We can solve the problem by using Newton's law of cooling, which gives the rate of temperature variation of the thermometer given the temperature of the air ([tex]30^{\circ} F[/tex])
[tex] \frac{dT}{dt} = k(T-30) [/tex]
Where T is the temperature in Fahrenheit and t is the time in minutes.
We also know the temperature after t=1 min and t=5 min:
(A) T(1)=80
(B) T(5)=55
The problem asks to find the temperature at time t=0.
Let's solve the differential equation. We can rewrite it as:
[tex] \frac{dT}{T-30} = k dt [/tex]
by integrating on both sides, we have
[tex]ln(T-30) = kt + C_1[/tex]
[tex]T-30 = e^{kt+C_1}[/tex]
[tex]T=30+Ce^{kt}[/tex]
where we called [tex]C=e^{C_1}[/tex].
Nowe we have to find k and C. We can use conditions (A) and (B) for this purpose. We have the following system of equations:
(A) [tex]80=30+Ce^k[/tex]
(B) [tex]55=30+Ce^{5k}[/tex]
From which we find
[tex]k=-0.173[/tex]
[tex]C=59.44[/tex]
And finally, we can substituting these values into the main expression of T, and calculating it at time t=0:
[tex]T(t)=30+59.44 e^{-0.173 t}[/tex]
and using t=0, we find
[tex]T(0)=89.44 ^{\circ} F[/tex]
which is the initial temperature of the thermometer.
