A piece of ice of mass 100 g and at temperature 0⁰C is put in 200 g of water at 25⁰C. How much ice will melt
as the temperature of water reaches 0⁰C ? (Take, latent heat of fusion of ice = 3.4 X 10⁵ Jkg^-1) 

Given, mass of ice-piece, m₁ = 100 g = 0.1 kg

Mass of water, m₂ = 200 g = 0.2 kg

When temperature of mixture is 0⁰C, then heat lost by water,

$Q=m_{2}c\left(\mathrm{\Delta }\theta \right)=0.2\times 4200(25-0)$     ($\because$ Specific heat of water = 4200 J)

$\Rightarrow$ Q = 21000J

Now, mass of melted ice,  $m\text{'}=\frac{Q}{L}=\frac{21000}{3.4\times {10}^5}\simeq 61.8\mathrm{g}$