Question

In a container of negligible heat capacity, mix 6kg of ice at -40 ° C with 3kg of steam at 120 ° C, determine the equilibrium temperature.

Answer 1

Answer:

100°C

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