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

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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.

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 = `mc(theta)`

= 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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