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# Elliot takes a roast out of the oven when the internal temperature of the roast is 165F

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Elliot takes a roast out of the oven when the internal temperature of the roast is 165°F. After 12 minutes, the temperature of the roast drops to 140°F.

The temperature of the room is 70°F.

How long does it take for the temperature of the roast to drop to 125°F?

Use the Newton's Law of Cooling equation, ​ T(t)=TA+(T0−TA)e^−kt

Jan 13, 2020

#1
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I got 21.86... is that right?

Jan 13, 2020
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T(t)  =  TA  + (T0 - TA)e^(-kt)

We know that   after 12 minutes

140  = 70  + ( 165 - 70)e^(-k* 12)       simplify

Subtract 70 from both sides

70 = ( 95)e^(-12k)       divide both sides by  25

70/95  = e^(-12k)       take the Ln of both sides

Ln (70/95) =  Ln e^(-12k)          and we can write

Ln (70/95)  = (-12k)  Ln e                    [ Ln e  =1, so we can ignore this  ]

Dvide both sides by -12

Ln (70/95) / -12  = k  ≈   .02545

So

We want to find this

125  =  70  + ( 165 - 70)e^(-.02545t)

55 = (95)e^(-.02545*t)

55/95  =  e^(-.02545*t)      take the Ln of both sides

Ln (55/95)  = Ln e ^(-.02545 *t)

Ln (55/95)  =  (-.02545 * t)  * Ln e

Ln (55/95) = -.02545 * t

Ln (55/95) / (-.02545)  = t  ≈ 21.475 minutes      ...... pretty  close  !!!!   Jan 13, 2020
edited by CPhill  Jan 13, 2020