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Math Help - Newton's Law of Cooling

  1. #1
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    Newton's Law of Cooling

    so this is so confusing i got lost.
    The Question:

    A cup of cofee placed in a room cools from 160F to 150F in one minute. if the room temperature is 65F, Find the temperature in ten minutes.

    so the formula for this is
    (change in temperature over one time period) = A ( temperature at the beginning of the - room temperature time period)

    150F - 160F = A (160 - 65)
    -10F = A(105)
    A = -1050F ???what this is where i got lost help please!
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  2. #2
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    Re: Newton's Law of Cooling

    Are you sure of that formula? The Newton's law of cooling I know is given by

    T(t)=T_s+(T_0-T_s)e^{-At}

    Here, T_s is the ambient temperature, T_0 the initial temperature of the coffee, and T(t) the temperature of the coffee at time t

    You need to solve for A first at t=1, T(1)=150. Then substitute t=10 for the solution to your question.
    Thanks from Ruun and topsquark
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  3. #3
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    Re: Newton's Law of Cooling

    i am still not understanding this

    where i have gotten to using the formula that my teacher gave us with this packet

    Xk - Xk-1 = A(Xk-1 - 65)
    150-160 = A(160 - 65)
    -10 or -0.10 = A(95)
    Xk = Xk-1 + (-0.10) Xk-1 + (-0.10)(-65)
    Xk = (1 + (-0.10)) Xk-1 + 6.5
    Xk = (0.9) Xk-1 + 6.5

    C = 0.9, K = 6.5 and An = 160
    Am = 65 + (95)(0.9)10
    65+(95)(9)
    855????

    what am i doing wrong?
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  4. #4
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    Re: Newton's Law of Cooling

    Quote Originally Posted by zhengcl86 View Post
    -10 or -0.10 = A(95)
    I'm not familiar with this method, but what's going on in this line. Either it's -10 or it's not. Why are you arbitrarily dividing by 100?

    -Dan
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  5. #5
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    Re: Newton's Law of Cooling

    wouldn't i be solving for A? i kinda got the formula messed up cause i was looking at an example and doing this other problem as well, probably why i got it mixed, sorry xD

    well solving for A in that line -10 = A(95)
    so A = -9.5
    guess i have to re-do this question, let me give it another try
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  6. #6
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    Re: Newton's Law of Cooling

    so i am using a different formula that i found online

    T(t) = T0 + ( T1 - T0) e^-k(t)

    T(t) = 65 +(160-65)e^-k(t)
    T(t) = 65+95 e^-k(t)
    time = 1 Temp = 150
    150 = 65+95 e^-k(1)
    85 = 95 e^-k
    .8947368421 = -k
    T(t) 65+95 e^-.8947368421(10)
    T(10) = 160 - 8.947368421
    T = 151

    which still makes no sense!! what am i doing wrong?
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  7. #7
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    Re: Newton's Law of Cooling

    Quote Originally Posted by zhengcl86 View Post
    so i am using a different formula that i found online

    T(t) = T0 + ( T1 - T0) e^-k(t)
    This was the formula I gave you in the second post.

    T(t) = 65 +(160-65)e^-k(t)
    T(t) = 65+95 e^-k(t)
    time = 1 Temp = 150
    150 = 65+95 e^-k(1)
    85 = 95 e^-k
    Yes.

    .8947368421 = -k
    How did you get this from the previous equation? It should be -k=\ln{\frac{85}{95}}=-0.111225...

    T(t) 65+95 e^-.8947368421(10)
    T(10) = 160 - 8.947368421
    I suggest you revise this second line. There are too many elementary errors to point out.
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  8. #8
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    Re: Newton's Law of Cooling

    ok so i have all the math done and it seems about right with the temperature in 10 minutes.

    so continued from 85/95 = e^k
    17/19 = e^k
    T(10)=95(17/19)^10+65
    T(10)=95(.3288161986)+65
    T(10)=31.23753887+65
    T = 96.2 F

    please correct me if this is wrong, it is part of a project due next thursday.
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  9. #9
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    Re: Newton's Law of Cooling

    Quote Originally Posted by zhengcl86 View Post
    ok so i have all the math done and it seems about right with the temperature in 10 minutes.

    so continued from 85/95 = e^k
    17/19 = e^k
    It should be e^{-k}, not e^{k}.

    T(10)=95(17/19)^10+65
    T(10)=95(.3288161986)+65
    T(10)=31.23753887+65
    T = 96.2 F
    Seems about right.
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