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Math Help - Newton's Law of cooling?? HELP!!

  1. #1
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    Exclamation Newton's Law of cooling?? HELP!!

    This problem is the law just backwards. There are also two unknown constants and i therefor have no idea how to manipulate the equation when two are missing...

    Question:
    On a hot day a thermometer is taken outside from an air-conditioned room where the temperature is 21 degrees C. After one minutes it reads 27 degrees C, and after two minutes it reads 30 degrees C. Use Newton's law of cooling to find the outdoor temperature.

    My work so far:
    dy/dx = K(y-?) ?=outdoor temperature y=thermometer reading

    21=(21-?) e^K(0)
    27=(27-?) e^K(1)
    30=(30-?) e^K(3)


    What should I do??? I really need guidance on this problem!!!!!!!!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Let T be the temp of the environment

    dy/dt = k (T - y)

    solving we obtain y(t) = T + c e^(-kt)

    Where c,K are constants and T is the temperature of the environment we want.

    We have :

    1) y(0) = T + c =21

    2) y(1) = T + c e^(-k) = 27

    3) y(2) = T + c e^(-2k) = 30

    using 1) c = 21 - T

    Using 2) T + (21-T)e^(-k) =27

    e^(-k) = (27-T)/(21-T)

    Now e^(-2k) = [e^(-k)]^2

    Using 3) we have [ (27-T)^2]/(21-T) = 30 - T

    (27-T)^2 = (30-T)((21-T)

    Multiply out and solve


    You should find T = 33 c = -12 and k = ln(2)
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  3. #3
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    At the begining, how do we know that K is negative?

    and also I got y=T - c e^kt because I added y to both sides from (T-y), and then subtracted everything back to the T side.

    ?
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  4. #4
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    Quote Originally Posted by klooless View Post
    This problem is the law just backwards. There are also two unknown constants and i therefor have no idea how to manipulate the equation when two are missing...

    Question:
    On a hot day a thermometer is taken outside from an air-conditioned room where the temperature is 21 degrees C. After one minutes it reads 27 degrees C, and after two minutes it reads 30 degrees C. Use Newton's law of cooling to find the outdoor temperature.

    My work so far:
    dy/dx = K(y-?) ?=outdoor temperature y=thermometer reading

    21=(21-?) e^K(0)
    27=(27-?) e^K(1)
    30=(30-?) e^K(3))
    This formula is incorrect. You seem to be using "T= (T- T_o)e^{K^t}". That is not correct. Since you don't show how you got that, I can't say precisely what you may have done wrong. If dy/dt= k(y-?) then \frac{dy}{y-?}= kdt. Integrating both sides, ln(y- ?)= kt+ C and taking exponentials of both sides of that, [tex]y- ?= e^{kt+ C}= C'e^{kt}[\math]

    y= C'e^{kt}+ ?
    Now you have 3 unknown parameters and three equations to solve for them:
    at t= 0, 21= C'e^{k0}+ ?= C'+ ?; at t= 1, 27= C'e^{k}+ ?; at t= 3, 30= C'e^{3k}+ ?.

    Subtracting two of those equations eliminates "?".
    30- 27= 3= C'(e^{3k}- e^{k})
    27- 21= 6= C'(e^{k}-1)

    Dividing the second equation by the first eliminates C'.
    \frac{6}{3}= 2= \frac{e^{k}- 1}{e^{3k}- e^{k}}
    Multiplying both sides by e^{3k}- e^{k} gives 2(e^{3k}- e^{k})= 2e^{3k}- 2e^{k}=  e^k- 1

    Let u= e^k and the equation becomes 2u^3- 2u= 2u- 1 or 2u^3- 4u+ 1= 0. Unfortunately, that has no rational roots.




    What should I do??? I really need guidance on this problem!!!!!!!!
    (Calculus26 got k negative, while you (and I) got k positive because he set up the problem as dy/dt= k(T- y) while I used your equation dy/dt= k(y- T).)
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Why is ?

    should be 30 = C'e^(2k) + T since at t = 2 y = 30 which I believe leads to quadratic I suggested
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