Steam at 100°C is passed into 25 g of water at 10°C. When water acquires a temperature of 60°C, the mass of water present will be: [Take specific heat of water = 1 cal g^-1 °C^-1 and latent heat of steam = 540 cal g^-1]

21 g

mL_vaporization + mc\(\Delta T\) = m'c\(\Delta T'\)

L_vaporization = 540 cal/g ; c = 1 cal/g ; \(\Delta T\) = 100 - 60 = 40 ; \(\Delta T'\) = 60 - 10 = 50 ; m' = 20 g

\(\Rightarrow m = 1.72 \text{ g}\)

Total mass = 20 g + 1.72 g = 21.72 g