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Chapter 11: Problem 2

A cylindrical storage tank has a radius of 1.22 m. When filled to a height of 3.71 m, it holds 14 300 kg of a liquid industrial solvent. What is the density of the solvent?

Short Answer

Expert verified
The density of the solvent is approximately 827.67 kg/m鲁.

Step by step solution

01

Understand the Problem

The problem involves a cylindrical tank filled with a liquid, and we need to find the density of the liquid. We know the radius of the tank (1.22 m), the height the liquid fills (3.71 m), and the mass of the liquid (14,300 kg).
02

Calculate the Volume of the Cylinder

The volume of a cylinder is calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Here, \( r = 1.22 \) m and \( h = 3.71 \) m. Therefore, \[ V = \pi (1.22)^2 (3.71) \].
03

Perform the Volume Calculation

Using the values \( r = 1.22 \) m and \( h = 3.71 \) m, the volume is calculated as:\[ V = \pi (1.22)^2 (3.71) \approx 17.27 \text{ cubic meters} \].
04

Density Formula

The density \( \rho \) is the mass divided by the volume of the substance. The formula is \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume. We know \( m = 14300 \) kg and from Step 3, \( V \approx 17.27 \) m鲁.
05

Calculate the Density

Substitute the known values into the density formula: \[ \rho = \frac{14300}{17.27} \approx 827.67 \text{ kg/m}^3 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Tank
A cylindrical tank is a container with a circular base and straight sides, much like a giant can. These tanks are often used to store liquids because their shape allows for efficient use of space and structural integrity.

Key characteristics of a cylindrical tank include:
  • The radius, which is the distance from the center of the circular base to the outer edge.
  • The height, which is the distance from the base to the top of the cylinder.
  • The volume, which is the amount of space inside the tank.
In our exercise, the cylindrical tank had a radius of 1.22 meters and a height filled to 3.71 meters. These dimensions are critical for calculating the volume of the tank, which leads us to our next key concept.
Volume Calculation
Calculating the volume of a cylinder is straightforward if you remember the formula: \[ V = \pi r^2 h \] Where:
  • \( V \) = Volume of the cylinder
  • \( r \) = Radius of the cylinder鈥檚 base
  • \( h \) = Height of the cylinder
  • \( \pi \) = Math constant approximately equal to 3.14159
For our cylindrical tank with a radius of 1.22 meters and a height of 3.71 meters, the volume calculation involves plugging these values into our formula: \[ V = \pi (1.22)^2 (3.71) \] After carrying out the multiplication and considering \( \pi \), we find that the tank's volume is approximately 17.27 cubic meters. Understanding this volume is essential as it helps us proceed to calculating the density of the liquid, which relates the mass to this calculated volume.
Mass and Volume Relationship
The relationship between mass and volume is fundamentally explained by a substance's density. Density provides a way to compare how much matter is in a given space. Mathematically, density \( \rho \) is expressed as: \[ \rho = \frac{m}{V} \] Where:
  • \( \rho \) = Density
  • \( m \) = Mass
  • \( V \) = Volume
In our scenario, the industrial solvent filling the cylindrical tank has a mass of 14,300 kg. We calculated its volume to be approximately 17.27 m鲁. Substituting these values into the density formula gives: \[ \rho = \frac{14300}{17.27} \approx 827.67 \text{ kg/m}^3 \] Knowing the density of a substance is useful in many real-world applications, such as designing storage tanks, ensuring structural integrity, and fulfilling industrial requirements.

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Most popular questions from this chapter

A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using \(1.013 \times 10^{5}\) Pa for the atmospheric pressure and 1200 \(\mathrm{kg} / \mathrm{m}^{3}\) for the density of the sauce, find the absolute pressure in the bulb when the distance \(h\) is \(\quad\) (a) 0.15 \(\mathrm{m}\) and (b) 0.10 \(\mathrm{m} .\)

If a scuba diver descends too quickly into the sea, the internal pressure on each eardrum remains at atmospheric pressure, while the external pressure increases due to the increased water depth. At sufficient depths, the difference between the external and internal pressures can rupture an eardrum. Eardrums can rupture when the pressure difference is as little as 35 kPa. What is the depth at which this pressure difference could occur? The density of seawater is 1025 \(\mathrm{kg} / \mathrm{m}^{3}\)

A pipe is horizontal and carries oil that has a viscosity of 0.14 \(\mathrm{Pa} \cdot \mathrm{s}\) The volume flow rate of the oil is \(5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\) . The length of the pipe is \(37 \mathrm{m},\) and its radius is 0.60 \(\mathrm{cm}\) . At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

A cylindrical air duct in an air conditioning system has a length of 5.5 \(\mathrm{m}\) and a radius of \(7.2 \times 10^{-2} \mathrm{m} . \mathrm{A}\) fan forces air \(\left(\eta=1.8 \times 10^{-5} \mathrm{Pa} \cdot \mathrm{s}\right)\) through the duct, so that the air in a room (volume \(=280 \mathrm{m}^{3} )\) is replenished every ten minutes. Determine the difference in pressure between the ends of the air duct.

A fountain sends a stream of water straight up into the air to a maximum height of 5.00 \(\mathrm{m}\) . The effective cross-sectional area of the pipe feeding the fountain is \(5.00 \times 10^{-4} \mathrm{m}^{2}\) . Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (Note: 1 gal \(=3.79 \times 10^{-3} \mathrm{m}^{3} . )\)
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