Steam at ${100}^0\mathrm{C} \mathrm{is} \mathrm{passed} \mathrm{into} 20 \mathrm{g} \mathrm{of} \mathrm{water} \mathrm{at} {10}^0\mathrm{C}. \mathrm{When} \mathrm{water} \mathrm{acquires} \mathrm{a} \mathrm{temperature} \mathrm{of} {80}^0\mathrm{C}$, the mass of water present will be

[Take specific heat of water = 1 cal ${{\mathrm{g}}^{-1}}^{ 0}{\mathrm{C}}^{-1} \mathrm{and} \mathrm{latent} \mathrm{heat} \mathrm{of} \mathrm{steam} = 540 \mathrm{cal} {\mathrm{g}}^{-1}$]