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 dr

Question

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?​

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Iris 1 week 2021-10-14T00:17:25+00:00 1 Answer 0

Answers ( )

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    2021-10-14T00:18:36+00:00

    Answer:

    27 mins

    Step-by-step explanation:

    Assuming the relationship between Temperature and  time follows Newton’s law of cooling:

    T (t) = T₀ + T₁ exp (rt)————————————————————– (1)

    Where  T₀ = Original temperature  of the room (°F) ; 72°F

                 T₁ = Temperature of the roast (°F); 165-72 =93°F

    Please note that we have to subtract the room to get the absolute       temperature of the roast.

                  r = Rate of cooling (°F/min)

                  t = time (min)

    T (t) = 72 + 93 exp (rt)————————————————————– (2)

    After 10 minutes, the temperature of the roast drops to 145°F.  Here t =10 mins and T(10) = 145 °F

    Substituting into (2) we have :

    T(10) = 72 +93 exp(10r)

    145 = 72 + 93 exp(10r)

    (145-72)/93 = exp (10r)

    Taking the natural logarithm of both sides, we have:

    In [(145-72)/93 ] = 10r

    -0.24214 =10r

    r  = -0.02421  °F/min

    Substituting into equation (2), the complete equation as a function of time becomes:

    T (t) = 72 + 93 exp (-0.024214t)——————————————— (3)

    To find the time taken to reach 120°F, we substitute into equation (3)

    120 = 72 +93 exp (-0.024214t)

    (120-72)/93= exp (-0.024214t)

    In[(120-72)/93] =-0.024214t

    -0.66139 =  -0.024214t

    t = -0.66139/ -0.024214

      =27.3143

     ≈ 27 mins

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