anonymous
  • anonymous
The temperature of a cup of hot tea varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 60°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
https://gyazo.com/8aa6a713d1c912933a9dcf9581c5483f
anonymous
  • anonymous
@jim_thompson5910 you taught me this a while back. I used your method and got 58 it would be nice if you can check it
anonymous
  • anonymous
work with the temperature differences find the exponential model that goes through the points \((0,40)\) and \((1,20)\)

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More answers

abbycross167
  • abbycross167
@satellite73 could you finish helping me with my question?
anonymous
  • anonymous
initially the difference is 40, after one minute it is 20
anonymous
  • anonymous
@satellite73 I am not sure what you're trying to get at I followed http://openstudy.com/users/ephemera#/updates/56468ffde4b0c3aa4a128f95
jim_thompson5910
  • jim_thompson5910
ok I'm looking it over now
jim_thompson5910
  • jim_thompson5910
@Ephemera can you post your work so far?
ganeshie8
  • ganeshie8
@Ephemera is your answer 58 ? the room temperature is 60, how can the temperature of tea go below room temperature ?
anonymous
  • anonymous
Site was rally glitchy for me. And I know what I was doing wrong @ganeshie8 I used the same exact formula when in that question they ask for time, here they ask for temperature. According to your reasoning (very good reasoning might I add) the answer can't be 58. Please allow me a couple minutes to try and do it properly.
anonymous
  • anonymous
Ok so this is what I have: \[\left| T-A \right| = e^c e^-kt\] T = 100 A= 80 t= 1 e^c = 60?
anonymous
  • anonymous
\[\left| 100-80 \right| = 60e ^{-k}\]
anonymous
  • anonymous
is it good so far?
anonymous
  • anonymous
@jim_thompson5910 @ganeshie8
anonymous
  • anonymous
I'd solve for k then use k to find T?
jim_thompson5910
  • jim_thompson5910
dt/(T-A) = -kdt ln(|T-A|) = -kt + C |T - A| = e^(-kt + C) |T - A| = e^C*e^(-kt) |T - 60| = e^C*e^(-kt) |100 - 60| = e^C*e^(-k*0) 40 = e^C e^C = 40 I'm not sure how you got e^C = 60
anonymous
  • anonymous
hmm
ganeshie8
  • ganeshie8
Remember, \(T\) is a function of time \(t\). Your function for temperature at any given time should be : \[T(t) = 60 + Be^{-kt}\]
anonymous
  • anonymous
What does B represent?
ganeshie8
  • ganeshie8
B is a constant
jim_thompson5910
  • jim_thompson5910
B = e^C
ganeshie8
  • ganeshie8
plugin (0, 100) in above equation and you may solve B
anonymous
  • anonymous
Same as e^c, I see it now
anonymous
  • anonymous
So we need to solve for k then T now, correct?
ganeshie8
  • ganeshie8
T is not a constant
ganeshie8
  • ganeshie8
T is the temperature at any given time
anonymous
  • anonymous
Are they not asking for T(4)?
ganeshie8
  • ganeshie8
Yes, they are
anonymous
  • anonymous
And I can't solve for that without k correct?
ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
That is what I meant.
ganeshie8
  • ganeshie8
find the constants first using the given initial conditions
anonymous
  • anonymous
I would use (1,80) to solve for k?
ganeshie8
  • ganeshie8
Yes, try finding B first though
anonymous
  • anonymous
I thought be was equal to 40
anonymous
  • anonymous
B*
anonymous
  • anonymous
Or would it be a different scenario for 80?
jim_thompson5910
  • jim_thompson5910
yes, B = 40
ganeshie8
  • ganeshie8
B is a constant, it doesn't change
anonymous
  • anonymous
\[k= -\ln(1/2)\]
jim_thompson5910
  • jim_thompson5910
or equivalently, k = ln(2)
anonymous
  • anonymous
Yes using log properties.
anonymous
  • anonymous
I'm getting T(4) = 700
jim_thompson5910
  • jim_thompson5910
way too high
anonymous
  • anonymous
exactly
anonymous
  • anonymous
\[T(4) = 60 + 40e^{4\ln2}\]
jim_thompson5910
  • jim_thompson5910
you should find that \[\Large T(t) = 60+40e^{-\ln(2)*t}\]
jim_thompson5910
  • jim_thompson5910
you lost a negative in your exponent
anonymous
  • anonymous
why would there be a negative?
anonymous
  • anonymous
-ln(1/2) ------> ln(1/2)^-1
jim_thompson5910
  • jim_thompson5910
\[\Large T(t) = 60+40e^{-kt}\] \[\Large T(t) = 60+40e^{-\ln(2)t}\]
anonymous
  • anonymous
Which is ln 2
anonymous
  • anonymous
Ohhhhh
anonymous
  • anonymous
The negative in the formula, got you
anonymous
  • anonymous
62.5 or 63
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
thanks for both your help
jim_thompson5910
  • jim_thompson5910
np

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