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iheartfood Group Title

You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling:

  • one year ago
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  1. iheartfood Group Title
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    • one year ago
  2. iheartfood Group Title
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    A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes

    • one year ago
  3. satellite73 Group Title
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    work with the difference in the temperature

    • one year ago
  4. iheartfood Group Title
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    idk how to do that... can u pls show me??

    • one year ago
  5. satellite73 Group Title
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    the initial difference is \(210-68=142\) after 10 minutes the difference is \(200-68=132\) one way to model this is (probably not what your math teacher had in mind \[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]

    • one year ago
  6. satellite73 Group Title
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    you want to know when it is 180 degrees, or when the temperature difference is \(180-68=112\) so set \[112=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\] and solve for \(t\)

    • one year ago
  7. satellite73 Group Title
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    you can also use \[T=142e^{kt}\] but that requires solving for \(k\)

    • one year ago
  8. iheartfood Group Title
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    kk but idk how to solve for t since its in exponent form :(

    • one year ago
  9. satellite73 Group Title
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    i get \[\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}\] \[\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}\] \[t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}\]

    • one year ago
  10. satellite73 Group Title
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    requires change of base

    • one year ago
  11. satellite73 Group Title
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    oh i didn't open the link, you are supposed to do it the hard way, solve for \(k\) first

    • one year ago
  12. iheartfood Group Title
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    so i get 32.50 ?? so my answer is B. 35 minutes ???

    • one year ago
  13. iheartfood Group Title
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    since i round ?

    • one year ago
  14. satellite73 Group Title
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    \[132=142e^{10k}\] \[\frac{132}{142}=e^{10k}\] \[\frac{\ln(\frac{132}{142})}{10}=k\]

    • one year ago
  15. iheartfood Group Title
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    so k=-.0073025135?

    • one year ago
  16. satellite73 Group Title
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    i got 32.5 as well remember to work with the temperature differences, not the actual temperature

    • one year ago
  17. iheartfood Group Title
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    ok so how do i know what my answer is based off of that??

    • one year ago
  18. jim_thompson5910 Group Title
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    T(t) = Ta + (To - Ta)e^(-kt) T(t) = Ta + (210 - Ta)e^(-kt) ... Plug in To = 210 (this is the temp of the object) T(t) = 68 + (210 - 68)e^(-kt) ... plug in Ta = 68 T(10) = 68 + (210 - 68)e^(-k*10) ... Plug in t = 10 200 = 68 + (210 - 68)e^(-k*10) ... Plug in T(10) = 200 (this is the temp of the object at 10 min) Now solve for k 200 = 68 + (210 - 68)e^(-k*10) 200 = 68 + (142)e^(-10k) 200-68 = (142)e^(-10k) 132 = (142)e^(-10k) 132/142 = e^(-10k) 0.929577 = e^(-10k) e^(-10k) = 0.929577 -10k = ln(0.929577) -10k = -0.0730256 k = -0.0730256/(-10) k = 0.00730256 So your function goes from T(t) = 68 + (210 - 68)e^(-kt) to T(t) = 68 + (210 - 68)e^(-0.00730256t) ---------------------------------------------------------------- Now plug in T(t) = 180 and solve for t T(t) = 68 + (210 - 68)e^(-0.00730256t) 180 = 68 + (210 - 68)e^(-0.00730256t) 180 - 68 = (210 - 68)e^(-0.00730256t) 112 = (210 - 68)e^(-0.00730256t) 112 = (142)e^(-0.00730256t) 112/142 = e^(-0.00730256t) 0.78873239 = e^(-0.00730256t) ln(0.78873239) = -0.00730256t -0.23732819 = -0.00730256t -0.23732819/(-0.00730256) = t 32.4993139 = t t = 32.4993139 So it will take about 32.4993139 minutes for the cup to cool to 180 degrees F

    • one year ago
  19. iheartfood Group Title
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    oh wow thanks!! :) but my answer choices are these: A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes so do i round up and get B. 35 min as my answer??

    • one year ago
  20. iheartfood Group Title
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    @jim_thompson5910 @satellite73 is B my answer??

    • one year ago
  21. iheartfood Group Title
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    so is 35 minutes my answer?? :O

    • one year ago
  22. jim_thompson5910 Group Title
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    maybe they're thinking in 5 min intervals

    • one year ago
  23. jim_thompson5910 Group Title
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    because the answer is 32.4993139 minutes

    • one year ago
  24. iheartfood Group Title
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    idk... so what is the answer then??

    • one year ago
  25. iheartfood Group Title
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    so probs B. 35 minutes then??

    • one year ago
  26. jim_thompson5910 Group Title
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    yeah probably

    • one year ago
  27. iheartfood Group Title
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    kk thanks!

    • one year ago
  28. iheartfood Group Title
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    kk thanks!

    • one year ago
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