Posted by Sarah on Saturday, November 17, 2012 at 12:27am.
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.)
q for H2O(g, 125°C) H2O(g, 100°C)
q for H2O(g, 100°C) H2O(l, 100°C)
q for H2O(l, 100°C) H2O(l, 0°C)
q for H2O(l, 0°C) H2O(s, 0°C)
CHEMISTRY - DrBob222, Saturday, November 17, 2012 at 12:45am
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.
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