Cooling problem

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colerelm

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Oct 24, 2011
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I'm having trouble understanding how to solve the following problem:

A pitcher of buttermilk initially at 25 degrees Celsius is to be cooled by setting it on the front porch, where the temperature is 0 degrees Celsius. Suppose that the temperature of the buttermilk has dropped to 15 degrees Celsius after 20 min. When will it be at 5 degrees Celsius?

I actually have the steps required to solve the problem in front of me but I don't get how this step was made:

1.443.jpg

Why does the differential equation reduce to that? Is that a formula I should know? Also, what does it mean by "Auxiliary equation"?
 
colerelm said:
I'm having trouble understanding how to solve the following problem:

A pitcher of buttermilk initially at 25 degrees Celsius is to be cooled by setting it on the front porch, where the temperature is 0 degrees Celsius. Suppose that the temperature of the buttermilk has dropped to 15 degrees Celsius after 20 min. When will it be at 5 degrees Celsius?

I actually have the steps required to solve the problem in front of me but I don't get how this step was made:

View attachment 2282

Why does the differential equation reduce to that? Is that a formula I should know? Also, what does it mean by "Auxiliary equation"?
Click to expand...

That uses Newton's law of Cooling - do you know that?
 
Newton's Law of Cooling says \(\displaystyle \frac{dT}{dt}=k(T-T_{m})\)

where \(\displaystyle T_{m}\) is the ambient temp. In this case, \(\displaystyle T_{m}=0\)

So, you have \(\displaystyle \frac{dT}{dt}=kT\)

Separate variables and integrate:

\(\displaystyle \int\frac{dT}{T}=\int kdt\)

\(\displaystyle \ln(T)=kt+c\)

\(\displaystyle T=e^{kt+c}=Ce^{kt}\)

Now, use your initial conditions to find C and k.

\(\displaystyle T(0)=25, \;\ T(20)=15\)

This will give the formula for the temperature, T, at any time t.

You can set T=5 and solve for t.
 
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