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Math Help - help with DE problem: Newton's law of cooling

  1. #1
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    help with DE problem: Newton's law of cooling

    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?

    u is the temperature reading
    u_o is the medium

    -k(u - u_o) = \frac{du}{dt}
    and by integrating
    u - u_o = Ce^{-kt}

    my solutions in solving the problem:
    "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
    70 - (-10) = Ce^{-k(0)}
    C = 80
    Now we already know the constant

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

    26 - (-10) = 80e^{-k(2)}
    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
    (u - 70) = Ce^{0.39925t}
    ....and i dont understand the problem


    the answer to this problem is 56 degrees Farenheit
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    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?

    u is the temperature reading
    u_o is the medium

    -k(u - u_o) = \frac{du}{dt}
    and by integrating
    u - u_o = Ce^{-kt}

    my solutions in solving the problem:
    "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
    70 - (-10) = Ce^{-k(0)}
    C = 80
    Now we already know the constant

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

    26 - (-10) = 80e^{-k(2)}
    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
    (u - 70) = Ce^{0.39925t}
    ....and i dont understand the problem


    the answer to this problem is 56 degrees Farenheit
    i didn't check, but i assume all your calculations are correct so far

    first you need to find what reading the thermometer has at 1:05 pm, you have the equation for that. use that as u(0), that is, the temperature at time 0. you want u(4). the thermometer is going to heat up. so we can use the equation you have, except make k positive (that should work)
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  3. #3
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    for the time at exactly before 5 minutes i used the equation:
    u - (-10) = 80e^{-0.39925(5)}
    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;
    (0.8675-70) = C(e^{0})
    ^
    so
    C = -69.1325

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

    edit: thanks jhevon!!!
    i fogot that it should be u(4) not u(9)
    Last edited by ^_^Engineer_Adam^_^; January 22nd 2008 at 10:02 PM.
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