Given : mass of water $\left(\mathit{m}\right)=140\mathrm{g}$ , initial temperature of water $={50}^{\circ }\mathrm{C}$
mass of ice $\left({\mathit{m}}^{\prime }\right)=60\mathrm{g}$, initial temperature of ice $=0^{\circ }\mathrm{C}$.
Let the final temperature $={\mathrm{T}}^{\circ }\mathrm{C}$
By the principle of method of mixture, Heat lost by water $=$ Heat gained by ice
$\mathit{m}\mathit{c}∆\mathit{t}={\mathit{m}}^{\prime }\mathrm{L}+{\mathit{m}}^{\prime }\mathrm{C}∆{\mathit{t}}^{\prime }$
$140\times 4.2\times \left(50-\mathrm{T}\right)=60\times 336$ $+60\times 4\cdot 2\times \left(\mathrm{T}-0\right)$
$588\left(50-\mathrm{T}\right)=20160+252\mathrm{T}$
$29400-588\mathrm{T}=20160+252\mathrm{T}$
$29400-20160=588\mathrm{T}+252\mathrm{T}$
$9240=840\mathrm{T}$