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Newton's law of cooling problem: Differential Eq

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data
    At 1:00 PM, a thermometer reading 70 degrees F is taken outside where the temperature is -10 degrees F (ten below zero). At 1:02PM, the reading is 26 degrees. At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F. What is the temperature reading at 1:09PM?


    2. Relevant equations
    [tex]\frac{du}{dt} = -k(u - u_o)[/tex]
    integrating we have
    [tex]u - u_o = Ce^{-kt}[/tex]

    u is the temperature reading
    u_o is the temperature of the atmosphere


    3. The attempt at a solution

    solutions by sentences:

    *"At 1:00 PM, a thermometer reading 70 degrees F is taken outside where the temperature is -10 degrees F (ten below zero)":
    at t = 0 , u = 70
    and u_o = 10
    [tex]70 - (-10) = Ce^{-k(0)}[/tex]
    C = 80
    Now we already know the constant

    *"At 1:02PM, the reading is 26 degrees":
    at t = 2; u = 26

    [tex]26 - (-10) = 80e^{-k(2)}[/tex]
    k = 0.39925


    my problem is that i dont understand when it is put indoors
    *"At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F":
    the constant of proportionality is the same
    [tex](u - 70) = Ce^{0.39925t}[/tex] <<< this might be the equation
    ....and i dont understand the problem anymore


    the answer to this problem is 56 degrees Farenheit
     
  2. jcsd
  3. Jan 22, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have calculated u(t) as long as the thermometer is outside. What will the temperature of the thermometer be when t= 5 (at 10:05)?

    Now you have a new problem to do: u is now whatever you got for the temperature at 10:05 and u0= 75. Fortunately k is exactly the same as before (it depends only on the thermometer) so you don't need to calculate that again.
     
  4. Jan 22, 2008 #3
    for the time at exactly before 5 minutes i used the equation:
    [tex]u - (-10) = 80e^{-0.39925(5)}[/tex]
    so
    u = 0.8675 degrees

    problem:
    "At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F":
    so here it is now indoors and the air temperature is 70
    for the time exactly after 5 minutes:


    when t = 0, u = 0.8675 degrees;
    [tex](0.8675-70) = C(e^{0})[/tex]
    ^
    so
    C = -69.1325

    "What is the temperature reading at 1:09PM?"
    t = 9min, u = ?
    [tex](u - 70) = -69.1325e^{-0.39925(9)}[/tex]
    u = 68.098 degrees

    my answer seems to be wrong.... can you please verify ?
     
    Last edited: Jan 22, 2008
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