Answer:
= 30.67°F
≅ 30.70°F
Explanation:
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
[tex]t[/tex]=duration of cooling
k = constant
[tex]46 = 0+(155-0)e^{-k\times 15}46 = 155e^{-15k}[/tex]
Now take natural log on both the sides
[tex]ln(46)=ln(155e^{-15k})ln(46)=ln(155)+ln(e^{-15k})ln(46)=ln(155)=-15k\\3.8286 - 5.0434 = -15k[/tex]
k = 0.081
[tex]T(t)=0+(155-0)e^{(0.081\times 20)}[/tex]
[tex]= 155e^{-1.62}[/tex]
= 155 (0.1979)
= 30.67°F
≅ 30.70°F
