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 g1 °C^-1 and latent heat of steam = 540 cal g^-1]

22.5 g

According to the principle of calorimetry
Heat lost = Heat gained

mL_vaporization + ms_w\(\Delta T\) = m_ws_w\(\Delta T\)

m×540 + m×1×(100-80)= 20×1×(80-10) 

m = 2.5 g

Total mass of water= 20 g + 2.5 g = 22.5 g