NEWTON'S LAW OF COOLING using differential equations

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SUMMARY

The discussion revolves around applying Newton's Law of Cooling to determine the time Sally removed a can of soda from a refrigerator. The initial temperature of the soda was 70°F, and after 15 minutes in a 40°F refrigerator, it cooled to 60°F. Using the differential equation dT/dt = -k(Tsoda - Tref) and the derived formula Tsoda = Tref + e^(-kt)(Tsoda - Tref), the cooling constant k was calculated as 0.027031. Two equations were established to solve for the time the soda was removed from the refrigerator, utilizing the known temperatures and time intervals.

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  • Basic knowledge of exponential functions
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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..
 
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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 T=40+e^{-kt'(60-40)}
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 60=70+e^{-k(45-t')(T-70)}
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:

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