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anonymous
 4 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:
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0work with the difference in the temperature

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0idk how to do that... can u pls show me??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the initial difference is \(21068=142\) after 10 minutes the difference is \(20068=132\) one way to model this is (probably not what your math teacher had in mind \[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you want to know when it is 180 degrees, or when the temperature difference is \(18068=112\) so set \[112=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\] and solve for \(t\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can also use \[T=142e^{kt}\] but that requires solving for \(k\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0kk but idk how to solve for t since its in exponent form :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i get \[\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}\] \[\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}\] \[t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0requires change of base

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i didn't open the link, you are supposed to do it the hard way, solve for \(k\) first

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i get 32.50 ?? so my answer is B. 35 minutes ???

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[132=142e^{10k}\] \[\frac{132}{142}=e^{10k}\] \[\frac{\ln(\frac{132}{142})}{10}=k\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got 32.5 as well remember to work with the temperature differences, not the actual temperature

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so how do i know what my answer is based off of that??

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1T(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) 20068 = (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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 @satellite73 is B my answer??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so is 35 minutes my answer?? :O

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1maybe they're thinking in 5 min intervals

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1because the answer is 32.4993139 minutes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0idk... so what is the answer then??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so probs B. 35 minutes then??
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