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- anonymous

Really need some help understanding this problem?!?!
The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT/dt = -k (T-A), where T is the water temperature, A is the room temperature, and k is a positive constant.
If the coffee cools from 180°F to 100°F in 10 minutes at a room temperature of 75°F, how long (to the nearest minute) will it take the water to cool to 80°F?

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- anonymous

- katieb

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- Empty

Ahhh this is a fun one. Any ideas, how far can you get on your own just doing anything? What have you tried?

- anonymous

Well in know that it has something to do with the separation of variables, which is just a simple way of solving for a basic differential equation. But the issue lies in interpreting the numbers given and replacing them into the equation to make it solvable. in short, I have nothing

- Empty

Well I think we can do this with an integrating factor or possibly some other way too, this is just an ODE not a PDE so it should be easier to solve.
Since we're rounding it's even possible to just estimate some things such as replacing \(\frac{dT}{dt}=\frac{ \Delta T}{\Delta t}\)

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- Empty

Ok so here's what we do, substitute in:
\[u=T-A\] differentiate to get: \[\frac{du}{dt} = \frac{dT}{dt}\]
this just simplifies the differential equation to become:
\[\frac{du}{dt} = -ku\]
which is separable:
\[\int \frac{du}{u}=\int -kdt\]
Now you can solve this for T(t), so do this now and I'll help you out. :)

- anonymous

found this http://learn.flvs.net/webdav/educator_apcalcab_v14/sfframe.htm?loc=http%3A%2F%2Flearn.flvs.net%2Feducator%2Fstudent%2Fframe_toolbar.cgi%3Fdhicks59*jdosio1*mpos%3D1%26spos%3D0%26option%3Dhidemenu%26slt%3DLjph6mFdLTT1E*4063*http%3A%2F%2Flearn.flvs.net%2Fwebdav%2Feducator_apcalcab_v14%2Findex.htm<iloc=http%3A%2F%2Ftool.studyforge.net%2Flti.php%3Fresource_type%3Dlesson%26resource_id%3D4715

- alekos

that link is no good

- alekos

do you know how to solve the integrals?

- anonymous

yes i do

- alekos

OK, so whats the next step?

- anonymous

this is what i'v got so far the left side is likely wrong

- anonymous

i mean right

- alekos

Thats all really good! You've found k = 0.14 and C = 105
so now you have the full equation and you can use this to find t when the temperature drops to 75 deg

- alekos

\[T = 105e ^{-0.14t} + 75\]

- alekos

I meant find t when the temp drops to 80. So substitute T=80

- anonymous

did that, was wrong. by the way T = 80 as final temp so it would be \[80 = 105e^(-.14t)+75\]

- alekos

yep. thats it. now solve for t

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