4 gms of steam at 100°C is added to 20 gms of water at 46°C in a container of negligible mass. Assuming no heat is lost to surrounding, find the mass of water in container at thermal equilibrium in grams.. Latent heat of vaporisation = 540 cal/gm. Specific heat of water = 1 cal/gm-°C.

Heat released by steam in conversion to water at 100°C is $Q_1$ = mL = 4 × 540  = 2160 cal.

Heat required to raise temperature of water from 46°C t 100°C is $Q_2=ms\Delta \theta =20\times 1\times 54=1080J,$

$Q_1>Q_{2}and\frac{Q_1}{Q_2}=2.$

Hence all steam is not converted to water only half steam shall be converted to water 

$\therefore$Final mass of water = 20 + 2  = 22 gm