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Thermal question

  1. Jun 10, 2005 #1
    Hi guys,

    I'm stuck on what should be a reasonably simple question...here it is:

    The quality of a certain supermarket product depends on a single thermally
    activated process. It can be stored safely for 3 days at 17 degrees C, but only one day at 37 degrees C. How long can it be stored at 0 degrees C?

    I know this has something to do with a boltzmann distribution, and possibly to do with putting it in the form of a lifetime...but i'm unsure of the lifetime method...is that where you say: t = 1/probability of decay ???

    At a guess I would from this have said that there is some activation energy E, and used boltzmann distribution integrated from E to infinity over all Energy to find the probability that a particle has enough energy to decay (or whatever).
    And then use t=1/p and the given values to compute E, and then plug it back in to find the t for the new T.

    Whether that is right or not, can someone step through the stages for me?

  2. jcsd
  3. Jun 10, 2005 #2
    if it is a thermally activated process, then the lifetime should be inversely proportinal to the speed of reaction:

    [tex]t_{lifetime}=A*exp (B/T)[/tex]
    where [tex]T[/tex] is the temperature.
    You can find out A and B from the given data.
  4. Jun 10, 2005 #3
    unfortunately I think the question demands some more working than that, even if it boils down to that. Since if lifetime=1/probability....well then it's obviously gonna be of the form Aexp(B/kT)...but i just need someone to step it through nice and clearly for me. Thanks
  5. Jun 11, 2005 #4

    Andrew Mason

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    Try this: Plot the energy distribution graphs of


    for T = 273, 290 and 310 K

    For each graph, consider the area to the right of the activation energy [itex]E_a[/itex]. That area divided by the entire area under the graph represents the relative population of molecules with sufficient energy to activate the process that degrades the product. For the last two graphs, the relative populations in that area to the right of [itex]E_a[/itex] differ by a factor of 3 (ie for T=310, the population of molecules with [itex]E \ge E_a[/itex] is 3 times that for T=290).

    [tex]\int_{E_a}^\infty Ae^{-E/kT} dE = kTAe^{-E_a/kT} - 0[/tex]


    [tex]\frac{310kAe^{-E/310k}}{290kAe^{-E/290k}} = 3[/tex]

    [tex]-\frac{E_a}{310k}} + \frac{E_a}{290k}} = ln(3 * 290/310) = [/tex]

    Solve that for [itex]E_a[/itex]

    Now knowing E_a, work out the proportional area to the right of E_a in the distribution for T = 290 K and T=273. The ratio of those two areas x 3 will give you the number of days the product can be stored safely.

    Last edited: Jun 11, 2005
  6. Jun 12, 2005 #5
    Actually, that more elaborate reasoning is not correct, but it clarifies the mechanism nevertheless. Andrew Mason forgot to normalize distribution,
    because he needs to assume the total amount of molecules to be constant. Thus A will be temperature dependent.

    But all this is irrelevant because we do not know the density of states anyway. (Andrew Mason assumed it to be constant) One may try different
    models, they will give you that reaction speed is proprtional to the

    speed[tex]\approx T^b*e^{E_a/kT}[/tex].
    usually the term with exponent dominates, so we can ignore the term [tex]T^b[/tex] with more weak temperature dependence. Then there is another problem- what happened with all these molecules which reacted? Should't they be excluded? You see, it is impossible to give the exact formula without knowing the exact mechnism. However, if we say thermally activated process, we can approximate the speead of reaction by a simple exponent.

    The underlaying mechanism is, as it was mentioned, that only molecules with the energy above the activation energy could participate in reaction.
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