Oil tanker problem needing area when thickness of oil is given

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September 20th 2010, 06:38 AM
bradycat

Oil tanker problem needing area when thickness of oil is given

Oil tanker carrying 2.0 X10^5 metric tons of crude oil of density .70g/cm^3 breaks up at sea, what area of ocean in km^2 will be polluted if the oil spreads out to the thickness of .10 mm?? Metric ton = 1000 kg.
Ans is 2900 km^2.

So change the metric ton to kg is 2x10^8 kg.
Now kinda stuck again. I hae no clue what it is asking and what formula to use again.
HELP PLEASE. again, this is our own study material that we need to work out, nothing to hand in and need this to study from.
Please help thanks
Joannehttp://www.chemistryhelpforum.com/ch...milies/eek.gif
I DON"T KNOW WHERE TO START.(Headbang)
September 20th 2010, 08:03 AM
undefined

Quote:

Originally Posted by bradycat

Oil tanker carrying 2.0 X10^5 metric tons of crude oil of density .70g/cm^3 breaks up at sea, what area of ocean in km^2 will be polluted if the oil spreads out to the thickness of .10 mm?? Metric ton = 1000 kg.
Ans is 2900 km^2.

So change the metric ton to kg is 2x10^8 kg.
Now kinda stuck again. I hae no clue what it is asking and what formula to use again.
HELP PLEASE. again, this is our own study material that we need to work out, nothing to hand in and need this to study from.
Please help thanks
Joannehttp://www.chemistryhelpforum.com/ch...milies/eek.gif
I DON"T KNOW WHERE TO START.(Headbang)

I believe you can imagine the spill as some arbitrary shape (can be neat and tidy like a circle or square, or can be just a blob) with a very small height, and you need to find the area of the shape. You are given enough information to find the volume, and you have the height, then you use V = Ah. Apart from this just be careful about units and significant figures (for rounding).
September 20th 2010, 11:54 AM
bradycat

Still a bit confused, but..........so do I multiply the metric ton by the thickness to get the volume??
I am having a hard time with going from the .7g/cm3 to to a km^2. Or am I thinking way to hard on this.
Can you help me out with the first calculation please?
Jo
September 20th 2010, 12:30 PM
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Quote:

Originally Posted by bradycat

Still a bit confused, but..........so do I multiply the metric ton by the thickness to get the volume??
I am having a hard time with going from the .7g/cm3 to to a km^2. Or am I thinking way to hard on this.
Can you help me out with the first calculation please?
Jo

density is mass per volume; the units tell you this. we can write D = m/V. Solve for V. Then plug in D and m. Then you have V=Ah. You want A, so solve for it, then plug in V and h.

if you have trouble converting units, i find helpful using conversion factors. The idea is you multiply by 1, thus you are sure to get an equivalent value. an example of a conversion factor is (1 ft / 12 in). You multiply by conversion factors strategically to get units to cancel as you desire.
September 20th 2010, 12:54 PM
bradycat

OK so I got a volume of 2.85 x 10^11 mL.
Now get area I need the V=Ah as you noted above.
2.85 x 10^11ml = A ( .10 mm)
This is where I am stuck. or do I have it correct????
September 20th 2010, 01:33 PM
undefined

Quote:

Originally Posted by bradycat

OK so I got a volume of 2.85 x 10^11 mL.
Now get area I need the V=Ah as you noted above.
2.85 x 10^11ml = A ( .10 mm)
This is where I am stuck. or do I have it correct????

Why did you convert to mL?

$\displaystyle (2.0\cdot10^5\ \text{metric tons})\left(\frac{1000\ \text{kg}}{1\ \text{metric ton}}\right) = 2.0\cdot10^8\ \text{kg}$

V = m / D (or just look at units). I will write many decimal places in the intermediate result and only round at the end

$<br /> (2.0\cdot10^8\ \text{kg})\left(\dfrac{1000\ \text{g}}{1\ \text{kg}}\right)\left(\dfrac{1\ \text{cm}^3}{0.70\ \text{g}}\right)\left(\dfrac{10\ \text{mm}}{1\ \text{cm}}\right)^3\approx 2.857143\cdot10^{14}\ \text{mm}^3$

A = V/h

$\displaystyle (2.857143\cdot10^{14}\ \text{mm}^3)\left(\frac{1}{0.10\ \text{mm}}\right)$

$=(2.857143\cdot10^{15}\ \text{mm}^2)\left(\frac{1\ \text{m}}{1000\ \text{mm}}\right)^2\left(\frac{1\ \text{km}}{1000\ \text{m}}\right)^2=2.8\cdot10^3\ \text{km}^2$
September 20th 2010, 04:52 PM
bradycat

Undefined: THANKS for the break down. I was lost and now I am found. I will study this, which I already have and understand it. I was forgetting the A=V/h. I was looking at this from density only, and that was it. I was lost after that point.
I appreciate your help and trust me, will memorize this to memory.