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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?
Answer:
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
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