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Using newton's law of cooling

  1. Mar 11, 2009 #1
    a cup of coffee is heated to 100° then left in a room with a constant temperature of 68° after a minute the temperature of the coffee has dropped to 95° after another minute to 92 degrees. how much time must pass before the temperature of the coffee has dropped to 70°?
     
  2. jcsd
  3. Mar 11, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi baileyyc! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Mar 12, 2009 #3
    this is how I tried to solve the problem:

    y'=k(y-68)
    dy/dt=k(y-68)
    ∫(1/y-68)dy=∫kdt
    e^ln|y-68|=e^kt+c1
    y-68=e^kt+c1
    y=68+ce^kt
    100=68+ce^k(0)
    100=68+c
    c=32

    when t=1, y=95 when t=2, y=92
    95=68+32e^k(1) 92=68+32e^k(2)
    27=32e^1k 24=32e^2k
    ln(27/32)=k ln(24/32)=2k
    k ~ -.1699 k ~ -.1438

    plugging k into the equation plugging k into the equation
    y=68+32e^-.1699t y=68+32e^-.1438t

    when y=70 when y=70
    70=68+32e^-.1699t 70=68+32e^-.1438t
    2=32e^-.1699t 2=32e^-.1438t
    ln(2/32)=-.1699t ln(2/32)=-.1438t
    t~16.32 minutes t~19.28 minutes

    I don't know which one is the answer.. can you help me?
     
  5. Mar 12, 2009 #4
    baileyyc,

    Are you sure the problem is stated correctly ? The reason I ask is because with those times and temperatures the thermal time constants differ by 18% and that is why you are getting two different answers.

    Thanks
    Matt
     
  6. Mar 12, 2009 #5
    yes I am typing it correctly
     
  7. Mar 12, 2009 #6

    tiny-tim

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    Hi baileyyc! :smile:

    I think this must be a trick question :frown:

    if the coffee is at 100º, then it must be boiling, so it'll have some latent heat, but you don't know how much.

    So I'd ignore the first minute, and calculate on the basis of the second minute only. :wink:
     
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