Maurice takes a roast out of the oven when the internal temperature of the roast is 165F after ten minutes , the temperature roast drops to 145

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Complete 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?

27.3143 minutes

Step-by-step explanation:

Using the Newton's law of cooling: The formula is given as:

T (t) = T₀ + T₁ e^ (rt)

Where

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

T₁ = Temperature of the roast (°F)

This is calculated as:

165 - 72 =93°F

r = Rate of cooling (°F/min)

t = time (min)

Substituting the values, we can obtain:

T (t) = 72 + 93 e ^rt

Step 1

We solve for r

From the question, we know that: the temperature dropped to 145°F after 10 minutes,

t =10 mins and T(10) = 145 °F

Substituting

T(10) = 72 +93 e^10r

145 = 72 + 93 e^10r

145 - 72 = 93 e^10r

Divide both sides by 93

(145-72)/93 = e^(10r)

We take the natural logarithm of both the left and right hand sides

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

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

-0.24214 =10r

r = -0.24214/10

r = -0.02421 °F/min

Substituting into

-0.02421 °F/min for r in the Equation below

T (t) = 72 + 93 e ^rt

T (t) = 72 + 93 e^ (-0.024214t)

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

120 = 72 +93 e^(-0.024214 ×t)

(120-72)= 93 e^ (-0.024214t)

Divide bother sides by 93

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

We take the natural logarithm of both sides

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

-0.66139 = -0.024214t

Divide both sides by -0.024214t

t = -0.66139/ -0.024214

t =27.3143 minutes