anonymous
  • anonymous
A potato with temperature 70 degrees F is placed into a oven with an internal temperature 400 degree after 30 minutes, the temperature of the potato has increased to 205 degrees F. Give the heating constant k (rounded to 6 decimal places), and find an expression for the temperature potato as a function of time.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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NotTim
  • NotTim
so useful... whoa...was gunna give the anseer there for some reason...
anonymous
  • anonymous
huh?
NotTim
  • NotTim
well, what's our initial constant? What would the entire thing equal?

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More answers

anonymous
  • anonymous
A potato with temperature 70 degrees F is placed into a oven with an internal temperature 400 degree. After 30 minutes, the temperature of the potato has increased to 205 degrees F.
NotTim
  • NotTim
uh
anonymous
  • anonymous
You need the Newton's formula for this one.\[T(t)=T_1+(T_0-T_1)e ^{-kt}\]T0 is the starting temperature of the object, T1 is the temperature of the environment.
anonymous
  • anonymous
As t gets larger, the second term fades away, which leads to the temperature of the object T(t) being T1 (the temperature of the surroundings.
anonymous
  • anonymous
So plug the data from the problem into the formula; you will have only one unknown value (k) that can be solved for.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Solve this....\[205=400+(70-400)e^{-30k}\]
anonymous
  • anonymous
what is the temperature of the potato after 80 minutes?
anonymous
  • anonymous
After what amount of time will the temperature of the potato reach 350 degrees F?
anonymous
  • anonymous
You can find that after you have solved for k.
anonymous
  • anonymous
The last unknown is k; once you have that, you just plug the time or temperature into the formula, and there will be only one unknown to solve for.
anonymous
  • anonymous
\[205=400+(70−400)e^{−30k}\implies \frac{-195}{-330}=e^{−30k}\]\[\implies \frac{\ln(\frac{195}{330})}{-30}=k\]
anonymous
  • anonymous
After you have k (use your calculator to get an approximation), you can now solve for unknown time or temperature easily.
anonymous
  • anonymous
so 80 replaces 205 in the equation to solve for the temperature after 80 minute?
anonymous
  • anonymous
No, you put 80 in for t, the time in the exponent. T(t) is the temperature of the object as a function of t (time). Previously, the time was 30 in the problem above. See it?

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