Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 19 - Thermal Properties - Questions and Problems - Page 800: 19.11a

Answer

$p = 18.699 \frac{g}{cm^{3}}$

Work Step by Step

Given: Gold at room temperature (20°C) density is $19.320 g/cm^{3}$ volume coefficient $(α_{v})= 3 α_{l}$ Required: density $p$ of gold at 800°C if only thermal expansion is considered Solution: Using Equation 19.4 and rearranging it: $ΔV= α_{v}ΔT{V_{0}}$ Using the basis of 1 $cm^{3}$ of material at 20°C, mass is 19.320 g, and the mass will remain constant upon heating to 800°C. Also, $α_{v} = 3α_{l}$, and from Table 19.1, $α_{l} = 14.2 \times 10^{-6} (°C^{-1}) $ Substituting, $V = V_{0} + ΔV = V_{0} + (α_{v}ΔT{V_{0}}) = V_{0}(1 + 3α_{l}ΔT)= (1 cm^{3})(1 + (3)[14.2 \times 10^{-6} (°C^{-1})] (800°C - 20°C) = 1.03323 cm^{3}$ $p = \frac{19.320 g}{1.03323 cm^{3} } = 18.699 \frac{g}{cm^{3}}$
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