Chapter 5.8, Problem 16AYU

To determine

To find: Warming Time of a Beer Stein A beer stein has a temperature of ${28}^{\circ }F$. It is placed in a room with a constant temperature of ${70}^{\circ }F$. After 10 minutes, the temperature of the stein has risen to ${35}^{\circ }F$. What will the temperature of the stein be after 30 minutes? How long will it take the stein to reach a temperature of ${45}^{\circ }F$?

Answer to Problem 16AYU

$u\left(t\right)={47.49}^{\circ }F$

$t=4.22\text{minutes}$

Explanation of Solution

Given:

A beer stein has a temperature of ${28}^{\circ }F$. It is placed in a room with a constant temperature of ${70}^{\circ }F$. After 10 minutes, the temperature of the stein has risen to ${35}^{\circ }F$.

Calculation:

Let us consider the temperature be $u_0$ and the temperature after time $t$ will be

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$ where $T$ is the surrounding temperature.

Assuming initial temperature is ${28}^{\circ }F$ in the surrounding temperature of ${70}^{\circ }F$.

Assuming final temperature is ${35}^{\circ }F$ after 10 minutes. Then

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$35=70+\left(28-70\right)e^{k10}$

$-35=-42e^{k10}$

$\frac{35}{42}=e^{k10}$

$0.83=e^{k10}$

Taking log on both sides we get

$\text{ln}0.83=\text{ln}{e}^{k10}$

$\text{ln}0.83=k10$

$k=\frac{\text{ln}0.83}{10}=-0.0182$

Then after 5 minutes temperature will be

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$u\left(t\right)=70-42e^{-0.0182t}$

$u\left(t\right)={45.67}^{\circ }F$

Then after 10 minutes the temperature will be

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$u\left(t\right)=70-27e^{-0.0182\left(10\right)}$

$u\left(t\right)={47.49}^{\circ }F$

Let us consider the final temperature to be ${45}^{\circ }F$.

Then

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$45=70-27e^{-0.0182t}$

$t=4.22\text{minutes}$