Lalorimetry & Thermal Expansion Example 8. Tkg ice at -20°C is mixed with 1 kg steam at 200°C. Then find equilibrium temperature and mixture content. Solution : Let equilibrium temperature is 100 °C heat required to convert 1 kg ice at -20°C to 1 kg water at 100ºC is equal to Hi = 1 x - x 20 + 1 x 80 + 1 x 1 x 100 = 190 Kcal heat release by steam to convert 1 kg steam at 200°C to 1 kg water at 100°C is equal to Hz = 1 * - * 100+ 1 x 540 = 590 Kcal 1 kg ice at -20°C = H1 + 1kg water at 100°C 1 kg steam at 200°C = H2 + 1kg water at 100°C .....(2) by adding equation (1) and (2) 1 kg ice at -20°C + 1 kg steam at 200°C = H2 + H2 + 2 kg water at 100°C. Here heat required to ice is less than heat supplied by steam so mixture equilibrium temperature is 100°C then steam is not completely converted into water. So mixture has water and steam which is possible only at 100°C mass of steam which converted into water is equal to 190-1xx100 540 so mixture content mass of steam 1 mass of water 1 27