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Math Help - Newton's Law of Cooling HELP

  1. #1
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    Newton's Law of Cooling HELP

    Jake boils water at home for a cup of tea. He takes his tea and sits down at his desk to study, where the temperature of the room is at 67 F. After how many minutes will the temperature of the tea reach 175 F. (Water boils at 212 F. Assume k = .0486) Use Newton's Law of Cooling and solve for t. Show your work.

    This is all I've got and I'm stuck here:

    T(t) = 67 + 145e^-.0486(t)

    Can someone help me out and show me how to do this? THANKS!
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  2. #2
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    Re: Newton's Law of Cooling HELP

    You are looking for the value of t so that T(t) = 175 = 67+145e^{-0.0486t}. That means \dfrac{109}{145} = e^{-0.0486t}. Take the natural log of both sides and divide both sides by -0.0486.
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  3. #3
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    Re: Newton's Law of Cooling HELP

    Quote Originally Posted by SlipEternal View Post
    You are looking for the value of t so that T(t) = 175 = 67+145e^{-0.0486t}. That means \dfrac{109}{145} = e^{-0.0486t}. Take the natural log of both sides and divide both sides by -0.0486.
    I'm confused, how did you get to:

    \dfrac{109}{145} = e^{-0.0486t}. ?

    I know how to take the natural logs and what not but I just need to know how you got to that part, thanks.
    Last edited by StephenSkinner; November 12th 2013 at 09:49 PM.
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  4. #4
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    Re: Newton's Law of Cooling HELP

    You have 175 = 67+145e^{-0.0486t}. Subtract 67 from both sides, then divide both sides by 145. I just solved the equation for the only term that has a t.
    Thanks from StephenSkinner
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  5. #5
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    Re: Newton's Law of Cooling HELP

    Quote Originally Posted by SlipEternal View Post
    You have 175 = 67+145e^{-0.0486t}. Subtract 67 from both sides, then divide both sides by 145. I just solved the equation for the only term that has a t.
    Oh so you must've meant to put 108 instead of 109. That's what I figured. So the answer is about 6 minutes. THANKS!
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  6. #6
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    Re: Newton's Law of Cooling HELP

    Yes, sorry about that. It is late.
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