# Newton's Law of Cooling air temperature

## 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?

## 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?

uart

## 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?

## 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.

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