If you pour a cup of coffee that is 200 F, and set it on desk in a room that is 68 F, and 10 minutes later it is 145 F,...?

If you pour a cup of coffee that is 200 F, and set it on desk in a room that is 68 F, and 10 minutes later it is 145 F, what temperature will it be 15 minutes after you originally poured it?


If t=0, P = 200

\[P_{t} = 132e^{-kt}+68\]

\[145 = 132e^{-10k}+68\]
\[\rightarrow k = 0.0539\]
\[P_{15} = 132e^{-.0539*15}+68\]
\[P_{15} = 126.81\]

so its 127??

Did I do it correctly? If not, please help me!

az_lender2015-04-27T07:11:32Z

Favorite Answer

dP/dt = -k(P-68)
ln(P-68) = -kt + C
P = 68 + ce^(-kt)
200 = 68 + ce^(-0) => c = 132, I agree.
145 = 68 + 132e^(-k*10)
77/132 = e^(-k*10) =>
ln(132/77) = 10*k =>
k = 0.0539, I agree.
Yeah, probably P15) = 127...

Aidan2015-04-27T07:07:38Z

wat. How do i math