# Newton's law of cooling problem: Differential Eq

## Homework Statement

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?

## Homework Equations

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

u_o is the temperature of the atmosphere

## 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

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## Homework Statement

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?

## Homework Equations

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

u_o is the temperature of the atmosphere

## 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
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.

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