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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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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A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes
work with the difference in the temperature

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

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