Newton's Law of Cooling air temperature

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SUMMARY

The discussion focuses on applying Newton's Law of Cooling to determine the temperature reading of a thermometer at 1:09 PM after being exposed to two different environments. Initially, the thermometer reads 70°F at 1:00 PM, then drops to 26°F at 1:02 PM in -10°F air, and is later brought back indoors at 70°F. The key equations used include u(t) = u∞ + (u0 - u∞) e^(-kt) and the constant k calculated as -0.39925. The final temperature reading at 1:09 PM is determined to be 58°F.

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Mastur
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Homework Statement


At 1:00PM, a thermometer reading 70oF is taken outside where the air temperature is -10oF. At 1:02PM, the reading is 26oF. At 1:05PM, the thermometer is taken back indoors, where the air is at 70oF. What is the temperature reading at 1:09PM?

Homework Equations


The Attempt at a Solution


At t=0, u=70F, temp.env.=-10F
At t=2, u=26F.
At t=5 u=? env=70F
At t=9, u=?

\frac{du}{dx}=k(u+10)

u+10=cekt
at t=0, c=80F

u+10=80ekt
at t=2, u=26, k=-0.39925

Giving me the u value which is 0.86729

After that, I don't know what to do next. :|

I actually tried letting u=0.86729 and k=-0.39925 when t=0. And after doing the same process, I ended up having u=60.608F but the answer should be 58F.

Is there any shortcut or pattern in obtaining the answer to this kind problems?
 
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Mastur said:

Homework Statement


At 1:00PM, a thermometer reading 70oF is taken outside where the air temperature is -10oF. At 1:02PM, the reading is 26oF. At 1:05PM, the thermometer is taken back indoors, where the air is at 70oF. What is the temperature reading at 1:09PM?

Homework Equations


The Attempt at a Solution


At t=0, u=70F, temp.env.=-10F
At t=2, u=26F.
At t=5 u=? env=70F
At t=9, u=?

\frac{du}{dx}=k(u+10)

u+10=cekt
at t=0, c=80F

u+10=80ekt
at t=2, u=26, k=-0.39925

Giving me the u(5) value which is 0.86729
Ok so far so good. That's correct :)
After that, I don't know what to do next. :|

I actually tried letting u=0.86729 and k=-0.39925 when t=0. And after doing the same process, I ended up having u=60.608F but the answer should be 58F.

Is there any shortcut or pattern in obtaining the answer to this kind problems?

Your method seems ok. Treat the second calculation as a separate problem with a new time variable starting at t=0 when the thermometer is bought back indoors.

The easiest way to sort out the constants is to use

u(t) = u_\infty + (u_0 - u_\infty) e^{-kt}

Where u_0=0.87 and u_\infty=70. Just remember that you're trying to find u(4) because you've re-started the time variable.
 
Last edited:

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