anonymous
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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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.
jamiebookeater
  • jamiebookeater
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Empty
  • 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
  • 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
  • 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
  • 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
  • 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
  • alekos
that link is no good
alekos
  • alekos
do you know how to solve the integrals?
anonymous
  • anonymous
yes i do
alekos
  • alekos
OK, so whats the next step?
anonymous
  • anonymous
this is what i'v got so far the left side is likely wrong
anonymous
  • anonymous
i mean right
alekos
  • 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
  • alekos
\[T = 105e ^{-0.14t} + 75\]
alekos
  • alekos
I meant find t when the temp drops to 80. So substitute T=80
anonymous
  • anonymous
did that, was wrong. by the way T = 80 as final temp so it would be \[80 = 105e^(-.14t)+75\]
alekos
  • alekos
yep. thats it. now solve for t

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