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.)
gas
liquid
solid
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
all three?
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.
To determine the phase or phases present when energy is removed from the sample, we need to calculate the amount of energy required to change the phase of water at 115°C.
1. Calculate the energy required to heat the initial sample from its current temperature to its boiling point:
q1 = mass * specific heat capacity * ΔT
q1 = 50.4 g * 4.2 J/g·°C * (100°C - 115°C)
q1 = -756 J
2. Calculate the energy required to completely vaporize the sample at its boiling point:
q2 = moles * ΔHvap
First, we need to convert the mass of the sample to moles.
Moles of water = 50.4 g / molar mass of water
Molar mass of water = 18 g/mol
Moles of water = 50.4 g / 18 g/mol = 2.8 moles
q2 = 2.8 mol * 40.7 kJ/mol
q2 = -113.96 kJ = -113,960 J
3. Calculate the energy required to cool the remaining steam to the final temperature:
q3 = mass * specific heat capacity * ΔT
q3 = 50.4 g * 2.0 J/g·°C * (100°C - 0°C)
q3 = -10,080 J
Now we can sum up the energies obtained in each step:
Total energy = q1 + q2 + q3
Total energy = -756 J - 113,960 J - 10,080 J
If the total energy is equal to or greater than 156 kJ, then all three phases (gas, liquid, and solid) will be present in the given sample.
-756 J - 113,960 J - 10,080 J = -124,796 J
Since the total energy obtained is less than -156,000 J, it means that the total energy removed from the sample is not enough to result in all three phases being present. Therefore, the only phase present when 156 kJ of energy is removed from the sample is the gas phase.
Thus, the answer is:
- Gas