A lump of ice (0.1 kg) at −10∘C is put in 0.15 kg of water at 20∘C. How much water will be found in the mixture when it has reached thermal equilibrium? (Cice=0.5kcalkg−1k−1,Lice=80kcalkg−1)
A
150 g
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B
31.25 g
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C
181.25 g
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D
210 g
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Solution
The correct option is C 181.25 g The heat released by water (20∘C→0∘C) Q1=mcΔT=0.15×1×103×(20−0)=5kcal
Heat absorbed by ice (−10∘C→0∘C) Q2=0.1.(0.5×103).(0−(−10))=0.5kcal
Heat that remains = 3 -0.5 = 2.5 k cal
The heat that remains can melt only 250080×103 kg of ice to water = 0.03125 kg
Total water = 150 g + 31.25 g
= 181.25 g
The equilibrium temperature will be 0∘C