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iheartfood

  • 2 years ago

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:

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  1. iheartfood
    • 2 years ago
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  2. iheartfood
    • 2 years ago
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    A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes

  3. satellite73
    • 2 years ago
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    work with the difference in the temperature

  4. iheartfood
    • 2 years ago
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    idk how to do that... can u pls show me??

  5. satellite73
    • 2 years ago
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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 \[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]

  6. satellite73
    • 2 years ago
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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 \(t\)

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

  8. iheartfood
    • 2 years ago
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    kk but idk how to solve for t since its in exponent form :(

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

  10. satellite73
    • 2 years ago
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    requires change of base

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

  12. iheartfood
    • 2 years ago
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    so i get 32.50 ?? so my answer is B. 35 minutes ???

  13. iheartfood
    • 2 years ago
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    since i round ?

  14. satellite73
    • 2 years ago
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    \[132=142e^{10k}\] \[\frac{132}{142}=e^{10k}\] \[\frac{\ln(\frac{132}{142})}{10}=k\]

  15. iheartfood
    • 2 years ago
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    so k=-.0073025135?

  16. satellite73
    • 2 years ago
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    i got 32.5 as well remember to work with the temperature differences, not the actual temperature

  17. iheartfood
    • 2 years ago
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    ok so how do i know what my answer is based off of that??

  18. jim_thompson5910
    • 2 years ago
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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

  19. iheartfood
    • 2 years ago
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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??

  20. iheartfood
    • 2 years ago
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    @jim_thompson5910 @satellite73 is B my answer??

  21. iheartfood
    • 2 years ago
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    so is 35 minutes my answer?? :O

  22. jim_thompson5910
    • 2 years ago
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    maybe they're thinking in 5 min intervals

  23. jim_thompson5910
    • 2 years ago
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    because the answer is 32.4993139 minutes

  24. iheartfood
    • 2 years ago
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    idk... so what is the answer then??

  25. iheartfood
    • 2 years ago
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    so probs B. 35 minutes then??

  26. jim_thompson5910
    • 2 years ago
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    yeah probably

  27. iheartfood
    • 2 years ago
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    kk thanks!

  28. iheartfood
    • 2 years ago
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    kk thanks!

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