Chapter 15, Problem 4PEA

To determine

The density of the wood to model the continental crust and the oceanic crust.

Answer to Problem 4PEA

The wood having density 0.74 g/cm³ would be appropriate to model continental crust. Wood having density 0.83 g/cm³ is required to model oceanic crust.

Explanation of Solution

Write the equation that can be used to calculate the density of the wood.

$\frac{{\rho }_{crust}}{{\rho }_{mantle}}=\frac{{\rho }_{wood}}{{\rho }_{oil}}$ (I)

After multiplying ${\text{ρ}}_{oil}$ on both side,

${\rho }_{oil}\left(\frac{{\rho }_{crust}}{{\rho }_{mantle}}\right)=\left(\frac{{\rho }_{wood}}{\cancel{{\rho }_{oil}}}\right)\cancel{{\rho }_{oil}}$ (II)

${\rho }_{wood}={\rho }_{oil}\left(\frac{{\rho }_{crust}}{{\rho }_{mantle}}\right)$ (III)

Here, ${\rho }_{crust}$ is the density of crust, ${\rho }_{mantle}$ is the density of mantle, ${\rho }_{wood}$ is the density of wood and ${\rho }_{oil}$ is the density of oil.

Conclusion:

i) Continental crust

Substitute $\text{0}\text{.91}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{oil}$, $\text{3}\text{.3}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{mantle}$, and $2.7\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{crust}$.in equation (III) to find the value of ${\rho }_{wood}$.

$\begin{array}{l}=\text{0}\text{.91}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\left(\frac{\text{2}\text{.7}\frac{\text{g}}{{\text{cm}}^{\text{3}}}}{\text{3}\text{.3}\frac{\text{g}}{{\text{cm}}^{\text{3}}}}\right)\\ =\text{0}\text{.91}\left(\frac{\text{2}\text{.7}}{\text{3}\text{.3}}\right)\frac{\text{g}}{{\text{cm}}^{\text{3}}}\left(\frac{\frac{\cancel{\text{g}}}{\cancel{{\text{cm}}^{\text{3}}}}}{\frac{\cancel{\text{g}}}{\cancel{{\text{cm}}^{\text{3}}}}}\right)\\ =\text{0}\text{.74}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\end{array}$

Therefore, the density of wood should be $\text{0}\text{.74}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ to model the continental crust.

ii) Oceanic crust

Substitute $\text{0}\text{.91}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{oil}$, $\text{3}\text{.3}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{mantle}$, and $3.0\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ for ${\rho }_{crust}$.in equation (III) to find the value of ${\rho }_{wood}$.

$\begin{array}{l}=\text{0}\text{.91}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\left(\frac{3.0\frac{\text{g}}{{\text{cm}}^{\text{3}}}}{\text{3}\text{.3}\frac{\text{g}}{{\text{cm}}^{\text{3}}}}\right)\\ =\text{0}\text{.91}\left(\frac{3.0}{\text{3}\text{.3}}\right)\frac{\text{g}}{{\text{cm}}^{\text{3}}}\left(\frac{\frac{\cancel{\text{g}}}{\cancel{{\text{cm}}^{\text{3}}}}}{\frac{\cancel{\text{g}}}{\cancel{{\text{cm}}^{\text{3}}}}}\right)\\ =\text{0}\text{.83}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\end{array}$

Therefore, the density of wood should be $\text{0}\text{.83}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$ to model the oceanic crust.