Cooling Problem

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July 12th 2009, 08:33 AM
DMDil

Cooling Problem

guys, i don't know how to solve this problem, i know i have to use dT/dt = -kT.

A pitcher of butter milk initially at 25
C is to be cooled by setting it on the front porch, where the temperature is 0C. Suppose that the temperature of the buttermilk has dropped to 15C after 20 min. When it will be at 5C?
July 12th 2009, 09:01 AM
galactus

$\frac{dT}{dt}=kT$

Separate variables:

$\frac{dT}{T}=kt$

Integrate:

$ln(T)=kt+C$

$T=C_{1}e^{kt}$

We are told that T(0)=25, so use this to find C:

$25=C_{1}e^{k(0)}\Rightarrow C_{1}=25$

$T=25e^{kt}$

Now, use the condition T(20)=15 to find k, then set the equation equal to 5 and solve for t.
July 12th 2009, 09:45 AM
DMDil

shouldn't that be dT/dt = -kT?
July 12th 2009, 10:22 AM
galactus

It doesn't much matter. You'll just get a negative instead of a positive solution.

The negative just means the temperature is dropping. Go ahead and use the negative in the same manner I outlined.
July 12th 2009, 11:00 AM
DMDil

Thanks alot for helping me, i am done solving that problem.

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