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Homework Help: Find the length of the side of a cube of iron

  1. Aug 29, 2004 #1
    Iron has a property such that a 1.00m^3 volume has a mass of 7.86 X 10^3kg(density equals 7.86 X 10^3kg/m^3). You want to manufacture iron into cubes and spheres.

    1.) Find the length of the side of a cube of iron that has a mass of 480 g.

    2.) Find the radius of a solid sphere of iron that has a mass of 480 g.

    please help me with these two questions, i attend class everyday and take notes, but i dont remeber the teacher talking about this type of question at all. i really want to learn how to do these questions. thanks in advance
  2. jcsd
  3. Aug 29, 2004 #2
    You already seem to know that density is mass/volume. So volume is mass/density.

    You have two types of object with different volumes. The volume of a cube is Length^3
    and the volume of a sphere is Pi(R^2).

    Bearing these in mind you should be able to rearrange and solve for L and R.

    Hope it helps. Incidently do you know what the answer should be?
  4. Aug 29, 2004 #3
    The volume of a sphere is [itex]\frac{4}{3} \pi R^3[/itex] with R the sphere's radius.
  5. Aug 29, 2004 #4
    Okay feeling a bit stupid now :yuck:

    That's what I meant to say, only my brain wasn't co-operating. Sorry about that, my bad. :biggrin:
  6. Aug 29, 2004 #5
    ok im still confuse, but let me try. ok for the first question, V=L^3 right? so...
    V= 7.86 X 10^3kg and L isnt defined? where does the 480 g come in?
  7. Aug 29, 2004 #6
    You can set this up as a ratio then apply the answer to your volume equations like so:

    \frac{7.86\times 10^3kg}{1m^3}=\frac{0.480kg}{x}

    Solving for x gives a volume. Since you now know the volume, and have volume equations for the sphere and the cube you should be able to solve for the length and the radius.

    [edit] The reason you can do this is because the 7.86 is the density of iron, which is a ratio of mass:volume. The size of the iron object will not effect the density of iron, so the ratio of mass:volume will be the same.

    Hope this helped.
    Last edited: Aug 29, 2004
  8. Aug 29, 2004 #7
    let me show you what i done.

    cross multiply, so i get 7.86000kgx=0.480kg
    x=.0610687 <-- is that the volume?

    that doesnt look right, i know im doing it incorrectly. why did you convert 480g to .480? and what should i do with 1m^3? am i suppose to do unit conversions first? im sorry, im totally lost
  9. Aug 29, 2004 #8
    How many grams are in 1 kilogram? Next what do you do with the m^3? You keep it becasue m^3 is a volume. Next, [itex]x=6.11\times 10^{-5}m^3[/itex] You need to watch the notation.

    We know [itex]x=6.11\times 10^{-5}m^3=V[/itex] and V=L^3 solve for L. You should get a ball park number of 4'ish cm (convert m to cm).
  10. Aug 29, 2004 #9
    To get L devide the mass 480g by the density and cube root what you get.

    Do something similar for part b) but with the equation for a volume of a sphere
  11. Aug 29, 2004 #10

    i got x=.0610687 when trying to find the volume. so shouldnt it be [itex]x=6.11\times 10^{5}m^3=V[/itex] instead of [itex]x=6.11\times 10^{-5}m^3=V[/itex]?

    cause [itex]x=6.11\times 10^{-5}m^3=V[/itex] =.0000611 right?

    so.....V=.0610687 = L^3
    L = .3937974428 m^3 <-- that doesnt seem to convert to around 4cm. what am i doing wrong?
    Last edited: Aug 29, 2004
  12. Aug 29, 2004 #11


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    Science Advisor

    You are told that the density of iron is 7.86 X 103kg/m3. You are also told that you have 480 g= 0.48 kg of iron. To convert from kg to m2, you need to cancel that "kg" so you need "kg" in the denominator and you need to get m3 into the numerator: INVERT 7.86x 103 kg/ m3: that is, divide 0.48 kg by 7.86x 103 kg/m3 to get 0.06107 x 10-3 m3 (did you forget the "10-3 in your solution) = 6.107x10-5 m3.
    What must x be in order that x3= 6.107 x 10-5?

    For part b, solve the equation [tex]\frac{4}{3}\pi r^3= 6.107x10^{-5}.
    Last edited by a moderator: Aug 29, 2004
  13. Aug 29, 2004 #12
    thank you everyone for the help, i finally get it!!! sorry for all the questions
  14. Aug 29, 2004 #13
    You don't need to say sorry. after all, they really wanted to help and that was why they were writing all these stuffs. am i right ? i hope so ....
  15. Sep 2, 2004 #14
    does anyone know the answer?
  16. Sep 2, 2004 #15
    1.) 3.94 cm
    2.) 2.44 cm

    where are you from? are you doing the same hw as i am? if so, maybe we can help each other out.
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