Chapter 7.4, Problem 30E

To determine

To estimate the initial temperature of aluminum beam.

Answer to Problem 30E

The initial temperature of aluminum beam is $5^{\circ }F$ .

Explanation of Solution

Given information:

An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at ${65}^{\circ }$ F. After $10$ min, the beam warmed to ${35}^{\circ }$ F and after another $10$ min its temperature was ${50}^{\circ }$ F.

Concept used:

Newton’s Law of Cooling

If $T$ is the temperature of the object at time $t$ , and $T_s$ is the surrounding temperature, then

  $\frac{dT}{dt}=-k\left(T-T_s\right)$

Its solution , by the law of exponential change ,is

  $T-T_s=\left(T_0-T_s\right)e^{-kt}$ ,

Where $T_0$ is the temperature at time $t=0$ .

Calculation:

Here , $T_s={65}^{\circ }F$

At $t=10\min$ , $T={35}^{\circ }F$

According to Newton’s Law of Cooling

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 35-65=\left(T_0-65\right)e^{-10k}\\ -30=\left(T_0-65\right)e^{-10k}\end{array}$

At $t=20\min$ , $T={50}^{\circ }F$

  $\begin{array}{l}T-T_s=\left(T_0-T_s\right)e^{-kt}\\ 50-65=\left(T_0-65\right)e^{-20k}\\ -15=\left(T_0-65\right)e^{-20k}\end{array}$

Equation 1: $\left(T_0-65\right)e^{-10k}=-30$

Equation 2: $\left(T_0-65\right)e^{-20k}=-15$

Now dividing equation 1 by equation2

  $\begin{array}{l}\frac{\left(T_0-65\right)e^{-10k}=-30}{\left(T_0-65\right)e^{-20k}=-15}\\ e^{10k}=2\\ 10k=\ln 2\\ k=\frac{\ln 2}{10}\end{array}$

Now putting the value of $k$ in Equation 1

  $\begin{array}{l}\left(T_0-65\right)e^{-10k}=-30\\ \left(T_0-65\right)e^{-10\times \frac{\ln 2}{10}}=-30\\ \frac{T_0-65}{e^{\ln 2}}=-30\\ T_0-65=-30e^{\ln 2}\\ T_0=-30e^{\ln 2}+65\\ T_0=-60+65=5^{\circ }F\end{array}$

Therefore, the initial temperature of aluminum beam is $5^{\circ }F$ .