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?

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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