# Newton's law of cooling problem: Differential Eq

1. Jan 22, 2008

### Edwardo_Elric

1. The problem statement, all variables and given/known data
At 1:00 PM, a thermometer reading 70 degrees F is taken outside where the temperature is -10 degrees F (ten below zero). At 1:02PM, the reading is 26 degrees. At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F. What is the temperature reading at 1:09PM?

2. Relevant equations
$$\frac{du}{dt} = -k(u - u_o)$$
integrating we have
$$u - u_o = Ce^{-kt}$$

u_o is the temperature of the atmosphere

3. The attempt at a solution

solutions by sentences:

*"At 1:00 PM, a thermometer reading 70 degrees F is taken outside where the temperature is -10 degrees F (ten below zero)":
at t = 0 , u = 70
and u_o = 10
$$70 - (-10) = Ce^{-k(0)}$$
C = 80
Now we already know the constant

*"At 1:02PM, the reading is 26 degrees":
at t = 2; u = 26

$$26 - (-10) = 80e^{-k(2)}$$
k = 0.39925

my problem is that i dont understand when it is put indoors
*"At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F":
the constant of proportionality is the same
$$(u - 70) = Ce^{0.39925t}$$ <<< this might be the equation
....and i dont understand the problem anymore

the answer to this problem is 56 degrees Farenheit

2. Jan 22, 2008

### HallsofIvy

Staff Emeritus
You have calculated u(t) as long as the thermometer is outside. What will the temperature of the thermometer be when t= 5 (at 10:05)?

Now you have a new problem to do: u is now whatever you got for the temperature at 10:05 and u0= 75. Fortunately k is exactly the same as before (it depends only on the thermometer) so you don't need to calculate that again.

3. Jan 22, 2008

### Edwardo_Elric

for the time at exactly before 5 minutes i used the equation:
$$u - (-10) = 80e^{-0.39925(5)}$$
so
u = 0.8675 degrees

problem:
"At 1:05PM the thermometer is taken back indoors, where the air is at 70 degrees F":
so here it is now indoors and the air temperature is 70
for the time exactly after 5 minutes:

when t = 0, u = 0.8675 degrees;
$$(0.8675-70) = C(e^{0})$$
^
so
C = -69.1325

"What is the temperature reading at 1:09PM?"
t = 9min, u = ?
$$(u - 70) = -69.1325e^{-0.39925(9)}$$
u = 68.098 degrees