How Much Sand Equals the Surface Area of a Cube?

[benEE2018]

Homework Statement

Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.5 × 103 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 0.9 m on an edge?

Homework Equations

density=mass/volume
surface area of sphere=4(pi)r^2
volume of sphere= (4/3)(pi)r^3

The Attempt at a Solution

My attempt at the solution was i plugged 0.9 into the surface area equation and solved for the value r which gives the surface area equaled to 0.9. I then used the r that i calculated for and plugged that into the volume equation to get the value of V to then plug that into the density equation to find the mass using the given density of the sand and the volume from what i calculated. I do not know where the 0.5μm and i think that is the problem i am having. FInal answer i got was 200kg?

 

[Simon Bridge]

i plugged 0.9 into the surface area equation and solved for the value r

The value 0.9(m) is the length of the edges of a cube - there is no "r" in the cube equation.

The individual grains are spheres.
Note: best practice is to work out the equations symbolically first - then put the numbers in.

 

[SteamKing]

Homework Statement

Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.5 × 103 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 0.9 m on an edge?

Homework Equations

density=mass/volume
surface area of sphere=4(pi)r^2
volume of sphere= (4/3)(pi)r^3

The Attempt at a Solution

My attempt at the solution was i plugged 0.9 into the surface area equation and solved for the value r which gives the surface area equaled to 0.9. I then used the r that i calculated for and plugged that into the volume equation to get the value of V to then plug that into the density equation to find the mass using the given density of the sand and the volume from what i calculated. I do not know where the 0.5μm and i think that is the problem i am having. FInal answer i got was 200kg?

The most common stumbling block students have is reading and understanding what the problem says.

My attempt at the solution was i plugged 0.9 into the surface area equation and solved for the value r which gives the surface area equaled to 0.9.

Why did you do this? This is what the problem states:

What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 0.9 m on an edge?

0.9 is not the surface area, and you do not list the formula for finding the surface area of a cube in the 'Relevant Equations'. Hint: areas are measured in square meters, not meters. The second most common mistake is that students overlook/do not understand the role units play in understanding physical quantities.

I do not know where the 0.5μm and i think that is the problem i am having.

This is where the 0.5μm comes in handy:

Grains of fine California beach sand are approximately spheres with an average radius of 50 μm ...

 

[benEE2018]

the equation for surface area of a cube is 6a^2 but i do not know how that plays into this because of the variable a unless r and a are interchangeable? i assume the problem asked for how much mass of sand is required to fill the surface area of 0.9 m^2 sorry i am so outta place in my explanation and replies. thanks

 

[SteamKing]

the equation for surface area of a cube is 6a^2 but i do not know how that plays into this because of the variable a unless r and a are interchangeable? i assume the problem asked for how much mass of sand is required to fill the surface area of 0.9 m^2 sorry i am so outta place in my explanation and replies. thanks

You're still not understanding. The surface area of the cube is not 0.9 m^2; according to the problem statement, the cube measures 0.9 m on each edge, which means the length of the edge is 0.9 m. The surface area of the cube can be found knowing the length of the edge.

You also seem to have trouble processing the fact that you are given multiple bits of information.

There are two different geometrical solids involved in this problem: spheres and cubes

'r' and 'a' are not interchangeable.

You've got to find the surface area of the cube before you can determine how many grains of sand have the same total surface area as the cube.

 

[benEE2018]

SteamKing, ok i found the surface area for the cube which is 6(0.9)^2 which comes out to 4.86. I am thinking that the question is asking how many grains of sands would fit into a cube with a surface area of 4.86, so to find the mass of each sand grain i would use the density=mass/volume given density, and the radius per sand grain to find the mass. WHen the mass is figured out, i would then divide the surface area of the cube by the mass of sand which would then give me how many grains of sand would be able to fit in a cube with a surface area of 4.86? I hope it doesn't come off as me just wanting the answer, i genuinely want to understand what is going in the problem if that's the attitude I've been giving off to you. thanks

 

[Simon Bridge]

I am thinking that the question is asking how many grains of sands would fit into a cube with a surface area of 4.86,...

Then your thinking is in error. The problem statement says:

What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a [the cube]?​

OK - you have the surface area of the cube. Let's call it A, so A=4.86 (what are the units?)
What is the surface area of one grain of sand? (Don't forget the units.)
What is the total surface area of 2 grains of sand? Three? Four?
How many grains of sand do you need so the total surface area is the same as A?

 

[benEE2018]

OK - you have the surface area of the cube. Let's call it A, so A=4.86 (what are the units?) m^2

What is the surface area of one grain of sand? (Don't forget the units.) SurfaceAreaofSnd= 4pi(50x10^-6)^2= 3.141592654 e-8 m^2What is the surface area of 2 grains of sand? Three? Four?, two=6.283185308e -8 m^2, three= 9.424777961e -8
four= 1.256637061e -7

How many grains of sand do you need so the combined surface area is the same as A? 4.86/ 3.141592654e -8?

 

[Simon Bridge]

Well done, can you complete the assignment from there?

 

[haruspex]

OK - you have the surface area of the cube. Let's call it A, so A=4.86 (what are the units?) m^2

What is the surface area of one grain of sand? (Don't forget the units.) SurfaceAreaofSnd= 4pi(50x10^-6)^2= 3.141592654 e-8 m^2


What is the surface area of 2 grains of sand? Three? Four?, two=6.283185308e -8 m^2, three= 9.424777961e -8
four= 1.256637061e -7

How many grains of sand do you need so the combined surface area is the same as A? 4.86/ 3.141592654e -8?

Right so far, but drop the excessive number of digits. You are only given a couple of significant digits on most of the data.
Now, what is the mass of that number of grains?

 

[SteamKing]

OK - you have the surface area of the cube. Let's call it A, so A=4.86 (what are the units?) m^2

What is the surface area of one grain of sand? (Don't forget the units.) SurfaceAreaofSnd= 4pi(50x10^-6)^2= 3.141592654 e-8 m^2


What is the surface area of 2 grains of sand? Three? Four?, two=6.283185308e -8 m^2, three= 9.424777961e -8
four= 1.256637061e -7

How many grains of sand do you need so the combined surface area is the same as A? 4.86/ 3.141592654e -8?

So far, so good. Now, what is the total mass of that many grains of sand? Remember, you know the density of the sand.

 

[benEE2018]

i am thinking to take (4.86m^2)/(3.141e -8m^2) to find out how many grains of sand is needed but when i do that the m^2 will both cancel out and i will be left with just a very large number (154698604.7). I am guessing that is the number of sand grains needed to fit a cube with a surface area of 4.86. the question is asking me how much mass of sand grains is needed so my answer has to be in kg but when i do the calculations i get a number alone because m^2 cancels out.

 

[benEE2018]

i am trying to connect surface area and volume but i don't know how they relate to each other. i know i am suppose to be using the density=mass/volume equation to find the mass but am having a difficult time relating the two to density.

 

[haruspex]

I am guessing that is the number of sand grains needed to fit a cube with a surface area of 4.86.

You are not guessing that's what it is. You intentionally calculated the number of grains:

i am thinking to take (4.86m^2)/(3.141e -8m^2) to find out how many grains of sand

You know the volume of one sand grain and its density, so what is the mass of one sand grain?

 

[SteamKing]

i am thinking to take (4.86m^2)/(3.141e -8m^2) to find out how many grains of sand is needed but when i do that the m^2 will both cancel out and i will be left with just a very large number (154698604.7). I am guessing that is the number of sand grains needed to fit a cube with a surface area of 4.86. the question is asking me how much mass of sand grains is needed so my answer has to be in kg but when i do the calculations i get a number alone because m^2 cancels out.

Your units are slightly lacking, which is causing your confusion.

When you calculated the surface area of a sphere 50 microns in diameter, you were calculating the surface area of a grain of sand, so the area of that grain should be 3.141e-8 m^2/grain of sand

Now, when you calculate how many grains of sand have the same surface area as the given cube, the units will be:

m^2 / (m^2/ grain of sand)

When the division is done, you are left with 'grains of sand' as the units of the answer.

 

[benEE2018]

thank you very much steamking, harubridge, and simonBridge i figured out the solution. thank you for being patient with me throughout the problem and i hope i can somehow repay u guys back if need be.