Equal masses of ice at $0\text{°C}$ and water at $100\text{°C}$ are mixed. The resultant temperature is

given that equal masses of ice at$0\text{°C}$  and water at $100\text{°C}$ are mixed

Let mass of ice = m

${\text{L}}_{\text{f}}=80\text{cal/g}$

initial temperature = $0\text{°C}$

resulting temperature = t

mass of water = m

initial temperature = $100\text{°C}$

final temperature = $\text{t°C}$

heat lost by water = heat gained by ice to become water at$0\text{°C}$  + heat gained by $0\text{°C}$ water to raise its temperature $\text{t°C}$

$\text{m}\times 1\times \left(100-\text{t}\right)=80\text{m}+\text{m}\times 1\times \left(\text{t}-0\right)$

$\text{m}\left(100-\text{t}\right)=\text{m}\left(80+\text{t}\right)$

$100-\text{t}=80+\text{t}$

$200=2\text{t}$

$\therefore \text{t}=10°\text{C}$