Newton's Law of Cooling

  • Thread starter D-Boy
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Homework Statement



A small object of unknown temperature was placed in a large room that had the fixed temperature 30 degrees C. After 10 minutes, the object's temperature is -10 degrees C, and after an additional 10 minutes, the object's temperature was -5 degrees C. What was the initial temperature of the object?

b = 30
When t = 10, y = -10
When t = 20, y = -5

Homework Equations



y(t) = b + Ce^(-kt)
dy/dt = -k(y - b)

The Attempt at a Solution



When I plug in the knowns of the variables I get either the equation:

-10 = 30 + Ce^(10k)
-or-
-5 = 30 + Ce^(20k)

I'm confused how I can find the constant k value and the constant C value when I'm not given the initial temperature. If someone can help me find either of the constant values I can probably work from there to finish the rest of the problem. Thanks
 
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Answers and Replies

  • #2
Gib Z
Homework Helper
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When I plug it in I get [tex]-10 = 30 + Ce^{-10k}, -5 = 30 + Ce^{-20k}[/tex]

Now, for each separate equation, take b to the other side, divide through by C, take a natural log, then divide by the coefficient of k. In short - Solve for k. You get two equations for k, set them equal to each other and simplify, you can solve for C. Once you solve for C, sub that into one equation and solve for k.
 

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