Law of cooling and Exponential growth

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Homework Help Overview

The discussion revolves around the application of the law of cooling to determine the time it takes for a roast turkey to cool from 185 degrees F to 100 degrees F in a room at 75 degrees F. The problem involves understanding the cooling process and the relevant equations governing it.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the cooling constant k and its derivation from the given temperature data. There are inquiries about the assumptions made regarding the temperature of the turkey after a specific time period.

Discussion Status

The discussion is ongoing, with participants sharing their methods for calculating k and questioning the assumptions related to the temperature of the turkey at 30 minutes. There is no clear consensus yet on the correct approach or solution.

Contextual Notes

There is uncertainty regarding the temperature of the turkey after 30 minutes, which is critical for accurately determining the cooling constant k. Participants are exploring different interpretations of the problem setup.

fk378
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Homework Statement


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. When will the turkey have cooled to 100 degrees F?


Homework Equations


dT/dt=k(T-Ts) where Ts=temperature of the surrounding
y(t)=y(0)e^kt



The Attempt at a Solution


I solved T(t)=75+110e^(ln .682)t/30

Solving for t:
100=75+110e^(ln .682)t/30
t=10

I know that t=10 isn't right, but I don't know what I did wrong. Any takers?
 
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How did you solve for k?

Just quickly I got:

\frac{1}{k}ln(\frac{25}{110})
 
To solve for k I used the fact that y(30)=150-75=75
So, 110e^30k=75
e^30k=.682
30k=ln .682
k= ln .682/30
 
dashkin111 said:
How did you solve for k?

Just quickly I got:

\frac{1}{k}ln(\frac{25}{110})

fk378 said:
To solve for k I used the fact that y(30)=150-75=75
So, 110e^30k=75
e^30k=.682
30k=ln .682
k= ln .682/30
And how did you know that y(30)= 75? You didn't tell us anything about the temperature of the turkey 30 minutes after being take out of the oven!
 

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