10 g of ice at$-\text{20°C}$  is added to 10 g of water at$\text{50°C}$ . The amount of ice and water that are present at equilibrium

given that

mass of ice${\text{m}}_{\text{i}}=10\text{g}$

initial temperature ${\text{t}}_{\text{1}}=-20°\text{C}$

Latent heat of fusion ${\text{L}}_{\text{f}}=80\text{cal/g}$

Specific heat of ice $=0.5\text{cal/g°C}$

water added ${\text{m}}_{\text{w}}=10\text{g}$

initial temperature$=50°\text{C}$

heat required by ice to become ice at $0°\text{C}$

${\text{Q}}_{\text{i}}={\text{m}}_{\text{i}}\times {\text{s}}_{\text{i}}\left(0-\left(-20°\text{C}\right)\right)$

$=10\times \text{1/2}\times 20=100\text{cal}$

heat required by the ice to convert in to water at $0°\text{C}$

${\text{Q}}_{\text{2}}={\text{m}}_{\text{i}}\times {\text{L}}_{\text{f}}$

$=10\times 80=800\text{cal}$

Let us account the heat that water can release by lowering its temperature to $0°\text{C}$

${\text{Q}}_{\text{3}}={\text{m}}_{\text{w}}\times {\text{s}}_{\text{w}}\left({\text{t}}_{\text{2}}-0°\text{C}\right)$

$=10\times 1\times 50=500\text{cal}$

Here the total heat required by Ice to become water at $0°\text{C}$ > heat that water can provide before getting into the thermal equilibrium

$800\text{cal}>500\text{cal}$

So final temperature of mixture will be$0°\text{C}$  

We can observe ice at $-20°\text{C}$ absorb 100 cal from water to become ice at $0°\text{C}$

And avail remaining amount of heat i.e., 400 cal to convert into water the mass of ice at$0°\text{C}$  that can convert as water

$400=\text{m}\times 80$

$\therefore \text{m}=5\text{g}$

Therefore at equilibrium

We get 5 g ice remains and 15 g of water