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First order ordinary differential equation

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    I haven't done ODEs in a few years and I am trying to do this equation:

    [itex]m_{Hg}C_{p,Hg}\frac{dT_{Hg}}{dt} = Q[/itex]

    [itex]Q = hA(T_{air} - T_{Hg})[/itex]

    [itex]T_{Hg}(t = 0) = 20[/itex]

    I need to find [itex]T_{Hg}(t=590)[/itex]

    2. Relevant equations

    3. The attempt at a solution

    [itex]h, A, m_{Hg}, C_{p,Hg}, T_{air}[/itex] are all constants

    [itex]\frac{dT_{Hg}}{dt} = \frac{hA(T_{air} - T_{Hg})}{m_{Hg}C_{p,Hg}}[/itex]

    [itex]\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = dt[/itex]

    [itex]\int\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = \int dt[/itex]

    [itex]\frac{-m_{Hg}C_{p,Hg}}{hA}\int\frac{1}{(-T_{air} + T_{Hg})} dT_{Hg} = \int dt[/itex]

    [itex]\frac{-m_{Hg}C_{p,Hg}}{hA} ln|T_{Hg} - T_{air}| = t + c[/itex]

    [itex]ln|T_{Hg} - T_{air}| = \frac{t + c}{\frac{-m_{Hg}C_{p,Hg}}{hA}}[/itex]

    [itex]T_{Hg} - T_{air} = e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}}[/itex]

    [itex]e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}} = c[/itex]

    [itex]T_{Hg} - T_{air} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}[/itex]

    [itex]T_{Hg} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}[/itex]

    [itex]T_{Hg}(t = 0) = 20[/itex]

    [itex]T_{air} = -7[/itex]

    [itex]h = 3.9[/itex]

    [itex]A = 0.00176[/itex]

    [itex]C_{p,HG} = 139.0908[/itex]

    [itex]m_{Hg} = 0.05[/itex]

    [itex]20 = ce^{0/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)[/itex]

    [itex]c = 27[/itex]

    [itex]T_{Hg} = 27e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}[/itex]


    [itex]T_{Hg}(t=590) = 27e^{590/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)[/itex]

    [itex]T_{Hg}(t=590) = 8.0822[/itex]
  2. jcsd
  3. Jun 17, 2011 #2


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    Man, you must like typing subscripts. Why not call your dependent variable T and write the equation in the form

    [tex]\frac{dT}{dt} = k(T_a - T)[/tex]

    in the first place and put the constants in for k at the end? It would be much easier to read, not to mention typing it all in. Anyway, it looks like you solved it OK, with the caveat that I didn't check all the details.
  4. Jun 17, 2011 #3


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    I think you made an error in using your initial condition (using the notation of LCKurtz) you had:
    -\ln |T_{a}-T|=kt+c
    Now if [itex]T(0)=T_{0}[/itex], then I would use that at this point here to tell you that [itex]-\ln |T_{a}-T_{0}|=c[/itex] and your equation becomes:
    -\ln |T_{a}-T|=kt-\ln |T_{a}-T_{0}|
    From here I think it is just algebra to obtain the correct answer.
  5. Jun 17, 2011 #4
    OK doing it with the substitutions:

    [itex]k = \frac{hA}{m_{Hg}C_{p,Hg}}[/itex]


    [itex] T = T_{Hg}[/itex]

    [itex]\frac{dT}{dt} = k(T_a - T)[/itex]

    [itex]\frac{1}{k(T_a - T)}dT = dt[/itex]

    [itex]\frac{-1}{k}\int\frac{1}{(-T_a + T)}dT = \int dt[/itex]

    [itex]\frac{-1}{k}ln|-T_a + T| = t + c[/itex]

    [itex]ln|-T_a + T| = -kt + c[/itex]

    [itex]-T_a + T = e^{-kt + c}[/itex]

    [itex]T = ce^{-kt} + T_a[/itex]

    Then initial conditions T(t=0) = 20

    [itex]20 = ce^{-k(0)} + (-7)[/itex]

    [itex]27 = c[/itex]

    [itex]T = 27e^{-kt} + T_a[/itex]
    Last edited: Jun 17, 2011
  6. Jun 17, 2011 #5


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    This line is incorrect:
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