- Download PDF
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.
1 Answer | Add Yours
Let the final temperature is x deg.C
The heat absorbed or diffused is given by `Q=mc(theta)`
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.
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 = `mc(theta)`
Heat diffused by water = 200/1000*4200*(50-x)
168x + 13440 = 840(50-x)
x = 28.3333
So the final temperature is 28.333 deg.C
We’ve answered 319,950 questions. We can answer yours, too.Ask a question