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Thermal Physics- Can you make good tea at a certain temperature

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    According to experts good tea can only be made at temperatures greater than 96 degrees celsius. If this is true, can you brew good tea at elevation 4.5km (Pressure = 6.2*[itex]10^{4}[/itex]Pa) . Given that latent heat of vaporisation for water is 2.4*[itex]10^{6}[/itex] J/kg and water has a molar mass of 18g.


    2. Relevant equations
    Clausius-clapeyron equation [itex]\frac{dp}{dT}[/itex] = [itex]\frac{L}{T(V2-V1)}[/itex]
    but V1 is negligable so [itex]\frac{dp}{dT}[/itex] = [itex]\frac{L}{TV2}[/itex]

    Ideal gas equation pV=nRT


    3. The attempt at a solution

    Tried to integrate so have [itex]\int\frac{1}{p}dp[/itex]=[itex]\frac{L}{R}[/itex][itex]\int\frac{1}{T^{2}}dT[/itex]
    from substituting in V = [itex]\frac{RT}{p}[/itex] from ideal gas equation.
    But dont really know what to do from here! I dont know what limits to put in for the integral or if im just going about this all wrong from the start!
    Ive been told the clausius-clapeyron statement must be used by my professor!
    Any suggestions would be helpful! Thankyou!
     
  2. jcsd
  3. Nov 20, 2012 #2

    gneill

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    Staff: Mentor

    Hi H.fulls, Welcome to Physics Forums.

    You might be interested having a look at the following arrangement of the clausius-clapeyron relationship:
    $$ ln\left(\frac{P_1}{P_2}\right) = \frac{H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$
    Where the P's are the vapor pressures and the T's the temperatures (in K).
     
  4. Nov 20, 2012 #3
    Thanks! :)

    I was getting this integral before but not sure what to do with it! And is [itex]H_{vap}[/itex] the same as my latent heat value? I have no idea what values to use for [itex]P_{1}[/itex], [itex]P_{2}[/itex], [itex]T_{1}[/itex] or [itex]T_{2}[/itex] :S
     
  5. Nov 20, 2012 #4

    gneill

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    Yes, Hvap is the latent heat of vaporization. The P's are the vapor pressures at temperatures T1 and T2. What do you know about the relationship between ambient atmospheric pressure and vapor pressure of a fluid at its boiling point?
     
  6. Nov 20, 2012 #5
    Ah okay :) Erm..nothing probably! But I seem to recall that the ambient pressure and pressure at boiling point are the same? maybe.... ? haha :)
     
  7. Nov 20, 2012 #6

    gneill

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    Yup.
     
  8. Nov 20, 2012 #7
    But if the pressures are the same does that not make the left hand side ln(1) =0! ?
     
  9. Nov 20, 2012 #8

    gneill

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    Why do say that the pressures are the same? The atmospheric pressure is different for the two cases.
     
  10. Nov 20, 2012 #9
    Ah okay so the P's denote just the atmospheric pressure? I know the pressure for not boiling, but I dont know the pressure for when it is boiling! So confused by this question!
     
  11. Nov 20, 2012 #10

    gneill

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    Staff: Mentor

    The P's represent vapor pressures. But, as you already mentioned previously, the vapor pressure at the boiling point is equal to the ambient air pressure... so they have the same values.
     
  12. Nov 20, 2012 #11
    so the vapour pressure at boiling point is the pressure I originally gave in my question?
     
  13. Nov 20, 2012 #12

    gneill

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    Yes, that's one of them (the pressure at the high altitude location for which you want to find the corresponding boiling temperature). What's the other pressure/temperature pair?
     
  14. Nov 20, 2012 #13
    Ahh okay.. is the other pair the boiling temperature and pressure for normal conditions? i.e 100 degrees and 101 KPa ?
     
  15. Nov 20, 2012 #14

    gneill

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    Staff: Mentor

    Yes, of course.
     
  16. Nov 20, 2012 #15
    Ahhhh okay! I finally understand! Thanks so much for your help :) !
     
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