Answer to Question #256908 in Molecular Physics | Thermodynamics for jay

Question

Two liquids with different densities (1500 kg/m³ and 500 kg/m³) were combined and poured into a 100-liter container to fill it. if the mixture's final density is 800kg/m³. Determine the weight of the mixture and the amounts of liquids used. 9.375 m/s² for local gravity.

Answer 1

Solution;

Given;

$\rho_1=1500kg/m^3$

$\rho_2=500kg/m^3$

$\rho_{12}=800kg/m^3$

$g=9.375m/s^2$

$v_{12}=100litres=0.1m^3$

Let volume of the liquids poured be $v_1$ and $v_2$

So the masses of the individual masses is;

$m_1=\rho_1v_1=1500v_1$

$m_2=\rho_2v_2=500v_2$

The density of the mixture is;

$\rho_{12}=\frac{m_1+m_2}{v}$

By substitution;

$800=\frac{1500v_1+500v_2}{0.1}$

Rewrite $v_2$ in terms of $v_1$

$800=\frac{1500v_1+500(0.1-v_1)}{0.1}$

$80=1500v_1+50-500v_1$

$30=1000v_1$

$v_1=0.03m^3$

$v_2=0.1-0.03=0.07m^3$

The masses of individual liquids will be;

$m_1=1500×0.03=45kg$

$m_2=500×0.07=35kg$

Total mass of the mixture is;

$m_{12}=m_1+m_2=80kg$

The weight of the mixture;

$W=m_{12}g=80×9.375=750N$