anonymous
  • anonymous
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:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes
anonymous
  • anonymous
work with the difference in the temperature

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
idk how to do that... can u pls show me??
anonymous
  • anonymous
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 \[T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}\]
anonymous
  • anonymous
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 \(t\)
anonymous
  • anonymous
you can also use \[T=142e^{kt}\] but that requires solving for \(k\)
anonymous
  • anonymous
kk but idk how to solve for t since its in exponent form :(
anonymous
  • anonymous
i get \[\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}\] \[\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}\] \[t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}\]
anonymous
  • anonymous
requires change of base
anonymous
  • anonymous
oh i didn't open the link, you are supposed to do it the hard way, solve for \(k\) first
anonymous
  • anonymous
so i get 32.50 ?? so my answer is B. 35 minutes ???
anonymous
  • anonymous
since i round ?
anonymous
  • anonymous
\[132=142e^{10k}\] \[\frac{132}{142}=e^{10k}\] \[\frac{\ln(\frac{132}{142})}{10}=k\]
anonymous
  • anonymous
so k=-.0073025135?
anonymous
  • anonymous
i got 32.5 as well remember to work with the temperature differences, not the actual temperature
anonymous
  • anonymous
ok so how do i know what my answer is based off of that??
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
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??
anonymous
  • anonymous
@jim_thompson5910 @satellite73 is B my answer??
anonymous
  • anonymous
so is 35 minutes my answer?? :O
jim_thompson5910
  • jim_thompson5910
maybe they're thinking in 5 min intervals
jim_thompson5910
  • jim_thompson5910
because the answer is 32.4993139 minutes
anonymous
  • anonymous
idk... so what is the answer then??
anonymous
  • anonymous
so probs B. 35 minutes then??
jim_thompson5910
  • jim_thompson5910
yeah probably
anonymous
  • anonymous
kk thanks!
anonymous
  • anonymous
kk thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.