coffee conundrum

[kflodias]

Alright, my little brother is having some trouble with his homework... not even my Dartmouth-educated uncle could figure this out. Here's all the information I have:

Some guy picked up coffee from a restaurant and it was 160 degrees F. He claims he picked it up at 7:58 am, held it in his hand for 5 minutes, then placed it in a cup-holder for five minutes... (His hand temperature was 92 degrees F and the car temperature was around 72 degrees F). Then, he proceeded to spill his nice warm beverage all over himself at around 8:08 am.
Now, in 4.5 minutes in a 72 degree enironment, the coffee could go from 160 degrees F to 149 degrees F. Still with me?
Anyway... this guy takes the all-American initiative of filing a lawsuit, however, he has no case unless the coffee was above 140 degrees F at the time it spilled.
My brother's teacher wants him to figure out the breakdown if the guy were holding the coffee for 4 minutes and had it in the cupholder in 6 (and 3:7, 2:8, 6:4, etc...) Also, what would happen if the guy picked up his coffee at 7:56 instead of 7:58? To anyone who could help, I would greatly appreciate it.

 

[stapel]

Trying to work through a "translator" who doesn't understand the material has never, in my experience, been successful. So please have the student reply with the exact (word-for-word, without the commentary) text of the exercise, the complete instructions, the formula(s) provided (and definitions of the variables), and a clear listing of everything he has tried so far.

Thank you.

Eliz.

 

[galactus]

Use Newton's Cooling Law.

I hope I am translating the problem correctly. The temperature of the coffee initially is 160 degrees. In an ambient temperature of 72 degrees it cools to 149 degrees in 4.5 minutes. How long until it cools to 140 degrees?.

\(\displaystyle \L\\T(0)=160\)

\(\displaystyle \L\\T(4.5)=149\)

\(\displaystyle \L\\\frac{dT}{T-72}=kdt\)

Integrate:

\(\displaystyle \L\\\int\frac{dT}{T-72}=\int{kdt}\)

\(\displaystyle \L\\ln|T-72|=kt+C_{1}\)

\(\displaystyle \L\\T=72+C_{2}e^{kt}\)

When t=0, T=160:

So, \(\displaystyle 160=72+C_{2}\)

\(\displaystyle C_{2}=88\)

\(\displaystyle \L\\T=72+88e^{kt}\)

But, \(\displaystyle \L\\149=72+88e^{4.5k}\)

\(\displaystyle \L\\k=-0.029673642805=\frac{2ln(7/8)}{9}\)

\(\displaystyle \H\\T=72+88e^{-0.029673t}\)

Set to 140 and solve for t:

That'll be the number of minutes 'til it cools to 140 degrees.

Does this help?.

EDIT: I'm sorry, I noticed you also want a cooling time while it's in his hand.
I am probably missing something, but we have no initial cooling to go off of for that portion. In other words, for the car temperature, we had that it cooled to 149 in 4.5 minutes. We have nothing like that for the hand time. Say, we had some info which said, "it cooled to 155 degrees in 3 minutes in his hand". Something like that. See what I mean?.