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