Question

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the temperature is 75 degrees F.

(a) If the temperature of the turkey is 150 degrees F after half an hour, what is the temperature after 45 minutes?
(b) When will the turkey have cooled to 100 degrees F?

Answer 1

(a) Using Newton's Law of Cooling,
`(dT)/(dt) = k(T - T_s)`, we have
`(dT)/(dt) = k(T - 75)` where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get
`(dT)/(T - 75) = k dt`. Integrate both sides to get
`\ln|T - 75| = kt + C`.

Since
`T(0) = 185`, we solve for C:

`|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110`
So we get
`\ln|T - 75| = kt + \ln 110`. Use T(30) = 150 to solve for k:

`\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= (1)/(30)\ln (75/110) = (1)/(30)\ln(15/22)`

So


`\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^(kt + \ln110) \Rightarrow \\ \\ |T - 75| = 110e^(kt) \Rightarrow T - 75 = \pm110e^((1/30)\ln(15/22)t) \Rightarrow \\ T = 75 \pm110e^((1/30)\ln(15/22)t)`

But choose Positive because T > 75. Temp of turkey can't go under.


`T(t) = 75 + 110e^((1/30)\ln(15/22)t) \\ T(45) = 75 + 110e^((1/30)\ln(15/22)(45)) = 136.929 \approx 137{}^(\circ)F`

(b)


`T(t) = 75 + 110e^((1/30)\ln(15/22)t) = 75 + 110(15/22)^(t/30) \\ 100 = 75 + 110(15/22)^(t/30) \\ 25 = 110(15/22)^(t/30) (25)/(110) = (15/22)^(t/30) \\ \ln(25/110) / ln(15/22) = t/30 \\ t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}`

Dogs of the AMS.