Changing the area (km) to volume (L) regarding h2o

**risteek** · October 4th 2011, 10:29 AM

I am so sorry if this is the wrong subforum, I think it fits here, but if not please go easy on me. It's been a long day.

I have a bunch of questions that are all similar for practice. Basically area to volume. I'm lost. Here's the first one we have:

Suppose a 2 cm thick layer of ice can be melted every day. Assuming a surface area of 4 km2, calculate the corresponding volume of ice melted each day in Litres.

I have about ten of them, if I could just understand how to do it then I'm sure I could plug the numbers in, however I am just lost on this. It's supposed to be review, but I don't really remember ever actually understanding how to do this. Any help is appreciated math gurus.

**e^(i*pi)** · October 4th 2011, 10:47 AM

Originally Posted by risteek

I am so sorry if this is the wrong subforum, I think it fits here, but if not please go easy on me. It's been a long day.

I have a bunch of questions that are all similar for practice. Basically area to volume. I'm lost. Here's the first one we have:

Suppose a 2 cm thick layer of ice can be melted every day. Assuming a surface area of 4 km2, calculate the corresponding volume of ice melted each day in Litres.

I have about ten of them, if I could just understand how to do it then I'm sure I could plug the numbers in, however I am just lost on this. It's supposed to be review, but I don't really remember ever actually understanding how to do this. Any help is appreciated math gurus.

The volume of a prism is cross sectional area * length

$V = Al$

$1000cm^3 = 1L$ so if you find out the volume in cm^3 then divide by 1000 to get an answer in Litres (note that $1km^2 = 10^{10}cm^2)$

edit: I have assumed that your rate of melting is constant

edit2: I have assumed that your ice is cuboid (or any other kind of prism - the important thing being that the surface area is constant)

**Soroban** · October 4th 2011, 11:25 AM

Hello, risteek!

What is the shape of the mass of ice?

Suppose a 2 cm thick layer of ice can be melted every day.
Calculate the corresponding volume of ice melted each day in Litres.

I will assume that we have a spherical ball of ice.

The original volume is: . $V _o \:=\:\tfrac{4}{3}\pi (r_o)^3$

The reduced volume is: . $V_1 \:=\:\tfrac{4}{3}\pi (r_1)^3$

The change in volume is:

. . $\Delta V \;=\;\tfrac{4}{3}\pi(r_o)^3 - \tfrac{4}{3}\pi(r_1)^3 \;=\;\tfrac{4}{3}\pi\left(r_o^3 - r_1^3\right)$

. . . . . . $=\;\tfrac{4}{3}\pi(r_o - r_1)(r_o^2 + r_or_1 + r_1^2)$

We are told that: . $r_o - r_1 \:=\:2$

Hence: . $\Delta V \;=\;\tfrac{4}{3}\pi\cdot 2\cdot(r_o^2 + r_ir_1 + r_1^2) \;=\;\tfrac{8}{3}\pi(r_o^2 + r_or_1 + r_1^2)\text{ cm}^3$

Therefore: . $\Delta V \;=\;\tfrac{1}{1000}\cdot\tfrac{8\pi}{3}(r_o^2 + r_or_1 + r_1^2) \;=\; \tfrac{\pi}{375}(r_o^2 + r_or_1+r_1^2){\text{ L.}$

**bjhopper** · October 4th 2011, 06:30 PM

assumed shape = right rectangular prism. Area *altitude =Volume
(2*10^3)^2* (2*10^-2)* (10^3)=8*10^7 liters

**risteek** · October 5th 2011, 03:48 PM

It doesn't say what the shape is actually, but I don't think it's a sphere.

Forgive my ignorance, math really is not my area of strength.
So I have to change the 2cm into km right? Or should I change the km into cm? Or should I not do that because it's km^2 and the other is just cm?

The length in e^(i*pi)'s reply is the 2 cm right?

So if it's length times area then it's what 2cm multiplied by the 4 km^2. So if I change the km to cm it's 40000000000cm^2. And that's multiplied by 2cm? So it's 80000000000cm^3? So then in Litres that would be 80000000L??

And in bjhopper's post... is altitude the same as length used in e^(i*pi)'s reply? And if so did you convert something or... ummm?

Thank you for the replies, even if I'm about 100% lost.

**skeeter** · October 5th 2011, 05:03 PM

note the question specifically asks for the volume in liters ... $1 \, L = 1000 \, cm^3$

$1 \, km = 10^5 \, cm$

$1 \, km^2 = 10^{10} \, cm^2$

$V = 2 \, cm \cdot 4 \times 10^{10} \, cm^2 = 8 \times 10^{10} \, cm^3 = 8 \times 10^7 \, L$

**risteek** · October 5th 2011, 05:24 PM

Ok so that makes a lot of sense.

Thank you!

Does it make a difference if the volume was supposed to be litres squared? I didn't notice it at first.... sorry.

**skeeter** · October 5th 2011, 05:26 PM

a liter itself is a unit of volume, not liters squared

**risteek** · October 5th 2011, 05:34 PM

That's what I thought, must have been a typo.

Ok so now if I have to calculate the mass and the density is 0.9167g/cm^3 I would do M=DxV right?
So that would be:

M=0.9167g/cm^3 X 8x10^10cm^3
M=73336x10^10

Yes?

**skeeter** · October 5th 2011, 05:49 PM

Originally Posted by risteek

That's what I thought, must have been a typo.

Ok so now if I have to calculate the mass and the density is 0.9167g/cm^3 I would do M=DxV right?
So that would be:

M=0.9167g/cm^3 X 8x10^10cm^3
M=73336x10^10

Yes?

$m = 7.3336 \times 10^{10} \, g$

**risteek** · October 5th 2011, 06:01 PM

It always was the little details that got me in trouble.... lol. Thank you so much for your help and goodnight.

**bjhopper** · October 6th 2011, 05:59 AM

Hi skeeter,
I know what you mean but your equation is in error. 8x10^10 cm^3 = 8x10^10 ml and this must be divided by 1000 to get liters

**skeeter** · October 6th 2011, 10:31 AM

Originally Posted by bjhopper

Hi skeeter,
I know what you mean but your equation is in error. 8x10^10 cm^3 = 8x10^10 ml and this must be divided by 1000 to get liters

I did that. See my post #6 above ...

$V = 2 \, cm \cdot 4 \times 10^{10} \, cm^2 = 8 \times 10^{10} \, cm^3 = 8 \times 10^7 \, L$

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