# Oil tanker problem needing area when thickness of oil is given

• Sep 20th 2010, 06:38 AM
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.
Joannehttp://www.chemistryhelpforum.com/ch...milies/eek.gif
I DON"T KNOW WHERE TO START.(Headbang)
• Sep 20th 2010, 08:03 AM
undefined
Quote:

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.
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).
• Sep 20th 2010, 11:54 AM
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
• Sep 20th 2010, 12:30 PM
undefined
Quote:

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.
• Sep 20th 2010, 12:54 PM
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????
• Sep 20th 2010, 01:33 PM
undefined
Quote:

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

$
(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

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

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