Calculate the total heat needed to convert 0.20 kg of ice at 0 C completely into steam at 100 C. (Lf = 3.34 x 10^5 J/kg, Lv = 2.26 x 10^6 J/kg, c_water = 4186 J/(kg K).) Solution: Three stages: (1) melt ice: Q1 = m Lf = (0.20)(3.34 x 10^5) = 66,800 J. (2) warm water 0 to 100 C: Q2 = m c delta_T = (0.20)(4186)(100) = 83,720 J. (3) vaporise: Q3 = m Lv = (0.20)(2.26 x 10^6) = 452,000 J. Total = 66,800 + 83,720 + 452,000 = 602,520 J, about 6.0 x 10^5 J. Vaporisation dominates.