# Cooling Problem

• Jul 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?

• Jul 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.
• Jul 12th 2009, 09:45 AM
DMDil
shouldn't that be dT/dt = -kT?
• Jul 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.
• Jul 12th 2009, 11:00 AM
DMDil
Thanks alot for helping me, i am done solving that problem.