What mass of steam at 100 C must be mixed with 150 gram of ice at its melting point, in a thermally insulated container, to produce liquid water at 50 C?

Heat of fusion of water at its normal freezing or melting temperature = 333 kJ/kg
Heat of vaporization (or steam) = 2256 kJ/kg
Specific heat of water = 4180 J/kg K

Question

What mass of steam at 100 C must be mixed with 150 gram of ice at its melting point, in a thermally insulated container, to produce liquid water at 50 C?

Heat of fusion of water at its normal freezing or melting temperature = 333 kJ/kg
Heat of vaporization (or steam) = 2256 kJ/kg
Specific heat of water = 4180 J/kg K

anonymous · Answer

Specific heat of water = 4.180 kJ/(kg K) to match all other units and use m =0.150 kg
I get 0.032981744421906693711967545638945 kg of water
temperture rise must equal temperature drop; equilibrium
.15 kg *333 kJ/kg + .15 kg * 4.180 kJ/(kg K) * 50 K = 81.3 kJ
81.3 kJ = (unknown mass) * (2256 kJ/kg) + (unknown mass) * 4.180kJ/(kg K) * 50 K
solve for unknown mass