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# Pies can only be delivered when their temperature is 120 degrees.When they are taken out of the oven, after two minutes their temperature is

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Pies can only be delivered when their temperature is 120 degrees.When they are taken out of the oven, after two minutes their temperature is 323 degrees, and after 5 minutes the temperature is 288 after how long will they be at 120 degrees

Dec 13, 2014

#1
+94526
+10

I'm going to use the following function to model this

T = abx      where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab2     →   323/b2  = a

And when x = 5   we have

288 = ab5  =  (323/b2)b5   = 323 b3     divide both sides by 323

288/323 = b3    take the 3rd root of each side

(288/323)^(1/3)  = b = about .9624

So, a = 323/(.9624)2 =  about 348.66

So, our function is

T = 348.66(.9624)x

So, to find out when they will be at 120 degrees, we have

120 = 348.66(.9624)x        divide both sides by 348.66

120/348.66  =  (.9624)x    take the log of each side

log ( 120/348.66 )  = log (.9624)x     and by a log property, we have

log ( 120/348.66 )  =  x log (.9624)    divide both sides by log(.9624)

log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes

Here's a graph of the function.......https://www.desmos.com/calculator/logb5i0pwa

Dec 14, 2014

#1
+94526
+10

I'm going to use the following function to model this

T = abx      where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab2     →   323/b2  = a

And when x = 5   we have

288 = ab5  =  (323/b2)b5   = 323 b3     divide both sides by 323

288/323 = b3    take the 3rd root of each side

(288/323)^(1/3)  = b = about .9624

So, a = 323/(.9624)2 =  about 348.66

So, our function is

T = 348.66(.9624)x

So, to find out when they will be at 120 degrees, we have

120 = 348.66(.9624)x        divide both sides by 348.66

120/348.66  =  (.9624)x    take the log of each side

log ( 120/348.66 )  = log (.9624)x     and by a log property, we have

log ( 120/348.66 )  =  x log (.9624)    divide both sides by log(.9624)

log ( 120/348.66 ) / log (.9624)= x = about 27.83 minutes

Here's a graph of the function.......https://www.desmos.com/calculator/logb5i0pwa

CPhill Dec 14, 2014
#2
+95177
0

thanks Chris.

Dec 14, 2014
#3
+95177
+5

I don't know Chris,

This might be the answer wanted but

Newton's law of cooling says that

$$\\\mbox{T is temperature, t is time, and S is the surrounding temperature}\\\\ \frac{dT}{dt}=-k(t-S)\\\\ I think this means\\\\ T(t)-S=(T_0-S)e^{-kt}\\\\$$

I don't think that we are given enough information to solve this question.

http://formulas.tutorvista.com/physics/newton-s-law-of-cooling-formula.html

Dec 14, 2014