Answer:
53.3324
Step-by-step explanation:
given that a thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 40° F.
By Newton law of cooling we have
T(t) = [tex]T+(T_0-T)e^{-kt}[/tex]
where T (t) is temperature at time t,T =surrounding temperature = 40, T0 =70 = initial temperature
After half minute thermometer reads 60° F. Using this we can find k
[tex]T(0,5) = 40+(70-40)e^{-k/2} = 60\\e^{-k/2} =2/3\\-k/2 = -0.4055\\k = 0.8110[/tex]
So equation is
[tex]T(t) = 40+(30)e^{-0.8110t}\\[/tex]
When t=1,
we get
[tex]T(1) = 40+(30)e^{-0.8110}\\\\=53.3324[/tex]