Answer:
The beam initial temperature is 5 °F.
Step-by-step explanation:
If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

where
is the ambient temperature,
is the initial temperature,
is the time and
is a constant yet to be determined.
The goal is to determine the initial temperature of the beam, which is to say 
We know that the ambient temperature is
, so

We also know that when
the temperature is
and when
the temperature is
which gives:


Rearranging,


If we divide these two equations we get


Now, that we know the value of
we can use it to find the initial temperature of the beam,

so the beam started out at 5 °F.