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anonymous

  • one year ago

Please help! You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling: T(t)=TA+(To−TA)e−kt A. 15 minutes B. 25 minutes C. 1 hour D. 45 minutes

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  1. anonymous
    • one year ago
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    plug in your numbers and solve for k. TA is the ambient temp, 72 and T0 is 205 the initial temp \(T(t) = 72+(205-72)e^{-kt}\) \(T(t) = 72+(178)e^{-kt}\) Since we know T = 195 at t = 10, you can solve for k \(195 = 72+(178)e^{-10k}\)

  2. anonymous
    • one year ago
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    Once you have k, plug it into the equation for T(t) and solve for t when T = 180

  3. anonymous
    • one year ago
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    Is k 0.037?

  4. anonymous
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    then you can do \[180=72+178e^{-0.03696t}\] BTW, the more digits you keep for k, the more accurate your t will be.

  6. anonymous
    • one year ago
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    Is it a?

  7. anonymous
    • one year ago
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    I got 13.5187 :/

  8. anonymous
    • one year ago
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    yeah that's what I got. I'm wondering if you're supposed to convert the units to celsius or seconds

  9. anonymous
    • one year ago
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    i have no idea. it doesn't say

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