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Question

10 g of ice at −20∘C is added to 20 g of water at 20∘C. Assuming that no heat is lost to the surrounding, find the temperature of the mixture. [Take, specific heat capacity of ice Ci=2100 J kg−1∘C−1, specific heat capacity of water Cw=4200 J kg−1∘C−1 & Latent heat of fusion of ice L=3.36×105 J kg−1]

A
0C
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B
1C
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C
2C
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D
3C
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Solution

The correct option is A 0∘CThe heat released by water when it cools down 20∘C to 0∘C is given by Q1=mCwΔT Q1=0.02 kg×4200 J kg−1∘C−1×20∘C=1680 J ....(1) The heat absorbed by ice to reach 0∘C from −20∘C is given by Q2=m′CiΔT Q2=0.01 kg×2100 J kg−1∘C−1×20∘C=420 J .....(2) Also: Heat required to just melt 10 g of ice is given by Q3=m′L=0.01 kg×3.36×105 J kg−1=3360 J .....(3) From the principle of calorimetry, Heat lost by water = Heat gained by ice. Q1=Q2+Q3 But from (1), (2) and (3), we can say that, Q1<Q2+Q3 [∵Q1<Q3 and Q1>Q2] Thus, we can conclude that all the ice will not be melt and temperature becomes 0∘C Hence, option (a) is the correct answer.

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