Question

I give medals! A cup of coffee which is initially at a temperature of 159 F is placed in a room which is at a constant temperature of 68 F. In 7 minutes, the coffee has cooled to 141 F. Find the Newton's Law of Cooling formula that models the coffee's temperature in minutes. Keep at least 4 decimal places in your formula for rounded values. Determine to the nearest minute how long it will take the coffee to cool to a temperature of 100 F

Answer 1

\[T(t)= S+(T _{0}-S)e ^{kt}\]
\[141=68+(159-68)e ^{7k}\]

Answer 2

this is only annoying because you have to work with the differences in the heated temp and the room temp
so instead of working with 159 and 141 you work with 91 and 73
then it is just like the last one

Answer 3

\[\frac{ 73 }{ 91 }=\frac{ 91e ^{7k} }{ 9 }\]
\[\frac{ \ln73 }{ 91 }=7k\]

Answer 4

\[k=\frac{ \ln(73/91) }{ 7}\]

Answer 5

Just wondering if I'm doing this correct so far

Answer 6

i assume you mean
\[\ln(\frac{73}{91})=7k\] right?

Answer 7

Yes!

Answer 8

oh yeah, i see from your next post
yes, you are right so far

Answer 9

Mmkay! I just gotta finish it up now

Answer 10

I got k= -0.0315

Answer 11

yes, i got that too
http://www.wolframalpha.com/input/?i=ln%2873%2F91%29%2F7

Answer 12

So, would the equation be \[T(t)=68+(T _{0}-68)e ^{-0.0315*t}\]

Answer 13

\(T_0=159\) so it is
\[\large T=68+91e^{-0.0.15t}\]

Answer 14

\(T_0\) is the initial temperature, not a variable in the formula

Answer 15

Wow, I'm an idiot! -_- thank you!

Answer 16

lol yw
ok form there? last part is to ste
\[\large 91e^{-0.0315t}=32\] and solve for \(t\)

Answer 17

*set

Answer 18

Got it! I just have to finish soling it!

Answer 19

I got t=33

Answer 20

kk i will too

Answer 21

me too

Answer 22

Sweet! I appreciate it! Thanks! :D

Answer 23

yw again
btw you did all the work