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Question

2kg ice at 20oC is mixed with 5kg water at 20oC. The final amount of water in the mixture would be ____.
Given the specific heat of ice = 0.5cal/gC; Specific heat of water = 1cal/gC; Latent heat of fusion for ice = 80cal/g.

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Solution

Given: 2 kg of ice at 20C is mixed with 5 kg of water at 20C. It is given that the specific heats of water and ice are 1cal/(gC) and 0.5cal/(gC) respectively and the latent heat of fusion of ice is 80cal/g.
To find the final mass of water in the mixture.
Solution:
As per the given criteria,
Mass of water, mw=5kg
Mass of the ice, mi=2kg
Temperature of ice, Ti=20C
Temperature of water, Tw=20C
Specific heat of ice, si=0.5cal/(gC)
Specific heat of water, sw=1cal/(gC)
latent heat of fusion of ice is L=80cal/g.
Let M be the final mass of the content
The amount of heat losed by water at 0 degreeis
Qw=mwss(ΔT)=5×1×20=100kcal
the amount of heat gained by ice to go to 0 degree is
Qi=misi(ΔT)=2×0.5×20=20kcal
heat left for absorption is
QwQi=10020=80kcal
This is equal to latent heat, i.e.,
mL=80m=80L=8080=1kg
This is the mass of water from the converted ice.
So the final mass of water in the vessel is , M=mw+m=5+1=6kg

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