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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

[tex]\frac{du}{dt} = -k(u - u_o)[/tex]

integrating we have

[tex]u - u_o = Ce^{-kt}[/tex]

u is the temperature reading

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

[tex]70 - (-10) = Ce^{-k(0)}[/tex]

C = 80

Now we already know the constant

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

at t = 2; u = 26

[tex]26 - (-10) = 80e^{-k(2)}[/tex]

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

[tex](u - 70) = Ce^{0.39925t}[/tex] <<< this might be the equation

....and i dont understand the problem anymore

the answer to this problem is 56 degrees Farenheit