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anonymous
 one year ago
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 (TA), 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?
anonymous
 one year ago
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 (TA), 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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Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh this is a fun one. Any ideas, how far can you get on your own just doing anything? What have you tried?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well 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
 one year ago
Best ResponseYou've already chosen the best response.0Well 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 \(\frac{dT}{dt}=\frac{ \Delta T}{\Delta t}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ok so here's what we do, substitute in: \[u=TA\] 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. :)

alekos
 one year ago
Best ResponseYou've already chosen the best response.1do you know how to solve the integrals?

alekos
 one year ago
Best ResponseYou've already chosen the best response.1OK, so whats the next step?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is what i'v got so far the left side is likely wrong

alekos
 one year ago
Best ResponseYou've already chosen the best response.1Thats 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
 one year ago
Best ResponseYou've already chosen the best response.1\[T = 105e ^{0.14t} + 75\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1I meant find t when the temp drops to 80. So substitute T=80

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did that, was wrong. by the way T = 80 as final temp so it would be \[80 = 105e^(.14t)+75\]

alekos
 one year ago
Best ResponseYou've already chosen the best response.1yep. thats it. now solve for t
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