Elliot takes a roast out of the oven when the internal temperature of the roast is 165°F. After 12 minutes, the temperature of the roast drops to 140°F.
The temperature of the room is 70°F.
How long does it take for the temperature of the roast to drop to 125°F?
Use the Newton's Law of Cooling equation, T(t)=TA+(T0−TA)e^−kt
+128631 T(t) = TA + (T0 - TA)e^(-kt)
We know that after 12 minutes
140 = 70 + ( 165 - 70)e^(-k* 12) simplify
Subtract 70 from both sides
70 = ( 95)e^(-12k) divide both sides by 25
70/95 = e^(-12k) take the Ln of both sides
Ln (70/95) = Ln e^(-12k) and we can write
Ln (70/95) = (-12k) Ln e [ Ln e =1, so we can ignore this ]
Dvide both sides by -12
Ln (70/95) / -12 = k ≈ .02545
So
We want to find this
125 = 70 + ( 165 - 70)e^(-.02545t)
55 = (95)e^(-.02545*t)
55/95 = e^(-.02545*t) take the Ln of both sides
Ln (55/95) = Ln e ^(-.02545 *t)
Ln (55/95) = (-.02545 * t) * Ln e
Ln (55/95) = -.02545 * t
Ln (55/95) / (-.02545) = t ≈ 21.475 minutes ...... pretty close !!!!
