# Density

## Practice

### practice problem 1

Your Mother gives you a kilogram of aluminum and a kilogram of lead. Both objects are solid, rectangular blocks.

- Which is
*more massive*on the surface of the*earth*? - Which is
*more massive*on the surface of the*moon*? - Which is
*heavier*on the surface of the*earth*? - Which is
*heavier*on the surface of the*moon*?

We will return to this question in the section on buoyancy. (The phrase "more massive" means "has more mass" not "fills more space".)

#### solution

Answer it.

### practice problem 2

A lucite cube has a mass of 142.5 g and a width of 4.9 cm. Determine its density in kg/m

^{3}.Lucite cube on electronic balance | Ruler next to Lucite cube |

#### solution

Start with the definition of density, replace *V* with *s*^{3} (the volume of a cube), substitute the given quantities in SI units, and solve.

ρ = | m |
= | m |
= | (0.1425 kg) | = 1200 kg/m^{3} |

V |
s^{3} |
(0.049 m)^{3} |

### practice problem 3

It has been known for several thousand years that the Earth is spherical (by educated people, at least). Sometime in the Second Century BCE the size of the Earth was determined (

Determine…

*r*_{earth}= 6,370 km). By the Nineteenth Century its mass was known (*m*_{earth}= 5.97 × 10^{24}kg). And in the early Twentieth Century the structure of the Earth was deduced. The earth has three main layers: crust, mantle, and core. The crust of the is the lightest and thinnest and, like the shell of an egg, contributes little to its overall mass. The mantle is a bit more dense, substantially thicker, and contains most of the Earth's mass. The core is the densest layer (but not the most massive) and is divided into a liquid outer core and a solid inner core. The relevant data for the interior of the Earth are summarized below.layer | depth range (km) | mean density (kg/m^{3}) |
consistency |
---|---|---|---|

crust | 0 ~ 20 | 2700 | solid |

mantle | 20 ~ 2890 | 4500 | plastic |

outer core | 2890 ~ 5160 | ? | liquid |

inner core | 5160 ~ 6370 | ? | solid |

- the average density of the entire earth
- the percent of the Earth's mass located in the mantle, and
- the average density of the core.

#### solution

- Density is the ratio of mass to volume. Use the given mass of the Earth and calculate its volume from the radius and the equation for the volume of a sphere.
ρ =_{earth}*m*_{earth}*V*_{earth}ρ =_{earth}3 *m*4π *r*^{3}_{earth}ρ =_{earth}3(5.97 × 10 ^{24}kg)4π(6.37 × 10 ^{6}m)^{3}ρ =_{earth}5510 kg/m ^{3} - You may want to do this is stages. Calculate the volume of the spherical shell that is the mantle.
*V*=_{mantle}4 π *r*^{3}−_{outer}4 π *r*^{3}_{inner}3 3 *V*=_{mantle}4 π ( *r*^{3}−_{outer}*r*^{3})_{inner}3 *V*=_{mantle}4 π [ (6,370,000 m) ^{3}− (3,480,000 m)^{3}]3 *V*=_{mantle}9.06164… × 10 ^{20}m^{3}Mass is density times volume. Use the density given in the data table and the mass calculated above.

*m*=_{mantle}ρ *V**m*=_{mantle}(4500 kg/m ^{3})(9.06164… × 10^{20}m^{3})*m*=_{mantle}4.07774… × 10 ^{24}kgDivide the mass of the mantle by the mass of the Earth to determine the percentage of earth's mass in its most substantial layer.

*m*_{mantle}= 4.07774… × 10 ^{24}kg*m*_{earth}5.97 × 10 ^{24}kg*m*_{mantle}= 0.683038… ≈ 70% *m*_{earth} - This part could also be solved in stages, but let's try a different approach. Start from the basic definition of density, replace mass and volume with their respective fractions of the Earth's total, then multiply by the average density. Recall that the volume of a sphere is proportional to the cube of its radius.
ρ =_{core}*m*_{core}*V*_{core}ρ =_{core}(1 − 0.683038…) *m*_{earth}⎛

⎝6370 km − 2890 km ⎞ ^{3}

⎠*V*_{earth}6370 km ρ =_{core}1.94396… ρ _{earth}ρ =_{core}(1.94396…)(5514.01… kg/m ^{3})ρ =_{core}10,700 kg/m ^{3}^{3}and the solid inner core from 12,300 to 12,600 kg/m^{3}. My suspicion is that the mean value is more like 10,900 kg/m^{3}, but I don't see any error in the solution presented above. I leave you with the classic excuse found in many college math textbooks, but with a slight modification. "The*dis*proof is left to the reader."

### practice problem 4

Write something completely different.

#### solution

Answer it.