Question

what mass of steam at 100°C must be mixed with 398 g of ice at its melting point, in a thermally insulated container to produce liquid water at 67.0°C? The specific heat of water is 4186 נ kg K. The latent heat of fusion is 333 krkg and the latent heat of vaporization is 2256 kJ/kg Nu 0.1725 kg the tolerance is +/-2%

Answer 1

Heat released by steam = Heat gained by ice Heat released by steam = Heat released by m kg of steam of 100 degree c to be m kg of water at 100 degree c + m kg of water of 100 degree c to be m kg of water of 67 degree c. H_1 = 2256 m + m times 4.186 times (100 - 67) = 2256 m + 138.138 m = 2394.138 m [specific heat of water = 4186 J/kg K = 4.186 kJ/kg K] Heat gained by ice = Heat absorbed by 398g (= 0.3982kg) ice to melt water of 0 degree c of 0.398 kg + Heat absorbed by 0.398 kg of water to be water of 67 degree c H_2 = 0.398 times 333 + 0.398 times 4.186 times (67 - 0) = 244.158 So, H_1 = H_2 = > 2394.138 m = 244.158 = > m = 244.158/2394.138 m = 0.102 kg