Chapter 7, Problem 52RE

To determine

To Find: How long did it take the pie to cool from there to ${70}^{\circ }F$

Answer to Problem 52RE

  $92.24\min$ did it take the pie to cool from there to ${70}^{\circ }F$

Explanation of Solution

Given information: A deep-dish apple pie, whose internal temperature was ${220}^{\circ }F$ when removed from the oven, was set out on a ${40}^{\circ }F$ breezy porch to cool. Fifteen minutes later, the pie’s internal temperature was ${180}^{\circ }F$

Formula used:

Newton’s law of cooling

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$

  $T$ is the temperature at time t

  $T_s$ is the surrounding temperature

  $T_0$ is the initial temperature

Calculation:

Newton’s law of cooling

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$

  $T$ is the temperature at time t

  $T_s$ is the surrounding temperature

  $T_0$ is the initial temperature

It is given that $T_0={220}^{\circ },T_s={40}^{\circ },T={180}^{\circ }$ when $t=15\min$

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ T_0={220}^{\circ },T_s={40}^{\circ },T={180}^{\circ }\\ t=15\min \\ {180}^{\circ }-{40}^{\circ }=\left({220}^{\circ }-{40}^{\circ }\right)e^{-15k}\\ {140}^{\circ }={180}^{\circ }{e}^{-15k}\\ k=\frac{\ln \frac{7}{9}}{-15}\\ k\approx 0.0167\\ \text{use the formula again}\\ T-T_s=\left(T_0-T_s\right)e^{-kt}\\ T_0={180}^{\circ },T_s={40}^{\circ },T={70}^{\circ }\\ {70}^{\circ }-{40}^{\circ }=\left({180}^{\circ }-{40}^{\circ }\right)e^{-kt}\\ {30}^{\circ }={140}^{\circ }{e}^{-kt}\\ t=\frac{\ln \frac{3}{14}}{-0.0167}\\ t=92.24\min \end{array}$