Let t be the common temperature above zero celsius. Heat lost by calorimeter and water added
=400×10−3×1000(5−t)+200×10−3×1000(10−t)
Heat gained by ice=400×10−3×5000(60+t)+400×10−3×80000+400×10−3×1000t
Since Heatlost=heatgained
∴400(5−t)+200(10−t)=200(60+t)+400×80×+400t
or t=−24.33∘C
This is contradictory to the assumption that t is above zero. Hence, the final temperature is either 0∘C or less than 0∘C. Let the temperature be t∘C (less than 0∘C).
Then 400×10−3×1000×5+200×10−3×1000×10+600×10−3×80000+600×10−3×500×t=400×500(60−t)
or 400×5+200×10+600×80+300t=200(60−t)
or 20+20+480+3t=120−2t
or t=−80∘C
This is absurd because the lowest temperature is −60∘C and we discard this second assumption also.
So the final temperature is 0∘C.
Let x grams of water be converted into ice.
Heatlost=400×10−3×100×5+200×10−3×1000×10+x×10−3×8000=4000+80x
Heatgained=400×10−3×500×60
or 4000+80x=12000
or x=100g
Hence, the final result is 500g of ice and 500g of water at 0∘C.