Given, mass of steam, mλ=5g Temperature of steam, Ts=100°C Mass of ice, mi=5g Temperature of ice, Ti=0°C Since, latent heat of vaporisation, lv=540‌cal∕g Latent heat of fusion, lj=80‌cal∕g Now, heat given by steam during fully conversion into water, at 100°C and Q1=mj×540=2700‌cal Heat taken by ice to get fully converted into water at 100°C. Q2=milf+mi×s×ΔT =5×80+5×1×100(∵s=1‌for‌water) =400+500=900 As, Q2<Q4 ∴ Temperature of mixture will be 100°C.