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Using newton's cooling law

  • Thread starter polaris90
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  • #1
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An object is taken out of the oven that is set at a
temperature of 370 F and taken to a room that is at a temperature70 F. After 5 minutes, the temperature of the object is250 F.
a) Use the Newton’s law of cooling to write a differential
equation for this situation.

I started with dy/dt = k(y-70) and after differentiating I ended up with y = Be^(kt) + 70
What would I use or how would I solve for B?

I went online but on every page I saw B as Tf-Ti. On another post(not here) someone mentioned to solve for B, but I don't know how to. I have the temperatures and can just plug it into Tf-Ti, in this case 300. I don't to use it that way as it appears like I'm missing or assuming something not given.
Should I use 370 as my value for B?
 

Answers and Replies

  • #2
Ray Vickson
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An object is taken out of the oven that is set at a
temperature of 370 F and taken to a room that is at a temperature70 F. After 5 minutes, the temperature of the object is250 F.
a) Use the Newton’s law of cooling to write a differential
equation for this situation.

I started with dy/dt = k(y-70) and after differentiating I ended up with y = Be^(kt) + 70
What would I use or how would I solve for B?

I went online but on every page I saw B as Tf-Ti. On another post(not here) someone mentioned to solve for B, but I don't know how to. I have the temperatures and can just plug it into Tf-Ti, in this case 300. I don't to use it that way as it appears like I'm missing or assuming something not given.
Should I use 370 as my value for B?
What is y(0)? What does that say about B and k? What is y(5)?

RGV
 
  • #3
45
0
y(0) = 370, therefore e^(kt) = 1 and B = 300
Now it's all just plugging in numbers. Thanks, I just didn't know how to get B.
 

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