Answer:
It will take 83.75 minutes
Explanation:
The initial temperature of the pop, T₀ = 45 degrees F
Temperature of the pop after 34 minutes, T = 60 degrees F
Time, t = 34 minutes
The temperature of the car, A = 108 degrees F
[tex]\begin{gathered} \frac{T-A}{T_0-A}=e^{-kt} \\ \\ \frac{60-108}{45-108}=e^{-34t} \\ \\ 0.76=e^{-34t} \\ \\ ln0.76=-34t \\ \\ -0.274=-34k \\ \\ k=-\frac{0.274}{-34} \\ \\ k=0.008 \end{gathered}[/tex]
If the final temperature, T = 75.75 degrees F
[tex]\begin{gathered} \frac{75.75-108}{45-108}=e^{-0.008t} \\ \\ 0.51=e^{-0.008t} \\ \\ ln0.51=-0.008t \\ \\ -0.67=-0.008t \\ \\ t=\frac{-0.67}{-0.008} \\ \\ t=83.75 \end{gathered}[/tex]
It will take 83.75 minutes
