1. The problem statement, all variables and given/known data 1Kg ice at -10° C is mixed with 1 Kg water at 100°C .Then find equilibrium temperature. 2. Relevant equations Specific heat of water = 1Cal/g/°C Specific heat of ice = 0.5 Cal/g/°C Latent heat of fusion of ice = 80 Cal/g 3. The attempt at a solution Heat required by ice to reach at 0°C= 1000x0.5x10 = 5000Cal Heat required by ice at 0°C to convert in water = 1000x80 = 80000Cal Heat released by water when it moves from 100°C to 0°C = 1000x1x100= 100000Cal Surplus heat = 15000 Cal This surplus heat increases the temp. of 2Kg water at 0°C 15000Cal = 2000x1x∆t Increase in temperature = 15000/2000 = 7.5°C Is my answer correct ?