157k views
2 votes
This exercise uses Newton's Law of Cooling.

A roasted 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 78°F.
(a) If the temperature of the turkey is 150°F after half an hour, what is its temperature after 45 min? (Round your answer to the nearest whole number.)

Incorrect: Your answer is incorrect.
°F

(b) After how many hours will the turkey cool to 100°F? (Round your answer to one decimal place.)

This exercise uses Newton's Law of Cooling. A roasted turkey is taken from an oven-example-1

1 Answer

1 vote

Answer:

(a) 137 °F

(b) 2.0 hr

Step-by-step explanation:

Given a turkey is removed from the oven at 185 °F to a room temperature of 78 °F and cools to 150 °F after 30 minutes, you want to know the temperature after 45 minutes, and the time to cool to 100 °F.

Newton's cooling equation

The temperature described by Newton's Law of Cooling can be written as the equation ...

T(t) = (initial temperature difference) × (cooling factor)^(t/(cooling period)) + (final temperature)

Here, the initial temperature difference is 185 -78 = 107 °F. The cooling factor is (150 -78)/(185 -78) = 72/107 in a period of 30 minutes.

Thus, we can write the temperature equation as ...

T(t) = 107×(72/107)^(t/30) . . . . . . . where t is time in minutes

(a) 45 minutes

When t=45, the temperature is ...

T(45) = 107×(72/107)^(45/30) +78 ≈ 137 . . . . . degrees F

After 45 minutes, the temperature is 137 °F.

(b) 100 °F

When T = 100 °F, the time is ...

100 = 107×(72/107)^(t/30) +78

22/107 = (72/107)^(t/30)

30×log(22/107)/log(72/107) = t ≈ 119.783 . . . . minutes

t = 119.783/60 = 1.996 . . . . hours

The turkey cools to 100 °F after 2.0 hours.

This exercise uses Newton's Law of Cooling. A roasted turkey is taken from an oven-example-1
User Kadi
by
7.9k points