Question

Newton's Law of Cooling:
Jane was late getting ready for the party, and the liters of soft drinks she bought were still at room temperature 73 degrees F with guest due to arrive in 15 minutes. If she put these in her freezer at -10 degrees F, will the drinks be cold enough (35 degrees F) for her guest? Assuem k is approximately -0.031

Answer 1

@experimentX

Answer 2

this is 11 th grade question ... i don't remember this ... but it must be first order simple differential equation ... must be pretty simple.

Answer 3

you are so funny :) ok what do I do with 35

Answer 4

T sub r is what

Answer 5

can you wait for a while ... I can't handle two threads for a while
If you have got any idea you can join us here
http://openstudy.com/study#/updates/4f858d73e4b0505bf085a07f

Answer 6

cool, take your time. Gotta go see you in a couple of hours. Other problem looks cool.

Answer 7

Damn ... site crashed .... all my work gone.

Answer 8

I'll put it simple way,

the Newton's cooling law states that

\( \LARGE \theta(t) = T_{1} + Ce^{-kt}\)

Theta is the temperature,
T1 is the temperate of the room.
C is constant from integration,
k is proportionality constant
and t is time.

It is given that \(\theta = 73F\) when t = 0

\( \LARGE 37F = -10F + Ce^{-k*0}\)

\( \LARGE C = ?? \)
From here calculate the value of C, and put it on your original equation

AND THERE IS ONE IMPORTANT THING, the value of temperature is in Fahrenheit, change it to Kelvin, and change time in seconds.

Answer 9

Cool, thanks. I think I like that equation much better.

Answer 10

I had to redo this ... my chrome crashed when i was about finish.