Originally Posted by

**^_^Engineer_Adam^_^** 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 is the temperature reading

u_o is the medium

$\displaystyle -k(u - u_o) = \frac{du}{dt}$

and by integrating

$\displaystyle u - u_o = Ce^{-kt}$

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

$\displaystyle 70 - (-10) = Ce^{-k(0)}$

C = 80

Now we already know the constant

"At 1:02PM, the reading is 26 degrees":

at t = 2; u = 26

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

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

$\displaystyle (u - 70) = Ce^{0.39925t}$

....and i dont understand the problem

the answer to this problem is 56 degrees Farenheit