Question

mr. oniskwski sure did love his turkey on thanksgiving. the problem was that when it came out of the oven it was too hot to eat. it is recommended the turkey be 180 degrees to be fully cooked. he likes to eat his turkey at a temperature of 100 degrees. he also keeps the heat on in his house to 65 degrees. if the decay constant is 0.125, how long does he have to wait to eat his turkey? round to the nearest hundredth.

Explanation:

Step1: Identify the cooling model

This is a Newton's Law of Cooling problem. The formula is \( T(t) = T_s + (T_0 - T_s)e^{-kt} \), where \( T(t) \) is the temperature at time \( t \), \( T_s \) is the surrounding temperature, \( T_0 \) is the initial temperature, and \( k \) is the decay constant.
Here, \( T_s = 65 \), \( T_0 = 180 \), \( T(t) = 100 \), \( k = 0.125 \). We need to solve for \( t \).

Step2: Substitute values into the formula

Substitute the known values into the formula:
\( 100 = 65 + (180 - 65)e^{-0.125t} \)

Step3: Simplify the equation

First, simplify the right - hand side:
\( 100-65=(180 - 65)e^{-0.125t} \)
\( 35 = 115e^{-0.125t} \)

Step4: Solve for the exponential term

Divide both sides by 115:
\( \frac{35}{115}=e^{-0.125t} \)
Simplify \( \frac{35}{115}=\frac{7}{23}\approx0.3043 \)
So, \( 0.3043 = e^{-0.125t} \)

Step5: Take the natural logarithm of both sides

Take the natural logarithm of both sides:
\( \ln(0.3043)=\ln(e^{-0.125t}) \)
Since \( \ln(e^x)=x \), we have:
\( \ln(0.3043)=- 0.125t \)

Step6: Solve for t

We know that \( \ln(0.3043)\approx - 1.209 \)
Then, \( t=\frac{-\ln(0.3043)}{0.125}=\frac{1.209}{0.125} = 9.672 \)
Rounding to the nearest hundredth, \( t\approx9.67 \)

Answer:

\( 9.67 \)