Question

3. An insulated container contains 2.7 kg of steam at 1.4 bar and 120°C. The top is opened and a 1.8 kg block of ice at 0°C and 4.5 kg of liquid water at 1.6°C are dumped into the container. What is the final temperature of the resulting vapor-liquid mixture in the container? Note: It takes 6.0095 KJ/mol to melt ice.

Answer 1

Given:

Steam:

Temperature = 120°C = T1

Pressure = 1.4 bar

Mass = 2.7 kg = m1

Specific heat of steam (from steam tables at 120°C) = 2.12 kJ/kgK = Cp1

Ice:

Temperature = 0°C = T2

Mass = 1.8 kg = m2

Moles = 1.8/18 = 0.1 kmoles = 100 moles = n

Water:

Temperature = 1.6°C = T3

Mass = 4.5 kg = m3

Specific heat of water (from literature) = 4.18 kJ/kgK = Cp3

Energy required to melt ice = 6.0095 kJ/mole = L

Total enery requird to melt ice = 6.0095*100 =600.95 kJ = n*L

Let T be final temperature of mixture

Energy balance:

Energy lost by steam = Energy required to melt ice + Energy required to heat water formed from melted ice + Energy required to heat water added

m1*Cp1*(T1-T) = n*L + m2*Cp3*(T-T2) + m3*Cp3*(T-T3)

2.7*2.12*(120-T) = 600.95 + 1.8*4.18*(T-0) +4.5*4.18*(T-1.6)

116.026 = 32.058*T

T = 3.619 °C

Temperature of the mixture = 3.619 °C