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Question
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=0.5 k cal kg−1k−1, Lice=80 k cal kg−1)
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=0.5 k cal kg−1k−1, Lice=80 k cal kg−1)
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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)=5 kcal
Heat absorbed by ice (−10∘ C→0∘ C)
Q2=0.1.(0.5×103).(0−(−10))=0.5 k cal
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
The heat released by water (20∘ C→0∘ C)
Q1=mcΔT=0.15×1×103×(20−0)=5 kcal
Heat absorbed by ice (−10∘ C→0∘ C)
Q2=0.1.(0.5×103).(0−(−10))=0.5 k cal
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
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