Here, Specific heat of water, sw=1calg−1°C−1 Latent heat of steam, Ls=540calg−1 Heat lost by m g of steam at 100ºC to change into water at 80ºC is Q1=mLs+mswΔTw =m×540+m×1×(100−80) =540m+20m=560m Heat gained by 20 g of water to change its temperature from 10ºC to 80ºC is Q2=msswΔTw=20×1×(80−10)=1400 According to principle of calorimetry, Q1=Q2 Therefore, 560 m = 1400 or m = 2.5 g Total mass of water present = (20+m) g = (20 + 2.5) g = 22.5g