Question

Se amesteca ulei ro 1=800kg/m cub cu glicerina p2=600kg/m3 obținându-se un amestec de volum de 2 litri și densitatea p4 700kg/m3
m1=? V1=?
m2=? v2=?
​.

Answer 1

Date initiale:

$\rho_1=\rho_{ulei}=800\frac{kg}{m^3}\\\rho_2=\rho_{glicerina}=600\frac{kg}{m^3}\\V_{amestec}=2l=2 dm^3=2*10^-3m^3=0.002m^3\\\rho_4=\rho_{amestec}=700\frac{kg}{m^3}$

Definitie:

Densitatea este definita ca fiind masa supra volum.

$\rho=\frac{m}{V}= > m=\rho V \;\;(ec\;1)$

Prelucrare date:

$V_{amestec}=V_1+V_2=0.002m^3\\m_{total}=m_1+m_2=\rho_4*V_{amestec}=700*0.002=1.4kg$

Scriem  V2 si m2 in functie de V1 si m1: (*)

$V_2=0.002-V1\\m_2=1.4-m1$

Din (ec 1) =>

$m_1=\rho_1*V_1\\m_2=\rho_2*V_2$

Introducem (*) in ec de mai sus:

$m_1=\rho_1*V_1\\1.4-m_1=\rho_2*(0.002-V_1)$

Adunand ecuatiile se obtine:

$1.4=\rho_1 V_1 + \rho_2*0.002-\rho_2 V_1\\1.4-\rho_2*0.002=V_1(\rho_1 -\rho_2)\\V_1=\frac{1.4-\rho_2*0.002}{\rho_1 -\rho_2}=\frac{1.4-0.002*600}{800-200}=\frac{0.2}{200}=0.001m^3$

Astfel se obtine si restul informatiilor necesare:

$m1=\rho_1*V_1=800*0.001=0.8kg\\m2=1.4-m_1=1.4-0.8=0.6kg\\V2=0.002-V1=0.002-0.001=0.001m^3$