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If \(4 \mathrm{~L}\) of a liquid with density \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) is mixed with \(6 \mathrm{~L}\) of a liquid with density \(940 \mathrm{~kg} / \mathrm{m}^{3},\) what is the density of the mixture?

Short Answer

Expert verified
The density of the mixture is \(972 \text{ kg/m}^3\).

Step by step solution

01

Convert Volume to Mass

Start by calculating the mass of each liquid. Use the formula \( \text{mass} = \text{density} \times \text{volume} \) for each liquid separately. Note that the volume should be converted from liters to cubic meters by using the factor \( 1 \text{ L} = 0.001 \text{ m}^3 \).
02

Calculate Mass of First Liquid

For the first liquid with a volume of \(4 \text{ L} = 0.004 \text{ m}^3\) and a density of \(1020 \text{ kg/m}^3\), the mass is calculated by \( \text{mass} = 1020 \times 0.004 = 4.08 \text{ kg} \).
03

Calculate Mass of Second Liquid

For the second liquid with a volume of \(6 \text{ L} = 0.006 \text{ m}^3\) and a density of \(940 \text{ kg/m}^3\), the mass is \( \text{mass} = 940 \times 0.006 = 5.64 \text{ kg} \).
04

Calculate Total Mass and Volume

Add the masses of the two liquids: \( 4.08 \text{ kg} + 5.64 \text{ kg} = 9.72 \text{ kg} \). Similarly, add the volumes: \( 0.004 \text{ m}^3 + 0.006 \text{ m}^3 = 0.01 \text{ m}^3 \).
05

Determine Density of the Mixture

The density of the mixture is found by dividing the total mass by the total volume: \( \text{density} = \frac{9.72 \text{ kg}}{0.01 \text{ m}^3} = 972 \text{ kg/m}^3 \).

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Key Concepts

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

Density Calculation
Understanding how to calculate density is a fundamental concept in fluid mechanics. Density, depicted by the Greek letter \( \rho \), is defined as the mass per unit volume of a substance. The formula for density is:
  • \( \rho = \frac{m}{V} \)
where \( m \) is the mass and \( V \) is the volume. Density helps us understand how compact or spread out the particles are in a fluid. It is essential to know the density of a fluid when you're working with different kinds of liquids as it affects the fluid's behavior.
By knowing a liquid's density, you can predict how it might react when mixed with other fluids, such as whether it will float or sink, influence pressure, or change the flow dynamics. Mastering density calculation is key to solving many practical problems in engineering and science.
Mass and Volume Conversion
To solve problems involving the mixing of different fluids, it’s crucial to manage mass and volume using consistent units. Here, normalizing units to the International System of Units (SI) is essential for keeping equations consistent and understandable.
Whenever you need to convert from liters to cubic meters, use the conversion factor:
  • \( 1 \text{ L} = 0.001 \text{ m}^3 \)
This step is important as most density values are provided in kilograms per cubic meter (\( \text{kg/m}^3 \)). Consistent units prevent errors and make complex calculations manageable.
Similarly, understanding that mass results from multiplying density and volume helps link these concepts together. Ensuring your volume is in cubic meters ensures that your final calculations for mass will be correct when dealing with density in the standard SI units.
Mixture Density
The density of a mixture of liquids is determined by calculating the cumulative mass and dividing it by the total volume. This concept shows how scientist can predict how various substances behave once combined. To find the mixture density, use:
  • Find the mass of each component liquid using \( \text{mass} = \text{density} \times \text{volume} \).
  • Add the masses to get the total mass of the mixture.
  • Sum the volumes to find the total volume.
  • Apply the formula \( \text{density}_{\text{mixture}} = \frac{\text{total mass}}{\text{total volume}} \).
This final step in determining the density of a mixture not only showcases mathematical skills but also provides insights into how different fluids interact when combined. Understanding mixture density is essential when predicting changes in properties and behaviors in real-world applications.
SI Units
The International System of Units, or SI units, is a globally recognized standard for measurements, streamlining scientific calculations everywhere. In fluid mechanics, common SI units include:
  • Mass: kilograms (\( \text{kg} \))
  • Volume: cubic meters (\( \text{m}^3 \))
  • Density: kilograms per cubic meter (\( \text{kg/m}^3 \))
Utilizing SI units ensures that calculations are comprehensible and consistent across scientific and engineering fields. They facilitate communication and understanding, minimizing confusion.
By mastering SI units, you gain the ability to seamlessly interpret physical phenomena, especially when dealing with conversions and calculations involving different properties of fluids. Thus, using SI units is not just about following a guideline but making the math more straightforward and the results universally understood.

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Most popular questions from this chapter

As the pressure in a liquid reservoir is increased from \(150 \mathrm{kPa}\) to \(28000 \mathrm{kPa}\), the volume of liquid in the reservoir decreases from \(1.500 \mathrm{~m}^{3}\) to \(1.450 \mathrm{~m}^{3}\). Estimate the bulk modulus of the liquid.

Give the dimensions and typical SI units of the following quantities commonly used in engineering: energy, force, heat, moment, momentum, power, pressure, strain, stress, and work.

A helium balloon is to be used to lift a lab rat that weighs \(1.5 \mathrm{~N}\). The density of the air is \(1.17 \mathrm{~kg} / \mathrm{m}^{3}\), atmospheric pressure is \(100 \mathrm{kPa}\), the air temperature is \(25^{\circ} \mathrm{C},\) and the balloon weighs \(0.5 \mathrm{~N}\). What mass of helium (in kilograms) must be put in the balloon to lift the rat?

A storage reservoir contains \(250 \mathrm{~kg}\) of a liquid that has a specific gravity of \(2 .\) What is the volume of the storage reservoir?

A force balance in a particular fluid flow is combined with Newton's second law to yield the equation $$ \rho \frac{\mathrm{d}^{2} z}{\mathrm{~d} t^{2}}+a \frac{\mathrm{d} z}{\mathrm{~d} t}+b z=c $$ where \(\rho, z,\) and \(t\) are dimensional variables with the following dimensions: \(\rho\left[\mathrm{ML}^{-3}\right],\) \(z[\mathrm{~L}],\) and \(t[\mathrm{~T}]\). (a) Determine the dimensions of the system parameters \(a, b,\) and \(c .\) (b) If standard SI units are to be used in the given equation and values of \(\rho, z,\) and \(t\) are provided in \(\mathrm{g} / \mathrm{cm}^{3}, \mathrm{~mm},\) and \(\mathrm{h}\), respectively, what conversion factors must be applied to these variables before they are used in the equation?

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