Chapter 10, Problem 27P

To determine

To Find: The likely identity of the metal for the given specifications.

Answer to Problem 27P

The element is iron .

Explanation of Solution

Given:

Apparent mass, $M{}_{\text{app}}=55.4\text{kg}$

Actual mass, $M{}_{\text{sample}}=63.5\text{kg}$

Formula used:

Write the expression for buoyant force.

  $\begin{array}{l}{F}_B=m_{\text{apparent}}g-m_{\text{metal}}g\\ m_{\text{metal}}g=m_{\text{apparent}}g-F_B\\ =m_{\text{apparent}}g-V_{\text{metal}}\times {\rho }_{\text{water}}\times g\\ m_{\text{metal}}=m_{\text{apparent}}-V_{\text{metal}}\times {\rho }_{\text{water}}\\ =m_{\text{apparent}}-\frac{m_{\text{metal}}}{{\rho }_{\text{metal}}}\times {\rho }_{\text{water}}\\ {\rho }_{\text{metal}}=\frac{m_{\text{metal}}}{m_{\text{metal}}-m_{\text{apparent}}}\times {\rho }_{\text{water}}\end{array}$

Where ${\rho }_{\text{metal}}$ is density of the material

  ${\rho }_{\text{water}}$ is water density,

  $m_{\text{metal}}$ is the actual mass of the metal,

  $m_{\text{apparent}}$ is its apparent mass,

  $F_B$ represents the buoyant force,

  $g$ is the acceleration due to gravity.

Calculation:

Now, substitute the given values in the formula,

  ${\rho }_{\text{metal}}=\frac{m_{\text{metal}}}{m_{\text{metal}}-m_{\text{apparent}}}\times {\rho }_{\text{water}}=\frac{63.5}{63.5-55.4}\times 1000=7839{\text{kg/m}}^{\text{3}}$

By comparing this density with given densities in the reference table, it is observed that the resulted element is iron.

Conclusion:

Therefore, the likely element is iron.