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Question

10 g of ice at 20C is added to 20 g of water at 20C. Assuming that no heat is lost to the surrounding, find the temperature of the mixture.
[Take, specific heat capacity of ice Ci=2100 J kg1C1, specific heat capacity of water Cw=4200 J kg1C1 & Latent heat of fusion of ice L=3.36×105 J kg1]

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 0C
The heat released by water when it cools down 20C to 0C is given by
Q1=mCwΔT
Q1=0.02 kg×4200 J kg1C1×20C=1680 J ....(1)
The heat absorbed by ice to reach 0C from 20C is given by
Q2=mCiΔT
Q2=0.01 kg×2100 J kg1C1×20C=420 J .....(2)
Also:
Heat required to just melt 10 g of ice is given by
Q3=mL=0.01 kg×3.36×105 J kg1=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 0C
Hence, option (a) is the correct answer.

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