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 meltas the temperature of water reaches 00C ? (Take, latent heat of fusion of ice = 3.4 X 105 Jkg-1) [UK PMT 2015]

Given, mass of ice-piece, m1 = 100 g = 0.1 kgMass of water, m2 = 200 g = 0.2 kgWhen temperature of mixture is 00C, then heat lost by water,Q=m2c(Δθ)=0.2×4200(25−0)     (∵ Specific heat of water = 4200 J)⇒ Q = 21000JNow, mass of melted ice,  m'=QL=210003.4×105≃61.8g