Heat given by water to reach $5^{\circ }\mathrm{C}+$ Heat given by copper vessel to reach $5^{\circ }\mathrm{C}=$ Heat taken by ice to melt at $0^{\circ }\mathrm{C}$ + Heat taken by melted ice to reach $5^{\circ }\mathrm{C}$
$\mathit{m}\mathit{c}∆\mathit{t}+\mathit{m}\mathit{c}∆\mathit{t}=\mathit{m}\mathrm{L}+\mathit{m}\mathit{c}∆\mathit{t}$
where $∆\mathit{t}=$ change in temperature
$\left[150\times 4.2\times \left({50}^{\circ }-5^{\circ }\right)\right]+[100\times 0.4\times$ $\left({50}^{\circ }-5^{\circ }\right)]$
$=\left(\mathit{m}\times 336\right)+\left[\mathit{m}\times 4.2\times \left(5-0\right)\right]$ $\left(150\times 4.2\times 45\right)+\left(100\times 0.4\times 45\right)$
$=\left(\mathit{m}\times 336\right)+\left(\mathit{m}\times 4.2\times 5\right)$
$28350+1800=336\mathit{m}+21\mathit{m}$
$357\mathit{m}=30150$
$\mathit{m}=\frac{30150}{357}$