5 g of ice at 0°C is mixed with 5 g of steam at 100°C. What is the final temperature of the mixture?

Heat required by ice to raise its temperature to 100°C is Q1=m1L1+m1c1Δθ1=5×80+5×1×100      = 400 + 500 = 900 calHeat given by steam, when condensed is Q2=m2L2=5×540=2700  calSince Q2>Q1, it means that the whole steam is not even condensed. Hence, temperature of mixture remains at 100°C.