+0  
 
0
313
3
avatar

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

Best Answer 

 #1
avatar+87301 
+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
 #1
avatar+87301 
+10
Best Answer

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
avatar+92781 
0

thanks Chris.  

Melody  Dec 14, 2014
 #3
avatar+92781 
+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

8 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.