4 gm of steam at 1000C is added to 20 gm of water at 460C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of water in container 540 cal/gm. Specific heat of water = 1 cal/gm- 0C.

Heat released by steam in conversion to water a 1000C is Q1 = mL = 4×540 = 2160 cal. Heat required to raise  tempreature of water from 460C to 1000C is Q2 = ms∆θ = 20×1×54 = 1080J     Q1 > Q2 and Q1Q2 = 2Hence all steam is not converted to water only half steam shall be converted to waterFinal mass of water = 20 + 2 = 22 gm