posted by Robert on .
Consider a 50.4 g sample of H2O(g) at 115°C. What phase or phases are present when 156 kJ of energy is removed from this sample? Specific heat capacities: ice, 2.1 J/g·°C; liquid, 4.2 J/g·°C; steam, 2.0 J/g·°C, ÄHvap = 40.7 kJ/mol, ÄHfus = 6.02 kJ/mol. (Select all that apply.)
I suggest you do this in steps, You know you want to remove 156,000 J of energy. Do one step, subtract from the 156,000 to see how much you have left to remove and go step by step.
Step 1. Remove heat from the steam at 115 C to steam at 100 C.
mass x specific heat steam x delta T = 50.4 g x 2.0 J/g*C x (100-115) = 1512 J. That leaves 156,000 - 1512 = 154,488 J remaining to be removed.
Step 2. Condense the steam at 100 to liquid water at 100.
mass x heat vap =
50.4 g x 40.7 kJ/mol x (1 mol/18 g) x (1000 J/kJ) = 113,120 J.
154,488 J - 113,120 J = 41,368 J remaining to be removed.
Continue stepwise through moving water from 100 to zero, then heat fusion to freeze it, then move it down to some final T from zero. I'll leave those steps for you. You want to remove 156,000 J in all.
The following calculations must be done:
(a) Cool the steam to 100 deg C
(2.0 J/g.°C)(50.4g)(115°C-100°C) =1512 J = 1.5 kJ
(b) Condense the steam at 100 deg C
40700J/mol ÷ 18g/mol = 2261 J/g
(50.4g)(2261J/g) =114000 J = 114 kJ
(c) Cool the water to 0 deg C
(4.2 J/g.°C)(50.4g)(100°C - 0°C) = 21168 J = 21 kJ
(d) Freeze the water at 0 deg C
6020 j/mol ÷ 18 g/mol = 334.4 J/g
(50.4g)(334.4 J/g) = 16856 J = 17 kJ
Now you decide what phase changes took place
The total heat released from steam to liquid, and then ice was:
1.5 + 114 + 21 + 17 = 153.5 kJ
That is less than the total heat released of 156 kJ. That is more than enough evidence to confirm (or not) your suggestion.