Question

540 g of ice at 0 degre is mixed with 540 g of water at 80 degree celcius. The temperature of final mixture is

Answer 1

Ice at 0°C has a total latent heat of  Q=m*L = 540*80 = 43200 cal.  

Now for water to come at 0°C and it's state to remain as water, we need to remove heat from it which is given by,  

Q= m*Cp*(∆T) = 540* 1* 80 = 43200 cal.  

So, we can see that  

Heat released by water to reach from 80°C to 0°C = heat required by ice to melt into water at 0°C.  

So, the ice will melt to water and it's temperature will be 0°C.  

Hence, the final state would be water at 0°C.