The Student Room Group

FAO: Current 2nd year Physicists

Hey guys, just wondered if any of you had tackled the Vacation Thermodynamics Problems set (#7) yet? If so, anyone feel like helping me with Q7.4:

"Some tea connoisseurs claim that a good cup of tea cannot be brewed with water at a temeprature less than 97degrees C (370K). Assuming this to be the case, is it possible for an astronomer, working on the summit of Mauna Kea in Hawaii (elevation 4194m) where the air pressire is 615mbar, to make a good cup of tea without the aid of a pressure vessel?"

I know that at lower pressures, the range over which a substance can remain in the liquid state is reduced, so I'm guessing we have to assess whether at this pressure the water is still in a liquid state or whether it has vaporized, so I'm thinking maybe about using the Clausius-Clapeyron eqtn, but I don't really know what I'm aiming for....do I need to use the latent heat of vaporisation of water, the pressure and then work out what the temperature would be and see if this is above or below 370K? If any of you guys could give me a hint, I'd appreciate it. Thanks :smile:
Reply 1
I don't know if you have this sort of table in Physics, but get a databook and look for the boiling temperature of water at 615mbar (or intrapolate it out.) If the temp turned out to be less than 370K, then the kettle would be empty before water at 370K could be obtained. Otherwise, the tea party at Mauna Kea could be as mad as in Alice's Adventures in the Wonderland :smile:
Reply 2
Thanks Ipsen, but I'm fairly sure that we're not supposed to just look up the boiling temperature....we're definitely expected to do some kind of mathematical calculation. Thanks anyways :smile:
hmmm... yeah I attempted the thing too... but I couldnt do it either. The clausius Clapeyron equation I do not get! I assume we are supposed to calculate the temperature it boils at at that pressure. and if its <97 you cant make a decent cup of tea. How one goes about doing this though is anybodys guess.

I love thermodynamics :rolleyes:
Reply 4
Thanks Stu for your help anyways :-( That problem sheet is horrible - I actually use to like thermodynamics and use to think I was fairly reasonable at it, but clearly not any more :frown:

Anyways, back to doing some more thermodynamics past questions <sighs>
Hoofbeat
Thanks Stu for your help anyways :-( That problem sheet is horrible - I actually use to like thermodynamics and use to think I was fairly reasonable at it, but clearly not any more :frown:

Anyways, back to doing some more thermodynamics past questions <sighs>


Tell me about it! I can do like 1 question on it! I don't like thermo, I've always found it annoying... theres too much freedom for my liking.

Enjoy your thermo... I shall be playing with boundary value problems and LVS - Im not sure which of us has the worse time :p:
Reply 6
Have you done the Boltzmann distribution? If yes, this could be a way to circumvent it:

a/ Assuming the energy E required for each H2O molecule to break off from the liquid is a constant, we can calculate it knowing the boiling temp at 1atm is 373K:

- If the density of H2O molecules in liquid form is a (this is fairly constant within the liquid range)
p = nRT
n = a*e^(E/kT)

- We know p, R, T and a (calculated from 1000kg of water = 1 cubic metre) so E is deductible.

b/ Now using: p = nRT and n = a*e^(E/kT) we have:
p = RT*a*e^(E/kT) (*)

Since p now is 615mbar, R, E and a are know, we can solve the equation (*) for T

I think this is a way to find out the boiling temperature of a liquid, although two big assumptions have to be introduced: constant E and constant a.
Reply 7
Umm....not sure I like it myself - but then I suppose the reason I'm sceptical is because I have some notes from a previous Physicist at my college and for this question (which he didn't finish) he started trying to do it with the Clausis-Clapeyron (he didn't get far enough for me to understand what he's doing) and from the comment the tutor wrote (which I can't remember off top of my head), it implied it was the right start to the question.

Also, if anyone else fancies looking at Q7 SectB off this paper (requires you to be on Oxford network or have VPN at home):
http://www.physics.ox.ac.uk/expapers/New%20Part%20A/2003/f03_na1.pdf

I'd be interested to hear some comments on part a) & b). I would have thought that because in both cases it's technically isothermal [surely a) is just a normal expansion and b) is Joule expansion] they give the same answers? But then reading some of the later part of the question, I'm under the impression that maybe they shouldn't be :-s

I was going to use the method that dw=-pdV=-dq

ds = dq/T
ds = pdV/T
p = nRT/V

etc
Hoofbeat
Umm....not sure I like it myself - but then I suppose the reason I'm sceptical is because I have some notes from a previous Physicist at my college and for this question (which he didn't finish) he started trying to do it with the Clausis-Clapeyron (he didn't get far enough for me to understand what he's doing) and from the comment the tutor wrote (which I can't remember off top of my head), it implied it was the right start to the question.

Also, if anyone else fancies looking at Q7 SectB off this paper (requires you to be on Oxford network or have VPN at home):
http://www.physics.ox.ac.uk/expapers/New%20Part%20A/2003/f03_na1.pdf

I'd be interested to hear some comments on part a) & b). I would have thought that because in both cases it's technically isothermal [surely a) is just a normal expansion and b) is Joule expansion] they give the same answers? But then reading some of the later part of the question, I'm under the impression that maybe they shouldn't be :-s

I was going to use the method that dw=-pdV=-dq

ds = dq/T
ds = pdV/T
p = nRT/V

etc


Im prety sure it does use Clausius Clapeyron. It says so in the summary at the top of the Q sheet. I just dont know how.
Reply 9
i did some thermodynamics last year. for this type of problem, dont you just have to use a given value for the enthalpy change of vaporization of water...can't you look that up? or are you actually going to have to work it out?

once you have that, it's easy to find the boiling temperature with that equation.
Willa
i did some thermodynamics last year. for this type of problem, dont you just have to use a given value for the enthalpy change of vaporization of water...can't you look that up? or are you actually going to have to work it out?

once you have that, it's easy to find the boiling temperature with that equation.


Well it doesnt appear to give us it. Surely we'd need a volume change as well, which we dont appear to have. :frown:
Reply 11
On the vacation sheet problem, this is how I did it: (I'm not sure if it's correct, but I'll have a tute on it with blundell, I suspect, so I'll ask the best way)

Along the line of co-existence, the gibbs functions have to be identical:
g_1(p,T) = g_2(p, t) => dg_1(p,T) = dg_2(p,T)

dg_1 = -S_1dT + v_1dp = dg_2 = -S_2dT + v_2dp

say 1 is liquid, 2 is gas. v_2 >> v_1 therefore we can ignore the v_1 term.

dp/dT = (s_2 -s_1)/v_2 = delta(s)/v_2

Latent heat = Tdelta(s)

=> dp/dT = L/TV (change v_2 -> V for simplicity of notation)

V = p/RT for one mole

dp/p = (L/R) dT/T^2

integrate all up => p=p_0exp(-L/RT)

L = 2.26E6 J/Kg

We know at STP, water will boil at 100 C. Therefore, all you need to do then is to substitute values in for the mountaintop at hawaii and find the temp needed. If it is less than 97, you're laughing.

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Sorry, forgot to mention: Naturally, my answer depends on knowing the latent heat of vapourisation of water. If you don't have that as assumed knowledge, then I don't know how to answer it; you have too many constants and not enough information.