= Density of a composite system
{title2=$\rho_{\rm total}=(m_A+m_B)/(V_A+V_B)$}
Consider two hypothetical parcels at the same <temperature> $300\,\mathrm K$, kept separated with negligible partition <mass> and excluded container <volume>. Parcel A has <mass> $2\,\mathrm{kg}$ and <volume (thermodynamics)> $0.002\,\mathrm{m^3}$; parcel B has <mass> $3\,\mathrm{kg}$ and <volume (thermodynamics)> $0.006\,\mathrm{m^3}$. They may be different materials. Their individual <mass densities> and the composite <mass density> are
$$
\rho_A=1000\,\mathrm{kg\,m^{-3}},\qquad
\rho_B=500\,\mathrm{kg\,m^{-3}},\qquad
\rho_{\rm total}=\frac{5\,\mathrm{kg}}{0.008\,\mathrm{m^3}}
=625\,\mathrm{kg\,m^{-3}}.
$$
The composite <mass density> is total material <mass> divided by total occupied <volume (thermodynamics)>; it need not be the local <mass density> anywhere. Equal <temperature> does not make these parcels replicas. In particular, the unweighted average $750\,\mathrm{kg\,m^{-3}}$ is wrong. The <volume (thermodynamics)> fractions are $1/4$ and $3/4$, giving
$$
\rho_{\rm total}=\tfrac14\rho_A+\tfrac34\rho_B
=625\,\mathrm{kg\,m^{-3}}.
$$
With <mass> fractions $2/5$ and $3/5$, the equivalent rule is a <weighted harmonic mean>:
$$
\frac1{\rho_{\rm total}}
=\frac{2/5}{\rho_A}+\frac{3/5}{\rho_B}
=0.0016\,\mathrm{m^3\,kg^{-1}}.
$$
This follows directly by writing $V_i=m_i/\rho_i$ and dividing total <volume (thermodynamics)> by total <mass>; it is also the <mass>-weighted average of <specific volumes>. An unweighted average of <mass densities> is valid when the parcel <volume (thermodynamics)> values are equal, not merely when their <temperatures> are equal.
If a partition is removed and different substances mix, the final <volume (thermodynamics)> need not equal the sum of initial <volume (thermodynamics)> values. A new <temperature>, <thermodynamic phase> or composition may also change the local <mass density>. Use the actual final <volume (thermodynamics)> in $\rho=m/V$; the pre-mixing weighted identity cannot supply it. Even unchanged <temperature> alone does not guarantee unchanged <volume (thermodynamics)>.
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