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Math Help - Changing the area (km) to volume (L) regarding h2o

  1. #1
    Junior Member risteek's Avatar
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    Smile 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.
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    Re: Changing the area (km) to volume (L) regarding h2o

    Quote Originally Posted by risteek View Post
    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)
    Last edited by e^(i*pi); October 4th 2011 at 11:35 AM.
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  3. #3
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    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.}

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    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
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    Junior Member risteek's Avatar
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    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.
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    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
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    Junior Member risteek's Avatar
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    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.
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    Re: Changing the area (km) to volume (L) regarding h2o

    a liter itself is a unit of volume, not liters squared
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    Junior Member risteek's Avatar
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    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?
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  10. #10
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    Re: Changing the area (km) to volume (L) regarding h2o

    Quote Originally Posted by risteek View Post
    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
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  11. #11
    Junior Member risteek's Avatar
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    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.
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    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
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    Re: Changing the area (km) to volume (L) regarding h2o

    Quote Originally Posted by bjhopper View Post
    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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