2. The average value of the continuous temperature function T(t) on the interval t = [to, tƒ] is given byrt f(1)A metal bar is being cooled and its average temperature over a two-hour period is sought. At 30minute intervals, from to= 0 until tf = 120 minutes, the following temperatures are observed:350°C, 154°C, 87°C, 80°C, 64°C.T=1tf - totoT(t) dt.(i) Use the trapezoidal rule to estimate the average temperature of the bar.(ii) Calculate the average of the five temperatures, then give an explanation for the difference be-tween the two averages.

(i) Trapezoidal rule∫abf(x)dx≈h2[f(a)+2∑i=1n−1f(a+ih)+f(b)]Where h=b−an is called step length and n…

Answered: 2. The average value of the continuous…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

2. The average value of the continuous temperature function T(t) on the interval t = [to, tƒ] is given by
rt f
(1)
A metal bar is being cooled and its average temperature over a two-hour period is sought. At 30
minute intervals, from to= 0 until tf = 120 minutes, the following temperatures are observed:
350°C, 154°C, 87°C, 80°C, 64°C.
T
=
1
tf - to
to
T(t) dt.
(i) Use the trapezoidal rule to estimate the average temperature of the bar.
(ii) Calculate the average of the five temperatures, then give an explanation for the difference be-
tween the two averages.

Expert Solution

Step 1: Evaluating integral using trapezoidal rule

(i) Trapezoidal rule

$\int_{a}^{b} f (x) d x \approx \frac{h}{2} [f (a) + 2 \sum_{i = 1}^{n - 1} f (a + i h) + f (b)]$

Where $h = \frac{b - a}{n}$ is called step length and n is the number of subintervals.

Given that

$t_{0} = 0, t_{f} = 120, h = 30$

$T (0) = 350, T (30) = 154, T (60) = 87, T (90) = 80, T (120) = 64$

Therefore

$\begin{aligned} \int_{t_{0}}^{t_{f}} T (t) d t & = \frac{30}{2} [T (0) + 2 T (30) + 2 T (60) + 2 T (90) + T (120) \\ = 15 (350 + 308 + 174 + 160 + 64) \\ = 15840 \end{aligned}$

Hence average temperature $\frac{15840}{120} = 132$

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