Example 2: a horizontal cylindrical drum is 2.00 m in diameter and 4.00 m in length. the drum is partially filled with benzene (density = 0.879 g/cm3). what is the mass (kg) of benzene when the liquid depth is 0.85 m?

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

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

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