Question

After heating up in a teapot, a cup of hot water is poured at a temperature of 210°F. The cup sits to cool in a room at a temperature of 70° F. Newton's Law of Cooling explains that the temperature of the cup of water will decrease proportionally to the difference between the temperature of the water and the temperature of the room, as given by the formula below: T = T. + (To - Tale To = the temperature surrounding the object To = the initial temperature of the object t = the time in minutes T= the temperature of the object after t minutes k = decay constant The cup of water reaches the temperature of 197°F after 1.5 minutes. Using this information, find the value of k, to the nearest thousandth. Use the resulting equation to determine the Fahrenheit temperature of the cup of water, to the nearest degree, after 4 minutes. Enter only the final temperature into the input box.

Answer 1

Newton's Law of Cooling:

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}$

$T(t) =$ Temperature given at a time

$t =$ Time

$T_{s} =$ Surrounding temperature

$T_{o}=$ Initial temperature

$e =$ Constant (Euler's number) ≈ 2.72

$k =$ Constant

Using this information, find the value of $k$, 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 $k$:

$T(t) =$ 197°, $t =$ 1.5 minutes, $T_{s} =$ 70° and $T_{o}=$ 210°  

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\197=70+(210-70)e^{-1.5k} \\197 -70 = (140)e^{-1.5k} \\127 =(140)e^{-1.5k}\\\frac{127}{140}=e^{-1.5k} \\ln(\frac{127}{140})=-1.5k\\-0.097=-1.5k\\0.0649 = k$

$k$ ≈ $0.065$

Let the temperature of the water cup after $t = 4$ minutes be $T(t) = x$

Now, let's plug the new time and $k$ constant in the equation and solve for $x$:

$T(t)=T_{s}+(T_{o}-T_{s})e^{-kt}\\\\\x=70+(210-70})e^{-0.065*4}\\\\x=70+(140})e^{-0.26}, -0.26=-\frac{26}{100}=-\frac{13}{50}$

$x=70+(140})e^{-\frac{13}{50}}$

$x=70+(140})e^{\frac{1}{\frac{13}{50}}$

$x=70+e^{\frac{140}{\frac{13}{50}}$

$x=70+{\frac{140}{\sqrt[50]{e^{13}}}$

$x = 70 +\frac{140}{1.3} \\x=70+107.947$

$x=177.95$ ≈ $178$

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