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

  2. anonymous
    • one year ago
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    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

  3. Empty
    • one year ago
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    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 \(\frac{dT}{dt}=\frac{ \Delta T}{\Delta t}\)

  4. Empty
    • one year ago
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    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. :)

  5. alekos
    • one year ago
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    that link is no good

  6. alekos
    • one year ago
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    do you know how to solve the integrals?

  7. anonymous
    • one year ago
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    yes i do

  8. alekos
    • one year ago
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    OK, so whats the next step?

  9. anonymous
    • one year ago
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    this is what i'v got so far the left side is likely wrong

  10. anonymous
    • one year ago
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    i mean right

  11. alekos
    • one year ago
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    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

  12. alekos
    • one year ago
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    \[T = 105e ^{-0.14t} + 75\]

  13. alekos
    • one year ago
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    I meant find t when the temp drops to 80. So substitute T=80

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

  15. alekos
    • one year ago
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    yep. thats it. now solve for t

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