2 Kg of ice at -20 degree C is mixed with 5Kg of water 20 degree C in an insulating vessel having negligible heat capacity. Calculate the final mass of water remaining in the comtainer. It is given thet the specific heats of water and ice are 1Kcal/kg/degree C and 0.5Kcal/kg/degree C while latent heat of fusion of ice is 80 kcal/kg

the amt of water remain is 6 kg

Heat required for ice to reach melting point=m*c(ice)*∆T =2*0.5*20 =20JAmt of heat required by 2 kg of ice to convert into water=m*L =2*80 =160JAmt of heat required by 5 kg of water to reach freezing point=m*c(water)*∆T=5*1*20=100JOut of these 100J,20J are required by the ice to reach melting point.Thus, Amt of heat left (Q1) is 100-20 =80JAmt of ice that melts (m1)=Q1/L=80/80=1Kg.Thus, final mass of water=5+1=6kg