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Thermal Flask, heat conduction

  1. Jan 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Find an expression for thermal conductivity using kinetic theory.

    Given a vacuum flask of these dimensions, find the heat loss per unit time.

    Estimate the time taken for the water to cool down to 40 deg.

    hvcqi8.png


    2. Relevant equations



    3. The attempt at a solution

    Part (a)
    [tex] k = \frac {1}{3} C_{mol} n λ <c> [/tex]

    Part (b)

    Estimate gap to be 0.5cm.

    [tex] n = \frac {PV}{RT} =\frac {(10^{-2})(3.3*10^{-4})}{(8.31)(293)} = 1.4 * 10^{-9} [/tex]

    Since the the molecular density is so low, we can effectively say the mean free path is the gap. [tex] λ ≈ 0.5 cm [/tex] (Not sure if this is right, but if the molecule would crash into the other end of the fask before it hits the molecule, then the mean free path should definitely be in the order of magnitude of the gap, right?)

    Heat Capacity of air should be same order of magnitude as water;
    [tex] C_{mol} ≈ 10^3 J K^{-1} [/tex]

    [tex] <c> ≈ 400 m s^{-1} [/tex]

    Estimate Temperature gradient ≈ ΔT/5cm) = 8000 K/m

    Together, [tex] k ≈ 9.3*10^{-7} [/tex] So, [tex] P = JA = 4.69 * 10^{-4} [/tex]


    Part (c)

    Time taken = [tex] \frac {Q}{Power} = 1.79*10^8 s = 6 years [/tex]
     
    Last edited: Jan 6, 2014
  2. jcsd
  3. Jan 7, 2014 #2
    The time taken is ridiculous, which is why I'm not sure if that's right.

    Also, heat is conducted away by radiation.
     
  4. Jan 7, 2014 #3

    TSny

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    According to the formula for ##k## in the statement of the problem, ##n## is the number of gas particles per unit volume. Looks like you're letting ##n## be the number of moles of gas in the flask.

    I'm not sure about what to do when λ is much greater than the gap. But your assumption seems to be reasonable. If you do assume that λ can be taken to be the thickness of the gap, then I think λ will cancel out by the time you calculate J.

    In the formula for ##k##, ##C_{mol}## is the heat capacity "of a molecule". Is that what your ##C_{mol}## represents? There is a well-known formula for the molar heat capacity at constant volume of an ideal diatomic gas. From that you can get the heat capacity per molecule.
     
  5. Jan 7, 2014 #4
    the correct value of n should be [tex]n = 2.47*10^{18} [/tex]

    From the ideal gas law: at constant volume, For N molecules of gas, [tex] dQ = dU = N C_{mol} dT = \frac {5}{2} Nk dT [/tex] So, [tex] C_v = \frac {5}{2}k = 3.45*10^{-23} [/tex]


    Using these data along with [tex] <c> = 400 m/s [/tex], we obtain:
    [tex] κ = 5.68*10^{-5} [/tex]
    [tex] Heat loss rate =7.14*10^{-4} J /s [/tex]
    [tex] time taken = \frac {energy}{heat loss rate} = \frac {mCΔT}{7.14*10^{-4}} = 3.7*10^{8} s [/tex]

    The time taken is astronomical!
     
    Last edited: Jan 8, 2014
  6. Jan 8, 2014 #5

    TSny

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    The number density ##n## is the number of molecules per unit volume: ##n= N/V## where ##N## is the number of molecules and ##V## is the volume. From the ideal gas law ##PV = Nk_BT## you should be able to calculate ##n## from the pressure, temperature, and Boltzmann's constant ##k_B##.

    The molar heat capacity at constant volume for an ideal diatomic gas is ##\frac{5}{2} R##. But that leads to the same result as you have for the heat capacity per molecule.

    I think ##\kappa## should be on the order of 10-4 or 10-5 W/(m##\cdot##K)
     
  7. Jan 8, 2014 #6
    I realised I calculated the wrong value of n. I have adjusted my answers above, and I get the following:

    [tex] k = 5.68*10^{-5}, time = 3.7*10^{8} s [/tex]
     
    Last edited: Jan 8, 2014
  8. Jan 8, 2014 #7

    TSny

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    I get a similar result.
     
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