Question

Find the change in entropy if 200g of ice at -2 ^oC is added to 700g of water at 80 ^oC that is in a 150g aluminum pot.

Answer 1

specific heat capacity of different material:

ice=2108 J/(kg.K)

water=4186 J/(kg.K)

aluminum=900 J/(kg.K)

latent heat of fusion of ice=334 kJ/kg

let final temperature is T degree celcius and T>0 so that all the ice have melted.

writing heat balance equation:

heat lost by aluminimum + heat lost by water=heat gained by ice

==>0.15*900*(80-T)+0.7*4186*(80-T)=0.2*2108*(0-(-2))+0.2*334*1000+0.2*4186*(T-0)

==>0.15*900*80+0.7*4186*80-0.2*2108*2-0.2*334*1000=0.15*900*T+0.7*4186*T+0.2*4186*T

==>T=(0.15*900*80+0.7*4186*80-0.2*2108*2-0.2*334*1000)/(0.15*900+0.7*4186+0.2*4186)=45.5 degree celcius

so heat lost=-Q=0.15*900*(80-T)+0.7*4186*(80-T)=1.0574*10^5 J

heat gained=Q=1.0574*10^5 J

so change in entropy=change in entropy of ice + change in entropy of water and aluminium

=(Q/temperature of ice)+(-Q/temperature of water)

=90.637 J/K