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A piece of ice of mass 40 g is added to 200 g of water at 50 deg C.Calculate the final temperature of water when all ice has melted.Specific heat...
A piece of ice of mass 40 g is added to 200 g of water at 50 deg C.Calculate the final temperature of water when all ice has melted. Specific heat capacity of water=4200 J kg-1K-1 Specific latent heat of fusion of ice=336 X 1000 J/kg I am in Class X and I need this answer today only.
Let the final temperature is x deg.C
The heat absorbed or diffused is given by 
What will happen here is that;
Ice will melt at 0 deg.C using its Specific latent heat of fusion
Then the melted ice at 0 deg will increase the temperature up to x deg C
The water will reduce temperature allowing ice to melt.
Therefore;
Heat absorbed by ice = Heat diffused by water
Heat absorbed by ice at latent heat = 336*1000*40/1000J= 13440J
Heat absorbed by ice when heating to x deg = 
= 40/1000*4200*(x-0)
= 168x
Heat diffused by water = 200/1000*4200*(50-x)
= 840(50-x)
So;
168x + 13440 = 840(50-x)
x = 28.3333
So the final temperature is 28.333 deg.C
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