Given, mass of water mw=300g Temperature of water, Tw=25∘C Mass of ice, ‌‌mi=100g Ti=0∘C Heat required to convert the ice at 0∘C to water at 0∘C
Q1=miLi=100×80‌‌[Li=80cal/g]
=8000cal....(i) Heat given by water from 25∘C water to 0∘C water
Q2=mwc∆T=300×1×(25−0)
⇒‌‌=7500cal........(ii) From Eqs. (i) and (ii), we observed that $Q_{2} Hence, total ice will not be melt, so final temperature of mixture will be 0∘C with some unmelted ice in mixture.