}^3，$容器内冰的质量$m_{冰}=ρ_{冰}V_{冰}=0.9\ \mathrm {g/cm}^3× 300\ \mathrm {cm}^3=270\ \mathrm {g}$
$(2)$剩余水的质量$m_{水}'=m_{冰}−m_{溢}=270\ \mathrm {g}-15\ \mathrm {g}=255\ \mathrm {g}，$剩余水的体积$V_{水}'=\frac {m_{水}'}{ρ_{水}}= \frac {255\ \mathrm {g}}{1.0\ \mathrm {g/cm}^3}=255\ \mathrm {cm}^3，$铁球的体积$V_{球}=V_{冰}−V_{水}'=300\ \mathrm {cm}^3−255\ \mathrm {cm}^3=45\ \mathrm {cm}^3，$铁球的质量$m_{球}=m_{总}−m_{水}'−m_{容器}=483\ \mathrm {g}−255\ \mathrm {g}−70\ \mathrm {g}=158\ \mathrm {g}，$$ρ_{铁}=7.9×10^3\ \mathrm {kg/m}^3=7.9\ \mathrm {g/cm}^3，$实心部分的体积$V_{球实}=\frac {m_{球}}{ρ_{铁}}=\frac {158\ \mathrm {g}}{7.9\ \mathrm {g/cm}^3}=20\ \mathrm {cm}^3，$故空心部分的体积$V_{空}=V_{球}-V_{球实}=45\ \mathrm {cm}^3−20\ \mathrm {cm}^3=25\ \mathrm {cm}^3$
