Newton's Law of Cooling:
[tex]T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}[/tex]
[tex]T(t) =[/tex] Temperature given at a time
[tex]t =[/tex] Time
[tex]T_{s} =[/tex] Surrounding temperature
[tex]T_{o}=[/tex] Initial temperature
[tex]e =[/tex] Constant (Euler's number) ≈ 2.72
[tex]k =[/tex] Constant
Using this information, find the value of [tex]k[/tex], to the nearest thousandth, then use the resulting equation to determine the temperature of the water cup after 4 minutes.
First, plug in the given values in the equation and solve for [tex]k[/tex]:
[tex]T(t) =[/tex] 197°, [tex]t =[/tex] 1.5 minutes, [tex]T_{s} =[/tex] 70° and [tex]T_{o}=[/tex] 210°
[tex]T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\197=70+(210-70)e^{-1.5k} \\197 -70 = (140)e^{-1.5k} \\127 =(140)e^{-1.5k}\\\frac{127}{140}=e^{-1.5k} \\ln(\frac{127}{140})=-1.5k\\-0.097=-1.5k\\0.0649 = k[/tex]
[tex]k[/tex] ≈ [tex]0.065[/tex]
Let the temperature of the water cup after [tex]t = 4[/tex] minutes be [tex]T(t) = x[/tex]
Now, let's plug the new time and [tex]k[/tex] constant in the equation and solve for [tex]x[/tex]:
[tex]T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\\\\x=70+(210-70})e^{-0.065*4}\\\\x=70+(140})e^{-0.26}, -0.26=-\frac{26}{100}=-\frac{13}{50} \\[/tex]
[tex]x=70+(140})e^{-\frac{13}{50}}\\\\[/tex]
[tex]x=70+(140})e^{\frac{1}{\frac{13}{50}}\\\\\\[/tex]
[tex]x=70+e^{\frac{140}{\frac{13}{50}}\\\\\\[/tex]
[tex]x=70+{\frac{140}{\sqrt[50]{e^{13}}}\\[/tex]
[tex]x = 70 +\frac{140}{1.3} \\x=70+107.947\\[/tex]
[tex]x=177.95[/tex] ≈ [tex]178[/tex]
Temperature of water after 4 minutes is 178°
sorry if there's any misspelling or wrong step but I hope my answer is correct ':3
