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 .
Enter your answer in the box. Round your answer to the nearest minute.
im not sure what to do, im very confused/:
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