Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 72


$5.43$ $396.721^{\circ}C$

Work Step by Step

Step 1. Given the function of the specific heat as $C_ v=8.27+10^{-5}(26T-1.87T^2)$ and the temperature range of $20^{\circ}C\leq T\leq 675^{\circ}C$, we can calculate the average value over this range as $\bar C_v=\frac{1}{675-20}\int_{20}^{675}C_v\ dT$ Step 2. Evaluating the above integral, we have $\bar C_v=\frac{1}{655}\int_{20}^{675}(8.27+10^{-5}(26T-1.87T^2)) dT=\frac{1}{655}(8.27T+10^{-5}(13T^2-\frac{1.87}{3}T^3))|_{20}^{675}=\frac{1}{655}[(8.27(675)+10^{-5}(13(675)^2-\frac{1.87}{3}(675)^3))-(8.27(20)+10^{-5}(13(20)^2-\frac{1.87}{3}(20)^3))]\approx\frac{1}{655}[3724.44-165.40]\approx5.43\ J/mol/ ^{\circ}C$ Step 3. To find the temperature when the above $\bar C_v$ happens, we need to solve the equation $C_ v=8.27+10^{-5}(26T-1.87T^2)=5.43$ in the interval given in step 1. We can solve it either algebraically or graphically and we can find a solution $T=396.721^{\circ}C$ as shown in the figure.
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