turkey taken out of oven 185 F.
room temp 75 F
30 min after turkey is 150 F
45 minutes after 115 F (according to my calculations) < this part wasn't given i had to do it on my own so it might be wrong.
how many minutes will it before before it is 100 F
thanks!!!!
room temp 75 F
30 min after turkey is 150 F
45 minutes after 115 F (according to my calculations) < this part wasn't given i had to do it on my own so it might be wrong.
how many minutes will it before before it is 100 F
thanks!!!!
-
Newton's Law of cooling states that the instantaneous rate of change of temperature is proportional to the difference between the temperature of a heated body and the environment.
let f(t) be the difference between turkey Temp T(t) and the ambient TsubAmbient (75)
df(t)/dt=dT(t)/dt-dTsubAmbient/dt=-k(T… where I have chosen to make the sign of k negative for convenience (temperature is dropping) Let y=f(t), then the solution to the differential equation can be written as
y(t) = (185-75)e^-kt=110e^-kt
and T(t) = TsubAmbient+y(t) = 75 +110e^-kt
To find k we have a point T(30)=75+110e^-30k=150 or 110e^-30k=75
or e^-30k=75/110=15/22
so -30k = log(15/22)
>>> import math
>>> math.log(10)
2.3025850929940459
>>> math.log(15/22.)
-0.38299225225610584
>>> _/30
-0.012766408408536861
>>>or k = .0128 (approximately)
at 45 minutes we have T(45) = 75+110e^-45*.0128=75 + 62 = 137
and the other one is
T(x)=75+110e^-.0128*x=100
or e^-.0128*x=25/110=5/22
>>> math.log(5/22.)
-1.4816045409242156
>>> _/-.0128
115.75035475970434
>>> or about 116 minutes to reach 100 degrees.
let f(t) be the difference between turkey Temp T(t) and the ambient TsubAmbient (75)
df(t)/dt=dT(t)/dt-dTsubAmbient/dt=-k(T… where I have chosen to make the sign of k negative for convenience (temperature is dropping) Let y=f(t), then the solution to the differential equation can be written as
y(t) = (185-75)e^-kt=110e^-kt
and T(t) = TsubAmbient+y(t) = 75 +110e^-kt
To find k we have a point T(30)=75+110e^-30k=150 or 110e^-30k=75
or e^-30k=75/110=15/22
so -30k = log(15/22)
>>> import math
>>> math.log(10)
2.3025850929940459
>>> math.log(15/22.)
-0.38299225225610584
>>> _/30
-0.012766408408536861
>>>or k = .0128 (approximately)
at 45 minutes we have T(45) = 75+110e^-45*.0128=75 + 62 = 137
and the other one is
T(x)=75+110e^-.0128*x=100
or e^-.0128*x=25/110=5/22
>>> math.log(5/22.)
-1.4816045409242156
>>> _/-.0128
115.75035475970434
>>> or about 116 minutes to reach 100 degrees.