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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:
 2 years ago
 2 years ago
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:
 2 years ago
 2 years ago

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iheartfood Group TitleBest ResponseYou've already chosen the best response.0
A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
work with the difference in the temperature
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
idk how to do that... can u pls show me??
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the 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 \[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you 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 \(t\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you can also use \[T=142e^{kt}\] but that requires solving for \(k\)
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
kk but idk how to solve for t since its in exponent form :(
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i get \[\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}\] \[\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}\] \[t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
requires change of base
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oh i didn't open the link, you are supposed to do it the hard way, solve for \(k\) first
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
so i get 32.50 ?? so my answer is B. 35 minutes ???
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
since i round ?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[132=142e^{10k}\] \[\frac{132}{142}=e^{10k}\] \[\frac{\ln(\frac{132}{142})}{10}=k\]
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
so k=.0073025135?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i got 32.5 as well remember to work with the temperature differences, not the actual temperature
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
ok so how do i know what my answer is based off of that??
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
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) 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
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
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??
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
@jim_thompson5910 @satellite73 is B my answer??
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
so is 35 minutes my answer?? :O
 2 years ago

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

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
because the answer is 32.4993139 minutes
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
idk... so what is the answer then??
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
so probs B. 35 minutes then??
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
yeah probably
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
kk thanks!
 2 years ago

iheartfood Group TitleBest ResponseYou've already chosen the best response.0
kk thanks!
 2 years ago
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