Question

A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume

Answer 1

Complete question:

A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.

Answer:

The mass of the air is 5.32 kg

The specific volume is 5.52 x 10⁻³ m³/kg

Explanation:

Given;

total volume of the container, $V_t$ = 5 m³

mass of granite, $m_g$ = 900 kg

density of granite, $\rho _g$ = 2,400 kg/m³

density of air, $\rho_a$ = 1.15 kg/m³

The volume of the granite is calculated as;

$V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3$

The volume of air is calculated as;

$V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3$

The mass of the air is calculated as;

$m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg$

The specific volume is calculated as;

$V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg$