Question #256908

Two liquids with different densities (1500 kg/mand 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.


1
Expert's answer
2021-10-27T08:50:39-0400

Solution;

Given;

ρ1=1500kg/m3\rho_1=1500kg/m^3

ρ2=500kg/m3\rho_2=500kg/m^3

ρ12=800kg/m3\rho_{12}=800kg/m^3

g=9.375m/s2g=9.375m/s^2

v12=100litres=0.1m3v_{12}=100litres=0.1m^3

Let volume of the liquids poured be v1v_1 and v2v_2

So the masses of the individual masses is;

m1=ρ1v1=1500v1m_1=\rho_1v_1=1500v_1

m2=ρ2v2=500v2m_2=\rho_2v_2=500v_2

The density of the mixture is;

ρ12=m1+m2v\rho_{12}=\frac{m_1+m_2}{v}

By substitution;

800=1500v1+500v20.1800=\frac{1500v_1+500v_2}{0.1}

Rewrite v2v_2 in terms of v1v_1

800=1500v1+500(0.1v1)0.1800=\frac{1500v_1+500(0.1-v_1)}{0.1}

80=1500v1+50500v180=1500v_1+50-500v_1

30=1000v130=1000v_1

v1=0.03m3v_1=0.03m^3

v2=0.10.03=0.07m3v_2=0.1-0.03=0.07m^3

The masses of individual liquids will be;

m1=1500×0.03=45kgm_1=1500×0.03=45kg

m2=500×0.07=35kgm_2=500×0.07=35kg

Total mass of the mixture is;

m12=m1+m2=80kgm_{12}=m_1+m_2=80kg

The weight of the mixture;

W=m12g=80×9.375=750NW=m_{12}g=80×9.375=750N



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