Newton's Law of Cooling help
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Newton's Law of Cooling help

[From: ] [author: ] [Date: 11-11-07] [Hit: ]
it becomes195=68+357e^(30k) or e^(30k)=127/357.k=-.100=68+357e^(-.dy/dt = 2y can be written in the form y =Ce^(kt) for some constants C and k.......
An apple pie comes out of the over at 425 degrees(f) and is placed on a counter in a 68 degree(f) room to cool. In 30 minutes it has cooled to 195 degrees(f). According to Newton's Law of Cooling, how many additional minutes must pass before it cools to 100 degrees(f)?

And can you answer this true/false question?

The general solution to dy/dt = 2y can be written in the form y = C(3^(kt)) for some constants C and k. Justify your answer.

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dT/dt=k(T-68)

T=68+ce^(kt), Since T=425 when t=0, it follows that c=357, so that T=68+357e^(kt)


To determine the value k, we use the fact that T=195 when t=30. Under this condition, it becomes195=68+357e^(30k) or e^(30k)=127/357.

30k=ln(127/357)

k=-.034452


100=68+357e^(-.034452t)

t=70 min


false
dy/dt = 2y can be written in the form y =Ce^(kt) for some constants C and k.
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