Certainly! To solve this problem, we will use Newton's Law of Heating, which is given by:
[tex]\[
T = T_a + (T_0 - T_a) e^{-kt}
\][/tex]
where
- [tex]\( T \)[/tex] is the temperature of the object after time [tex]\( t \)[/tex]
- [tex]\( T_a \)[/tex] is the surrounding temperature
- [tex]\( T_0 \)[/tex] is the initial temperature of the object
- [tex]\( k \)[/tex] is the decay constant
- [tex]\( t \)[/tex] is the time in minutes
We are given the following values:
- [tex]\( T_0 = 38^\circ F \)[/tex]
- [tex]\( T_a = 72^\circ F \)[/tex]
- After 15 minutes ([tex]\( t = 15 \)[/tex]), the temperature of the turkey is [tex]\( 47^\circ F \)[/tex]
### Step 1: Determining the decay constant [tex]\( k \)[/tex]
To find [tex]\( k \)[/tex], we first substitute the known values into the formula and solve for [tex]\( k \)[/tex].
[tex]\[
47 = 72 + (38 - 72) e^{-k \cdot 15}
\][/tex]
Simplify the equation:
[tex]\[
47 = 72 - 34 e^{-15k}
\][/tex]
Rearrange to isolate the term with [tex]\( e \)[/tex]:
[tex]\[
47 - 72 = -34 e^{-15k}
\][/tex]
[tex]\[
-25 = -34 e^{-15k}
\][/tex]
[tex]\[
\frac{25}{34} = e^{-15k}
\][/tex]
Take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[
\ln\left(\frac{25}{34}\right) = -15k
\][/tex]
[tex]\[
k = -\frac{\ln\left(\frac{25}{34}\right)}{15}
\][/tex]
Calculating the numerical value gives us approximately:
[tex]\[
k \approx 0.02
\][/tex]
### Step 2: Determining the temperature after 120 minutes
Now we use the decay constant [tex]\( k \approx 0.02 \)[/tex] to find the temperature [tex]\( T \)[/tex] after 120 minutes.
[tex]\[
T = T_a + (T_0 - T_a) e^{-kt}
\][/tex]
Substitute the values into the formula:
[tex]\[
T = 72 + (38 - 72) e^{-0.02 \cdot 120}
\][/tex]
Simplify the expression inside the exponential function:
[tex]\[
T = 72 + (38 - 72) e^{-2.4}
\][/tex]
Calculate the exponential term:
[tex]\[
e^{-2.4} \approx 0.0907
\][/tex]
Now compute the temperature:
[tex]\[
T \approx 72 + (38 - 72) \cdot 0.0907
\][/tex]
[tex]\[
T \approx 72 + (-34) \cdot 0.0907
\][/tex]
[tex]\[
T \approx 72 - 3.08
\][/tex]
[tex]\[
T \approx 69^\circ F
\][/tex]
### Final Answer
1. The decay constant [tex]\( k \)[/tex] is approximately [tex]\( 0.02 \)[/tex] to the nearest thousandth.
2. The temperature of the turkey after 120 minutes is approximately [tex]\( 69^\circ F \)[/tex] to the nearest degree.
