Question

What is the likely identity of metal (see Table 10–1) if a sample has a mass of 63.5 g when measured in air and an apparent mass of 55.4 g when submerged in water?

Short answer

The likely identity of the metal is iron or steel.

Step-by-step solution

Understanding the apparent weight of the body

The apparent weight of a body submerged in a fluid is the difference between its actual weight and buoyant force. It is expressed in terms of Newton.

Identification of the given data

The mass of the metal sample, when measured in air, is m_metal =63.5 g.

The apparent mass of the metal sample when measured in water is m_apparent = 55.4g

Determination the identity of the metal

As discussed above,

\begin{aligned}{W_{{\rm{apparent}}}}={W_{{\rm{metal}}}}-{F_{\rm{B}}}\\{m_{{\rm{apparent}}}}\cdot g={m_{{\rm{metal}}}}\cdot g-{F_{\rm{B}}}\end{aligned}

Here, F_B is the buoyant force, is the apparent mass when submerged in water, is the mass of metal in the air, and g is the acceleration due to gravity.

But we know that the buoyant force is the weight of a mass of water occupying the volume of the metal sample.

Therefore,

Here, ρ_H2Ois the density of water and ρ_metal is the density of the metal.

Calculations to determine the desired result

Substitute the values of densities in the above formula.

\begin{aligned}{\rho_{{\rm{metal}}}}=\frac{{63.5\;{\rm{g}}}}{{\left({63.5\;{\rm{g}}-55.4\;{\rm{g}}}\right)}}\times{\rho_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\end{aligned}

Now we know that the density of water is 1000kg/m³. Thus,

\begin{aligned}{\rho_{metal}}=\frac{{63.5\;{\rm{g}}}}{{\left({63.5\;{\rm{g}}-55.4\;{\rm{g}}}\right)}}\times\left({1000\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}\right)\\{\rho_{metal}}=7840\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}\end{aligned}

Hence, based on the density, the metal is probably iron or steel (after referring to Table 10-1).