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Newton's Law of Cooling, find room temperature

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose that the temperature of a pan of warm water obeys Newton's law of cooling. The water (47 degrees Celsius) was put in a room and 10 minutes later the water's temperature was 40 degrees Celsius. After another 10 minutes, the temperature of the water was 34 degrees Celsius. Find the room's temperature.

    2. Relevant equations
    (H-Hs)=(H0-Hs)ekt
    dH/dt = -k(H-Hs)

    Hs is the room's temperature
    H0 is the initial temperature of the water
    H is the temperature of the water at time t

    3. The attempt at a solution

    I'm rather at a loss at how to do this problem. I've tried setting up a system of equations and tried finding the room temperature, but that doesn't work out. So if algebra won't do it, then I'm assuming something calculus-y should. I'm just at a lost as what that is.
     
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 18, 2010 #2

    ehild

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    You have the temperatures at 10 minutes and at 20 minutes. The exponential terms are e10k and e20k which is the square of e10k . Plug in the temperature data and use that the second equation is just the square of the first one.

    ehild
     
  4. Sep 19, 2010 #3
    It doesn't work, it only leads to a polynomial that can't be factored.

    If it helps at all Newton's law of cooling is dH/dt = -k(H-Hs), the equation above is derived from it.
     
  5. Sep 19, 2010 #4
    It didn't lead to an unfactorable polynomial when I tried it. . . I ended up substituting M=e10k, which made it easier to solve (or less intimidating, anyways).
     
  6. Sep 19, 2010 #5

    ehild

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    It works, just try.

    Let be z=e10k. Then e20k=z^2.

    You have two equations:

    40-Hs=(47-Hs)z
    34-Hs=(47-Hs)z^2

    What about squaring the first equation and dividing by the second one?


    ehild
     
  7. Sep 19, 2010 #6
    OK..

    40-Hs=(47-Hs)z
    34-Hs=(47-Hs)z^2

    I square the first equation
    (40-Hs)^2=(47-Hs)^2*z^2

    and divide by the second, so I get:
    (40-Hs)^2/(34-Hs) = (47-Hs)

    ...FFFFFUUUUUUUUUUUUUUUUUUUUU

    I was doing this all along, but I was writing some of my signs wrong... so yeah. Anyway thanks for the help, answer is -2.
     
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