# Thread: Changing the area (km) to volume (L) regarding h2o

1. ## Changing the area (km) to volume (L) regarding h2o

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.

2. ## Re: Changing the area (km) to volume (L) regarding h2o

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)

3. ## Re: Changing the area (km) to volume (L) regarding h2o

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

4. ## Re: Changing the area (km) to volume (L) regarding h2o

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

5. ## Re: Changing the area (km) to volume (L) regarding h2o

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.

6. ## Re: Changing the area (km) to volume (L) regarding h2o

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

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

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

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

7. ## Re: Changing the area (km) to volume (L) regarding h2o

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.

8. ## Re: Changing the area (km) to volume (L) regarding h2o

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

9. ## Re: Changing the area (km) to volume (L) regarding h2o

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?

10. ## Re: Changing the area (km) to volume (L) regarding h2o

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

11. ## Re: Changing the area (km) to volume (L) regarding h2o

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

12. ## Re: Changing the area (km) to volume (L) regarding h2o

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

13. ## Re: Changing the area (km) to volume (L) regarding h2o

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 ...

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