Question

Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.0000 kg when weighed in air.

Short answer

A piece of aluminum's true mass (in vacuum) is 4.0019kg.

Step-by-step solution

Understanding apparent mass and true mass

The difference between the true and apparent mass of a body is the air mass displaced by the aluminum. The difference is due to the effect of air buoyancy.

Identification of the given data

The apparent mass of the aluminum in the air is m_apparent= 4.0000 kg.

Determination of the true mass of aluminum

As discussed above,

\begin{aligned}{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={m_{{\rm{air}}}}\\{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={\rho_{{\rm{air}}}}\cdot{V_{{\rm{Al}}}}\\{m_{{\rm{actual}}}}-{m_{{\rm{apparent}}}}={\rho_{{\rm{air}}}}\frac{{{m_{{\rm{actual}}}}}}{{{\rho_{{\rm{Al}}}}}}\end{aligned}

Here, m_actual is the true/actual mass of the aluminum, ρ_air is the density of air, ρ_AI is the density of aluminum, and V_AI is the aluminum piece’s volume.

\begin{aligned}{m_{{\rm{actual}}}}-{\rho_{{\rm{air}}}}\frac{{{m_{{\rm{actual}}}}}}{{{\rho_{{\rm{Al}}}}}}={m_{{\rm{apparent}}}}\\{m_{{\rm{actual}}}}\left({1-\frac{{{\rho_{{\rm{air}}}}}}{{{\rho_{{\rm{Al}}}}}}}\right)={m_{{\rm{apparent}}}}\\{m_{{\rm{actual}}}}=\frac{{{m_{{\rm{apparent}}}}}}{{\left({1-\frac{{{\rho_{{\rm{air}}}}}}{{{\rho_{{\rm{Al}}}}}}}\right)}}\end{aligned}

Calculations to determine the required result

Now, considering the standard values of ρ_air and ρ_AI we have,

ρ_air =1.29 kg/m³

ρ_AI = 2.70 x 10 kg/m³

Substitute these values in the above formula.

\begin{aligned}{m_{{\rm{actual}}}}=\frac{{4.0000\;{\rm{kg}}}}{{1-\frac{{1.29\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}}{{2.70\times{{10}^3}\;{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}}}}}\\{m_{{\rm{actual}}}}=4.0019\;{\rm{kg}}\end{aligned}

Hence the true mass of aluminum is 4.0019 kg.