1. The problem statement, all variables and given/known data Suppose molten (liquid) lead, mass ml = 9.83 kg, is at its melting point. The lead is poured into water of mass mw = 585 g and initial temperature Tw = 16.2 C. Find mwb, the amount of the water that boils. NOTE: Reaching the boiling point is not enough; the question asks for the amount of water that vaporizes as well. ASSUME: no water evaporates; it only vaporizes at the boiling point. 2. Relevant equations Constants: M.P Lead = 327 C, MP Water = 100 C Latent Heat Lead = 33 kJ/kg Latent Heat Q = mL Specific Heat Eq'n: Q=mcΔT 3. The attempt at a solution mL + mcΔT = mcΔT + mL 9.83 kg (33 kJ/kg) + 9.83 kg(0.130 kJ/kg)(327C - 100C) = 0.585(4.186)(100 -16.2) + m(2260 kJ/kg) Solving for m, we get m = 0.215 kg. Did I go wrong in my reasoning here? Many thanks in advance for your help!