Steam at 100°C  is passed into 20 g of water at 10°C. When water acquires a temperature of 80°C, the mass of water present will be [Take specific heat of water = 1 cal g-1oC and latent heat of steam = 540 cal g-1 ]

Here,  Specific heat of water, sw=1 cal g−1∘C−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+mswΔTw=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=mwswΔTw=20×1×(80−10)=1400 According to principle of calorimetry  Q1=Q2∴ 560m=1400 or  m=2.5gTotal mass of water present =(20+m)g=(20+2.5)g=22.5g