Consider a 58.4 g sample of H2O(g) at 125°C. What phase or phases are present when -162 kJ of energy is removed from this sample? Specific heat capacities: ice, 2.1 J g-1 °C-1; liquid, 4.2 J g-1 °C-1; steam, 2.0 J g-1 °C-1, ΔHvap = 40.7 kJ/mol, ΔHfus = 6.02 kJ/mol. (Select all that apply.)
gas
liquid
solid
q for H2O(g, 125°C) H2O(g, 100°C)
kJ
q for H2O(g, 100°C) H2O(l, 100°C)
kJ
q for H2O(l, 100°C) H2O(l, 0°C)
kJ
q for H2O(l, 0°C) H2O(s, 0°C)
kJ
Is that stuff at the bottom of your post a hint on how to work the problem. Just follow that. For example, how much energy is released when steam is moved from 125 C to 100 C?
That will be mass steam x specific heat steam x (25) = q1.
58.4 g x 2.0 J/g*C x 25 = 2,920 J.
You had 162,000 J to start; now you have 162,000-2920 = 158,080 J to go.
q2 to condense steam at 100 C to liquid water at 100 C.
mass steam x heat vap = 58.4 x (1 mol/18g) x 40.7 kJ/mol = 132.049 kJ or 132,049 J.
158,080-132,049 = 26,031 to go.
I'll leave it for you here.
To determine the phase or phases present when energy is removed from the sample, we need to calculate the amount of energy required to go through each phase change.
1. First, let's determine how much energy is required to cool the gaseous phase from 125°C to 100°C.
q = m * C * ΔT
q = 58.4 g * 2.0 J g^(-1) °C^(-1) * (100°C - 125°C)
q = -584 J
2. Next, let's determine how much energy is required for the phase change from gaseous to liquid form at 100°C (vaporization).
q = ΔHvap * n
First, we need to calculate the number of moles of water:
n = m / M
n = 58.4 g / 18.0 g mol^(-1)
n = 3.24 mol
q = 40.7 kJ/mol * 3.24 mol
q = -131.7 kJ
3. Now, let's determine how much energy is required to cool the liquid water from 100°C to 0°C.
q = m * C * ΔT
q = 58.4 g * 4.2 J g^(-1) °C^(-1) * (0°C - 100°C)
q = -2434.4 J
4. Lastly, let's determine how much energy is required for the phase change from liquid to solid form at 0°C (freezing).
q = ΔHfus * n
q = 6.02 kJ/mol * 3.24 mol
q = -19.5 kJ
Now, let's add up the total amount of energy:
Total energy = q1 + q2 + q3 + q4
Total energy = -584 J + (-131.7 kJ) + (-2434.4 J) + (-19.5 kJ)
Total energy = -132,217.9 J
Since -162 kJ of energy is removed from the sample, which is equivalent to -162,000 J, we can determine the phases based on the total energy value.
-132,217.9 J < -162,000 J
Since the total energy is less than the energy removed, it means that the sample will go through all the phase changes until the energy is completely removed.
Therefore, all three phases will be present: gas, liquid, and solid.