A sample of rough rice (paddy) has a bulk density of $560\text{ kg/m}^3$ and a true (particle) density of $1380\text{ kg/m}^3$. The porosity of the bulk paddy sample is ________ % (round off to two decimal places).
Answer Key:
59.42
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Void Fraction Definition**
In a unit total volume $V_{\text{total}} = 1\text{ m}^3$:
Total mass of grain: $M = \rho_b \times 1\text{ m}^3 = 560\text{ kg}$.
Volume occupied by solid grain particles:
$V_{\text{solid}} = \frac{M}{\rho_t} = \frac{560\text{ kg}}{1380\text{ kg/m}^3} \approx 0.40580\text{ m}^3$
Volume of void space:
$V_{\text{void}} = V_{\text{total}} - V_{\text{solid}} = 1.0 - 0.40580 = 0.59420\text{ m}^3$
Porosity:
$\varepsilon = \frac{V_{\text{void}}}{V_{\text{total}}} \times 100\% = 59.42\%$
**Method 2: Standard Formula**
$\varepsilon = \left(1 - \frac{\rho_b}{\rho_t}\right) \times 100\% = \left(1 - \frac{560}{1380}\right) \times 100\% = (1 - 0.405797) \times 100\% = 59.42\%$