Answer:
45.50° F
Step-by-step explanation:
As per Newton's law,
[tex]T(t) = T_{s}+(T_{0} + T_{s})e^{-kt}[/tex]
When T(t) is the final temperature
[tex]T_{s}[/tex] = Temperature of surrounding
[tex]T_{0}[/tex] = Initial temperature
t =duration of cooling
k = constant
[tex]103=0+(155-0)e^{-k\times 5}[/tex]
[tex]103=155e^{-5k}[/tex]
Now take natural log on both the sides
[tex]ln(103)=ln(155e^{-5k})[/tex]
[tex]ln(103)=ln(155)+ln(e^{-5k})[/tex]
[tex]ln(103)=ln(155)=-5k[/tex]
4.6347 - 5.0434 = -5k
[tex]k=\frac{0.40872}{5}[/tex]
k = 0.0817
[tex]T(t)=0+(155-0)e^{(0.0817\times 15)}[/tex]
= [tex]155e^{-1.226}[/tex]
= 155 (02935)
= 45.49 ≈ 45.50° F
