Question

120g of ice at 0c is mixed with 100 of water at 80c L=80cal/g final temp of mixture is

Answer 1

Explanation:

Here, Mass of water mw=100g

Mass of ice, mi=10g

Specific heat of water,Sw=1calg−1oC−1

Latent heat of fusion of ice,Lfi=80calg−1

Let T be the final temperature of the mixture.

Amount of heat lost by water

=mwsw(△T)w=100×1×(50−T)

Amount of heat gained by ice

=miLfi+misw(△T)i=10×80+10×1×(T−0)

According to principle of calorimetry:

Heat lost = Heat gained

100×1×(50−T)=10×80+10×1×(T−0)

500−10T=80+T

11T=420orT=38.2oC