NEET 2014 Physics and Chemistry

Specific heat of water sw=1 cal g−10C−1
Latent heat of steam, Ls=540 cal g−1
Heat lost by mg of steam at 100∘C to change into water at 80∘C is
Q1=mLs+msw∆Tw
=m×540+m×1×(100−80)
=540m+20m=560m
Heat gained by 20g of water to change its temperature from 10∘C to 80∘C is
Q2=mwsw∆Tw
=20×1×(80−10)=1400
According to principle of calorimetry, Q1=Q2
∴‌‌560m=1400 or m=2.5g
Total mass of water present =(20+m)g=(20+2.5)g=22.5g