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

1. Aug 29, 2004

### CellCoree

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. Aug 29, 2004

### Beer-monster

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?

3. Aug 29, 2004

### da_willem

The volume of a sphere is $\frac{4}{3} \pi R^3$ with R the sphere's radius.

4. Aug 29, 2004

### Beer-monster

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.

5. Aug 29, 2004

### CellCoree

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?

6. Aug 29, 2004

### faust9

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.

 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
7. Aug 29, 2004

### CellCoree

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

8. Aug 29, 2004

### faust9

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, $x=6.11\times 10^{-5}m^3$ You need to watch the notation.

We know $x=6.11\times 10^{-5}m^3=V$ and V=L^3 solve for L. You should get a ball park number of 4'ish cm (convert m to cm).

9. Aug 29, 2004

### Beer-monster

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

10. Aug 29, 2004

### CellCoree

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

cause $x=6.11\times 10^{-5}m^3=V$ =.0000611 right?

so.....V=.0610687 = L^3
$$L=\sqrt[3]{.0610687}$$
L = .3937974428 m^3 <-- that doesnt seem to convert to around 4cm. what am i doing wrong?

Last edited: Aug 29, 2004
11. Aug 29, 2004

### HallsofIvy

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
12. Aug 29, 2004

### CellCoree

thank you everyone for the help, i finally get it!!! sorry for all the questions

13. Aug 29, 2004

### Leong

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 ....

14. Sep 2, 2004