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

Guest Dec 13, 2014

#1**+10 **

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

T = ab^{x} where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab^{2} → 323/b^{2} = a

And when x = 5 we have

288 = ab^{5} = (323/b^{2})b^{5} = 323 b^{3} divide both sides by 323

288/323 = b^{3} 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

#1**+10 **

Best Answer

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

T = ab^{x} where T is the temperature after some "x" minutes have elapsed

So we have, when x = 2

323 = ab^{2} → 323/b^{2} = a

And when x = 5 we have

288 = ab^{5} = (323/b^{2})b^{5} = 323 b^{3} divide both sides by 323

288/323 = b^{3} 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

#3**+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

Melody
Dec 14, 2014