NEWTON'S LAW OF COOLING using differential equations

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 5K views
ardi
Messages
1
Reaction score
0
1. LAW OF COOLING PROBLEM!HELP PLEASE! :)

At 1:00pm, Sally puts into a refrigerator a can of soda that has been sitting of temperature 70degF. The temperature in the refrigerator is 40degF. fifteen minutes later 1:15pm, the temperature of the soda has fallen to 60degF. At some time later, Sally removes the soda form the refrigerator to the room, where at 2:00pm, the temperature of the soda is 60degF. At what time did Sally remove the soda from the refrigerator?

2.dT/dt= -k(Tsoda-Tref)
3. Tsoda=Tref + e^-kt(Tsoda-Tref)
at 1:15pm
60= 40 + e^-k(15)(70-40)
k= .027031

i don't know what to do next.. help please..
 
Physics news on Phys.org
well I would do this with 2 equations and 2 unknowns.
We know that at 1:15 the soda is 60F and in the fridge, and then she takes it out of the fridge at some time later we will call t' , and while it is in the fridge for t' it will cool to an unknown temp we will call T
so [itex]T=40+e^{-kt'(60-40)}[/itex]
now we know that we have 45 minutes between known temperatures so out time for our next equation will be 45-t', because this will be how much time the soda has to warm up in the room and we know the final temp that it needs to be and that is 60.
so our second equation is [itex]60=70+e^{-k(45-t')(T-70)}[/itex]
so know we have 2 equations and 2 unknowns and we should be able to solve it.
I might have my 60 and 70 backwards but you get the idea.
 
Last edited: