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# Maurice takes a roast out of the oven

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Maurice takes a roast out of the oven when the internal temperature of the roast is 165°F. After 10 minutes, the temperature of the roast drops to 145°F.

The temperature of the room is 72°F.

How long does it take for the temperature of the roast to drop to 120°F?

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

im not sure what to do, im very confused/:

Dec 5, 2019

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T(t)  =  TA  +  ( T0 -  TA)e^(kt)

TA  = room temp

T0  =  original temp of  roast

We need to find  k  , thusly.....we know that   that when t  = 10...we have.....

145  = 72  + ( 165 - 72)e^(k*10)       subtract 72 from both sides

73  =  (93) e^(k*10)      divide both sides by  93

(73/93)  = e^(k *10)       take the natural log   of each side

Ln  ( 73/93)  = Ln e ^(k *10)        and by a Ln property we can write

Ln (73/93)  = (k *10) Ln e          Ln e  = 1....so.....we can ignore this and we have

Ln (73/93)  = k * 10                 divide both sides by 10

Ln (73/93) / 10  = k   ≈   -0.0242

So....to find the time  it takes to  cool to  120°  we have

120  = 72 +  (93)e^(-0.0242 * t)       subtract  72 from both sides

48 =  93e^(-0.0242* t)     divide both sides by  93

(48/93)  =  e^(-0.0242 * t)     take the Ln of both sides

Ln  (48/93)  = Ln e^ (-0.0242 * t)

Ln ( 48/93)  =  (-0.0242 * t) * Ln e         (ignore the  Ln e)

Ln (48/ 93)  = -0.0242 * t       divide both sides by  -0.0242

Ln (48 / 93) / [ -0.0242 ]  = t  ≈  27.33  minutes =   27 minutes   Dec 5, 2019
edited by CPhill  Dec 5, 2019