anonymous
  • anonymous
The temperature of a roast varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the water temperature, A is the room temperature, and k is a positive constant. If a room temperature roast cools from 68°F to 25°F in 5 hours at freezer temperature of 20°F, how long (to the nearest hour) will it take the roast to cool to 21°F?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
https://gyazo.com/23b71615e032fca0d654bb5437697556
anonymous
  • anonymous
@jim_thompson5910 Help always appreciated :)
jim_thompson5910
  • jim_thompson5910
were you able to find the T(t) function? T = temperature t = time

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More answers

anonymous
  • anonymous
No
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{dT}{dt} = -k(T-A)\] \[\Large \frac{dT}{T-A} = -k dt\] \[\Large \int\frac{dT}{T-A} = \int-k dt\] I'll let you finish up with the integration. Tell me what you get
anonymous
  • anonymous
@jim_thompson5910 12?
jim_thompson5910
  • jim_thompson5910
how are you getting 12 ?
anonymous
  • anonymous
Would I integrate both sides?
jim_thompson5910
  • jim_thompson5910
yeah what do you get when you do so
anonymous
  • anonymous
I'm confused, how would I integrate the left side?
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{dT}{dt} = -k(T-A)\] \[\Large \frac{dT}{T-A} = -k dt\] \[\Large \int\frac{dT}{T-A} = \int-k dt\] \[\Large \ln(|T-A|) = -k*t + C\] \[\Large |T-A| = e^{-k*t + C}\] \[\Large |T-A| = e^{-k*t}*e^{C}\] \[\Large |T-A| = e^{C}*e^{-k*t}\] what comes next?
anonymous
  • anonymous
Having two equations, one positive and one negative.
baru
  • baru
T: roast temp (25 deg after 5 hrs) A: freezer temp (20 deg) so T-A is positive, so you can ignore the modulus sign substitute T,A and t=5
baru
  • baru
you need to find k and c, so you need another equation, so substitute t=0 and T=68
anonymous
  • anonymous
Final answer is 9

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