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Question
120 g of ice at 0∘C is mixed with 100 g of water at 80∘C . Latent heat of fusion is 80 cal/g and specific heat of water is 1 cal/g∘C. The final temperature of the mixture is
(correct answer + 1, wrong answer - 0.25)
(correct answer + 1, wrong answer - 0.25)
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Solution
The correct option is A 0∘C
Heat rejected by 100 g of water at 80∘C when it's temperature becomes 0∘C is
Q=m×s×Δθ=(100)(1)(80)=8000 cal
But this heat can melt m=QL=800080=100 g of ice at 0∘C only.
Final mixture consists of 200 g of water and 20 g of ice at 0∘C.
Heat rejected by 100 g of water at 80∘C when it's temperature becomes 0∘C is
Q=m×s×Δθ=(100)(1)(80)=8000 cal
But this heat can melt m=QL=800080=100 g of ice at 0∘C only.
Final mixture consists of 200 g of water and 20 g of ice at 0∘C.
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