1Kg ice at -10° C is mixed with 1 Kg water at 100°C .Then find equilibrium temperature.
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
The Attempt at a Solution
Heat required by ice to reach at 0°C= 1000x0.5x10
Heat required by ice at 0°C to convert in water = 1000x80
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
Is my answer correct ?