Question

If you have 10.0 grams of water vapor at 120.0℃, then how much energy in the form of heat must be removed to cool the water vapor down to ice at -10.0℃?

Hvap is 44.0 kJ/mol, Hfus is 6.01 kJ/mol, c for H2O(g) is 1.87 J/goC, c for H2O(l) is 4.18 J/goC, c for H2O(s) is 2.01 J/goC

Answer 1

The amount of water as vapour = 10 grams

The heat / energy will be involved for following processes

a) Q1 = Energy released in cooling the vapour from 120 to 100 ℃.

b) Q2 = Energy released in conversion of vapour to liquid

c) Q3= energy released in cooling liquid from 100 to 0 ℃

d) Q4 = Energy released in conversion of liquid to ice

e) Q5 = Energy released in cooling ice from 0 ℃ to -10 ℃

total heat = Q1 + Q2 + Q3+ Q4 + Q5

Let us calculate each energy

Q1 = Mass X specific heat of steam X change in temperature

      = 1.87 J/g·°C X 10g X (120°C - 100°C)
Q1 = 374 J

Q2 = Mass X Hvap

Hvap = 44 / 18 KJ / g = 2.44J / g

Q2 = 10 X 2.44 KJ = 24.4 KJ = 24400 J

Q3 = MAss X specific heat of water X change in temperature
Q3 = 4.18 J/g·°C X 10g X (100°C - 0°C) = 4180 Joules

Q4 = Heat of fusion X moles of water

Moles of water = Mass / Mol wt = 10 / 18 = 0.56 moles

Q4 = 6.01 X 0.56 = 3.36 KJ = 3360 J

Q5 = Mass of water X specific heat of ice X change in temperature

= 10 X 2.01 X (0-(-10) = 201 Joules

Q(total) = Q1 + Q2 + Q3 + Q4 + Q5 = 32515 Joules (this much of energy be removed)