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Newton's Law of Cooling ( Thanksgiving Turkey)?
3.) The cooling turkey problem
A.) The temperature of your turkey, when it is removed from the oven, is 165 degrees. The temperature of the room in which it is placed immediately after its removal is 70 degrees.
B.) Let the turkey sit for about 15 minutes, and take its temperature again. The temperature of your turkey after 15 minutes is 175 degrees.
C.) Use the information from parts a and b to find a function that models the cooing process of your turkey.
D.) sketch an accurate graph of our function, being sure to label all important aspects of the graph.
E.) How long until the turkey reaches the temperature of 120 degrees?
F.) According to Roasting a Turkey for Dummies, the turkey should rest for 20 minutes after removal from the oven. What temperature would your turkey be at this time?
Please help, I haven't been able to solve this problem. I know I'm supposed to use Newton's Law of Cooling.
u(t)=T+(usub0-T)e^kt k<0
1 Answer
- ?Lv 78 years ago
Hello David: Apply Newton's Law of Cooling to the cooling of the turkey :
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After 15 minutes, the turkey should have cooled downward from 165 F; not 175 F !!! I will use 150 F for temperature after 15 minutes.
Newton's Law Of Cooling gives :
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d ( T - TINF ) / dt = ( - KNLC ) ( T - TINF )
d ( T -TINF ) / ( T - TINF ) = ( - KNLC ) ( dt )
Integration gives :
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ln [ T - TINF / T0 - TINF ] = ( - KNLC ) ( t ) <----------------
T = TINF + ( T0 - TINF ) e^- (KNLC ) ( t ), <---------------------------
T0 = initial turkey temperature = 165 F
TINF = surroundings temperature = 70 F
Use the turkey temperature at two times ( 0 min and 15 min ) to get KNLC :
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ln [ 150 - 70 / 165 - 70 ] = ( - KNLC ) ( 15 )
ln ( 80 / 95 ) = ( - KNLC ) ( 15 )
- KNLC = ln [ 80/95 ] / 15 = - 0.01146 min^1
KNLC = 0.01146 min^-1 <-----------------------------------
To get time for turkey to be 120 F , you have :
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ln [ 120 - 70 / 165 - 70 ] = ( - 0.01146 ) ( t120 )
t120 = 56.01 min <-------------------------------------------------------
For t = 20 minutes, you have :
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T20 = 70 + ( 165 - 70 ) ( e^ (-0.01146 ) ( 20 )
T20 = 70 + ( 95 ) ( e^-0.2292 ) = 145.5 F <-----------------------------------
The graph will have axes of [ T - TINF / T0 - TINF ] versus time, t , and should be
a smooth curve.
Using semi-log paper with [ T - TINF / T0 - TINF ] on the logarithmic scale and t in minutes
along the conventional scale should give a straight line.