Question

540 g of ice at 0 ^C is mixed with 540 g of water at 80^C , The resultant temperature is

Answer 1

Latent heat to ice at 0°c to convert into saturated water at 0°c + sensible heating of this saturated water from 0°c to final temp T°c = cooling of hot water from 80°c to final temp T°c

=> (540*80) + {540*1*(T-0)} = {540*1*(80-T)}

=> 540*T = -540*T

=> 1080*T = 0

=> T = 0

So final temp of mixture is 0°c and mixture is saturated water.