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3 years ago

14

I think it is Neither but i am not sure so plz help

2 answers:

arlik [135]3 years ago

8 0

Answer:

Neither

Step-by-step explanation:

used desmos

look at screenshot

Vlad [161]3 years ago

7 0

Answer:

NEITHER I THINK

Step-by-step explanation:

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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

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$k =$ 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 $x$ :

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

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

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

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

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

$x=177.95$ ≈ $178$

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

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