Respuesta :
To find the mass of benzene, we must first solve for the volume of benzene inside the cylindrical drum. To calculate the volume of the partially filled cylindrical drum, refer to the working formula below:
Volume = L {(R^2)cos-1[(R-H)/R)] - (R-H)(2RH-H^2)^0.5}
Where:
L = length of the drum = 2 meters
R = radius of the drum = 4 meters
H = height of the liquid = 0.85 meter
substituting the given to the formula, we get
Volume = 5.087 m3
To solve for the mass of benzene, we must multiply the volume with the density.
Density = 0.879 g/cm3 or 879 kg/m3
Mass of benzene = Volume x Density
Mass of benzene = 5.087 x 879 = 4,471.5 kg
ANSWER: Mass of benzene = 4,471.5 kg
Volume = L {(R^2)cos-1[(R-H)/R)] - (R-H)(2RH-H^2)^0.5}
Where:
L = length of the drum = 2 meters
R = radius of the drum = 4 meters
H = height of the liquid = 0.85 meter
substituting the given to the formula, we get
Volume = 5.087 m3
To solve for the mass of benzene, we must multiply the volume with the density.
Density = 0.879 g/cm3 or 879 kg/m3
Mass of benzene = Volume x Density
Mass of benzene = 5.087 x 879 = 4,471.5 kg
ANSWER: Mass of benzene = 4,471.5 kg
The mass (kg) of benzene is about 4470 kg
Further explanation
This problem is about Density.
Density is the ratio of mass to the volume of the object.
[tex]\large {\boxed {\rho = \frac{ m }{ V } } }[/tex]
ρ = density of object ( kg / m³ )
m = mass of object ( kg )
V = volume of object ( m³ )
Given:
Diameter of Cylinder = d = 2.00 m
Radius of Cylinder = R = d/2 = 2.00/2 = 1.00 m
Length of Cylinder = L = 4.00 m
Liquid Depth = H = 0.85 m
Density of Benzene = ρ = 0.879 g/cm³ = 879 kg/m³
Unknown:
mass of benzene = m = ?
Solution:
This problem is about Liquid Volume in Partially Filled Horizontal Tanks
Firstly we will calculate the volume of Benzene by using following formula:
[tex]V = A \times L[/tex]
[tex]V = ( \texttt{Area of Sector - Area of Triangle} ) \times L[/tex]
[tex]V = [ R^2 \cos^{-1}(\frac{R - H}{R}) - (R - H)\sqrt {(2RH - H^2)} ] L[/tex]
[tex]V = [ 1^2 \cos^{-1}(\frac{1 - 0.85}{1}) - (1 - 0.85)\sqrt {(2(1)(0.85) - 0.85^2)} ] 4[/tex]
[tex]V = [ \cos^{-1} (0.15) - 0.15 \sqrt{ 0.9775} ] 4[/tex]
[tex]V \approx \boxed {5.0877 ~ m^3}[/tex]
[tex]m = \rho \times V[/tex]
[tex]m = 879 \times 5.0877[/tex]
[tex]m \approx \boxed {4470 ~ kg}[/tex]
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Answer details
Grade: High School
Subject: Mathematics
Chapter: Density
Keywords: Temperature , Density , Iron , Sphere , Volume , Mass