A 1-m^3 container is filled with 400 kg of granite stone, 200 kg of dry sand, and 0.2 m^3 of

A 1-m^3 container is filled with 400 kg of granite stone,...

A $1-\mathrm{m}^{3}$ container is filled with $400 \mathrm{~kg}$ of granite stone, $200 \mathrm{~kg}$ of dry sand, and $0.2 \mathrm{~m}^{3}$ of liquid $25^{\circ} \mathrm{C}$ water. Using properties from Tables $\mathrm{A} .3$ and A. 4 , find the average specific volume and density of the masses when you exclude air mass and volume.

A $1-\mathrm{m}^{3}$ container is filled with $400 \mathrm{~kg}$ of granite stone, $200 \mathrm{~kg}$ of dry sand, and $0.2 \mathrm{~m}^{3}$ of liquid $25^{\circ} \mathrm{C}$ water. Using properties from Tables $\mathrm{A} .3$ and
A. 4 , find the average specific volume and density of the masses when you exclude air mass and volume.

Key Concepts

Specific Volume

This is a fundamental property defined as the volume per unit mass of a substance. It is calculated by dividing the total volume by the total mass, which provides insight into how much space is occupied by a given mass. In multiphase systems, the average specific volume is determined by considering the volumetric contributions of all phases involved.

Density

Density is defined as the mass per unit volume and is a critical property for characterizing substances. It is essentially the inverse of specific volume and indicates how compact or concentrated a material is. In a mixture, the overall density is obtained by dividing the combined mass of all constituents by the combined volume contributing to the mass.

Weighted Averaging in Multiphase Mixtures

When dealing with mixtures composed of different materials or phases, each with its own mass and volume, the overall properties such as specific volume and density are computed using weighted averages. This involves summing the masses and volumes of the individual components (excluding any negligible phases) to accurately reflect the combined characteristics of the system.

Exclusion of Non-Contributing Phases

In certain calculations, components that do not significantly add to the mass or volume—such as the small amount of air in some contexts—are excluded to simplify the analysis. This ensures that the calculated properties, like average specific volume and density, accurately represent only the substances of interest.

Transcript

00:01 So let's first of all see what is given in the problem, including the properties from the table a .3 and a .4, right? so we have the mass of the grain that is equal to 400 kilogram, right? and the mass of the sand that is equal to 200 kilograms, right? then we have the volume of water that is equal to 0 .2 meter cube.

00:26 Right.

00:27 And then we have obtained the value of the value of the water.

00:30 The density of bernight that is equal to 2 ,750 kilogram per meter cube, right? and then we have the density of sand that is equal to 1 ,500 kilogram per meter cube.

00:44 And then we have the density of water that is equal to 997 kilogram per meter cube, right? so in order to find out the specific volume and the other quantity, what we need to do first of all is that we need to, we need to to calculate the total volume, right, and total mass.

01:09 So for that, we need the total mass of the water first, right? so the mass of the water is simply determined from the expression of the density of water divided by the volume of water.

01:21 We have the values we just need to plug them in and find out that the mass of the water is 199 .4 kilogram.

01:28 Right now that we have the mass of the water, we can calculate the total mass.

01:31 So that is m sub t and is equal to the sum of the mass of the water plus mass of the granite plus mass of the sand again, we have all these three parameters values we just need to plug them in and calculated and we guess equal to 799 4 kilograms right now that we have the total mass we need to calculate the total volume so for that we need the volume of grainite right and it can be easily calculated from the expression of the mass of the the granite divided by the density of the granite.

02:04 Again, you have the values.

02:06 You just need to plug them in and calculated.

02:08 And we get that's equal to 0 .145 meter cube...

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