Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: anthell on October 01, 2014, 10:23:22 AM
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If i understand it correctly
The specific heat capacity of water / mole is : 75.3 J
and the latent heat of evaporation is : 44 KJ
But is specific heat capacity of water and latent heat of evaporation affected by pressure?
Here is what im trying to figure out.
1. heat water to 100 C @ STP and it begins to boil
2. Move the water to a vacuum chamber / bell jar and depressurize it to the point that its boiling point is just under 25 C which is around 0.442 PSIA
3. The rapid boiling causes cooling ( evaporation / boiling cools )
i. How much water would be left?
ii. will the amount of evaporating water be enough to chill it down to under 25 C such that it
stops boiling?
iii .what kind of temperature difference is needed so that the chilling effects would be insufficient to chill to low enough temperature such that it boils completely?
Please help me out
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But is specific heat capacity of water and latent heat of evaporation affected by pressure?
Not significantly, I don't think.
If you put boiling water at 100°C under a vacuum it will boil violently and water will go everywhere, so it's probably not a good experiment.
Suppose you depressurise it very slowly, so that the pressure is always just slightly below the equilibrium boiling pressure at the water temperature.
Now suppose you get to a temperature T after n moles of water have evaporated (of the original N moles). To evaporate a further dn moles requires latent heat of 44000*dn J. Assuming this heat must come from the remaining water, its temperature changes by dT, and this heat change is given by -75.3*(N-n)*dT J. From these expressions you can derive a differential equation for dT/dn, which shows how temperature changes with evaporation.
Can you now calculate the answer to question i?
Can you see from the equation that (with our assumptions) the answer to ii must be yes, however low a final temperature you choose, and the answer to iii is that it is impossible?
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I see where you are getting hmm....
How about this way of looking at it?
What is wrong with this logic?
Let
Previous and higher boiling point be T1
new and lower boiling pointbe T2
number of moles be N
specific heat capacity of water be H
Latent heat of evaporation be V
The "potential heat energy for evaporation" is equal to ( T1 - T2 ) * H * N
Energy required to boil it completely is equal to N * V
Thus
If the potential heat energy due to change in boiling point is higher or equal to the energy required to boil it completely it would be able to do it.
( T1 - T2 ) * N * H = V * N
T1 - T2 = V / H
T1 - T2 = 584.3
We need 584.3 Kelvin of temperature difference between the old and new boiling point to boil it completely
What's wrong with this logic ?
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The "potential heat energy for evaporation" is equal to ( T1 - T2 ) * H * N
This is not true. Once some water has been evaporated it escapes into the atmosphere and is not available to release heat to evaporate more water. That heat has to come from the water that is left unevaporated, and that is constantly decreasing. That is why you have to do it by the differential method that I suggested.
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Ohhh that's true.... I get it now
It make sense... once evaporated... the super heated steam would have difficulties to transfer heat to let the left over water...
Thank you very much...
But here is the next problem... I'm pretty bad with math... not to mention differential equations... mind helping me with that?
Hehe thx
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Well, the grumpy reply would be, this is a chemistry forum, and we try to help people with their chemistry problems; if the problem is that they can't do maths, they should go away and learn it.
But to give a start: From the expressions I gave above, we can write
dT = -584.3*dn/(N-n)
Can you integrate the LHS from T = T1 to T2, and the RHS from n = 0 to n2?
Meanwhile, can you understand physically the answers I gave to questions 2 and 3 - why you can never evaporate all the water if the latent heat of evaporation must come solely from the remaining liquid?
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Yea I can see your point thx hehe
I guess external heating source would be needed for full evaporation