- #1
Edwardo_Elric
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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 don't 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 don't understand the problem anymore
the answer to this problem is 56 degrees Farenheit