Help BIF Calculation thickness of annual layer of ore

[kndietz]

Please I need help ASAP! This is the problem:
Cowen cites the deposition rate of BIF in the Hamersley Basin of Australia as 30 million metric tonnes per year. He says the ore is 55% iron (volume); iron has a density of 8 g/cm^3. Assume the rest of the ore is made of silica, which has a density of 2.6 g/cm^3. Using these data, calculate the volume of ore that was deposited aannually at Hamersley (recall that density = mass per unit volume). The Hamersley Basic is roughly circular ((we were told to consider it cylinder)), about 400 km in diameter. How thick was an annual layer of ore?

My teacher said we need to get average density, then volume, finally thickness. But I still have no idea how to go about doing this homework!

 

[Evo]

I'm going to move this for you, but you still need to show what you have done to solve this yourself.

 

[kndietz]

I have tried to start calculating the density but I am stumped on how to do this without mass, and in order to find volume you need density...
I tried converting from cm^3 to m^3, .08 and .026, and then I tried to make this into the 55% and 45% they represent... 4.4 and 1.17, combined is 5.57.
I really don't know if this is the right place to start or not, or where to go from here.

 

[Borek]

Imagine you have 100kg of BIF. Can you calculate its volume from a given data?

More elegant approach calls for assuming you have m kg of BIF and solving for the volume/density using this unknown - you will find m cancels out in the end.

 

[DeeNos]

I understand that calculating the thickness of an annual layer of ore is a complex problem that requires several steps. I will guide you through the process and provide the necessary calculations.

First, we need to calculate the average density of the ore at Hamersley Basin. This can be done by using the given information that the ore is 55% iron and 45% silica. We can therefore calculate the average density as follows:

Average density = (55% x 8 g/cm^3) + (45% x 2.6 g/cm^3)
= 4.4 g/cm^3 + 1.17 g/cm^3
= 5.57 g/cm^3

Next, we need to calculate the volume of ore that was deposited annually at Hamersley. To do this, we will use the deposition rate of 30 million metric tonnes per year and convert it to volume using the average density we calculated earlier. The conversion factor for metric tonnes to cm^3 is 1 metric tonne = 1,000,000 cm^3.

Volume of ore deposited annually = (30 million metric tonnes x 1,000,000 cm^3) / 5.57 g/cm^3
= 5.395 x 10^9 cm^3

Now, we need to consider the shape of the Hamersley Basin, which we were told to consider as a cylinder. We can calculate the volume of a cylinder by multiplying the area of the base (πr^2) by the height (h). Since the diameter of the basin is given as 400 km, the radius (r) can be calculated as 200 km or 200,000 m. The height (h) will be the thickness of the annual layer of ore that we are trying to calculate.

Volume of cylinder = πr^2 x h
5.395 x 10^9 cm^3 = π x (200,000 m)^2 x h
h = (5.395 x 10^9 cm^3) / (π x 4 x 10^10 m^2)
= 1.71 x 10^-4 m

Therefore, the thickness of an annual layer of ore at Hamersley Basin is 1.71 x 10^-4 m or 0.171 mm.

I hope this explanation helps you understand the process of calculating the thickness of