Question

Calculate the amount of work required to change 50g ice
at 50°C into water of 100°C.​

Answer 1

Answer:

Given: mass of ice, mi=50g

            mass of water, mw=50g

            temperature of ice, Ti=0oC

            temperature of water, Tw=100oC

We have,

            specific heat of water, cw=1 cal/g/oC

            latent heat of ice, Lw=89 cal/g

Let ToC be the final temperature of mixture.

Now energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at ToC= heat lost by water to reach at ToC

            miLi+micw(T−0)=mwcw(100−T)

            50×80+50×1(T−0)=50×1(100−T)

            4000+50T=5000−50T

            100T=5000−4000=1000

            T=1001000=10oC