Answer:
100°C
Step-by-step explanation:
The heat gained by the ice equals the heat lost by the steam, so the total heat transfer equals 0.
Heat lost by the steam as it cools to 100°C:
q = mCΔT
q = (3 kg) (2.00 kJ/kg/K) (100°C − 120°C)
q = -120 kJ
Total heat so far is negative.
Heat lost by the steam as it condenses:
q = -mL
q = -(3 kg) (2256 kJ/kg)
q = -6768 kJ
Heat absorbed by the ice as it warms to 0°C:
q = mCΔT
q = (6 kg) (2.11 kJ/kg/K) (0°C − (-40°C))
q = 506.4 kJ
Heat absorbed by the ice as it melts:
q = mL
q = (6 kg) (335 kJ/kg)
q = 2010 kJ
Heat absorbed by the water as it warms to 100°C:
q = mCΔT
q = (6 kg) (4.18 kJ/kg/K) (100°C − 0°C)
q = 2508 kJ
The total heat absorbed by the ice by heating it to 100°C is 5024.4 kJ.
If the steam is fully condensed, it loses a total of -6888 kJ.
Therefore, the steam does not fully condense. The equilibrium temperature is therefore 100°C