Answer
$p = 7.685 \frac{g}{cm^{3}}$
Work Step by Step
Given:
Iron with room temperature density of $7.870 g/cm^{3}$
volume coefficient $(α_{v})= 3 α_{l}$
Required:
density $p$ of iron at 700°C
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 7.870 g, and the mass will remain constant upon heating to 700°C. Also, $α_{v} = 3α_{l}$, and from Table 19.1: $α_{l} = 11.8 \times 10^{-6} (°C^{-1}) $
Substituting:
$V = V_{0} + ΔV = V_{0} + (α_{v}ΔT{V_{0}}) = V_{0}(1 + 3α_{l}ΔT)= (1 cm^{3})(1 + (3)[11.8 \times 10^{-6} (°C^{-1})] (700°C - 20°C) = 1.02407 cm^{3}$
$p = \frac{7.870 g}{1.02407 cm^{3} } = 7.685 \frac{g}{cm^{3}}$
