# Thread: help with DE problem: Newton's law of cooling

1. ## help with DE problem: Newton's law of cooling

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?

u_o is the medium

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

my solutions in solving the problem:
"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
$(u - 70) = Ce^{0.39925t}$
....and i dont understand the problem

the answer to this problem is 56 degrees Farenheit

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?

u_o is the medium

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

my solutions in solving the problem:
"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
$(u - 70) = Ce^{0.39925t}$
....and i dont understand the problem

the answer to this problem is 56 degrees Farenheit
i didn't check, but i assume all your calculations are correct so far

first you need to find what reading the thermometer has at 1:05 pm, you have the equation for that. use that as u(0), that is, the temperature at time 0. you want u(4). the thermometer is going to heat up. so we can use the equation you have, except make k positive (that should work)

3. 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

edit: thanks jhevon!!!
i fogot that it should be u(4) not u(9)

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# at a certain time, a thermometer reading 70Â°F is taken outdoors where the temperature is 15Â°F. Five minutes later, the thermometer reading is 45Â°F.

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