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A roast turkey is taken from an oven when its temperature has reached 185°F and is...

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gulliopr | Student, Undergraduate | eNotes Newbie

Posted October 11, 2012 at 6:48 PM via web

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

A roast turkey is taken from an oven when its temperature has reached 185°F and is placed on a table in a room where the temperature is 75°F. (Round your answers to the nearest whole number.)

(a) If the temperature of the turkey is 150°F after half an hour, what is the temperature after 40  minutes?

141  °F


(b) When will the turkey have cooled to 110°?
________________min. ?

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted October 12, 2012 at 1:51 AM (Answer #1)

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Newton's law of cooling states that the rate of change in the temperature of an object is proportional to the difference between the temperature of the object and the ambient temperature.

Let y be the temperature of the turkey. Then `y'=k(y-75)`

`(dy)/(dt)=k(y-75)`

`(dy)/(y-75)=kdt`  Integrating we get:

`ln|y-75|=kt+C_1` Since y>75 we can lose the absolute value signs. Now exponentiate each side:

`y-75=e^(kt+C_1)=e^(kt)e^(C_1)=Ce^(kt)`

So `y=Ce^(kt)+75`

We are given two data points: (0,185) and (30,150) where x is in minutes and y is in degrees F.

Using (0,185) we see that `185=C+75==>C=110`

Using (30,150) and C=110 we get `150=110e^(30k)+75`

`75=110e^(30k)`

`e^(30k)=15/22`

`30k=ln(15/22)`

`k=(ln(15/22))/30~~.-.013`

So `y=110e^(-.013t)+75`

(a) If t=40 then `y=110e^(-.52)+75~~141`

So the temperature is approximately `141^@`

(b) If y=110 then `110=110e^(-.013t)+75`

`35=110e^(-.013t)`

`e^(-.013t)=7/22`

`-.013t=ln(7/22)`

`t=(ln(7/22))/-.013~~88` minutes

So it will take approximately 88 minutes.

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