Mass — Question 8

Question 8

The same bulk mass can acquire different surface mass

A clean solid cube has edge $a=10.0\ \mathrm{cm}$ and material density $8.00\ \mathrm{g\,cm^{-3}}$. Compare two separate preparation routes using the same initial amount of material:

• Route A keeps one cube intact.

• Route B divides it into 1000 equal cubes with no material loss, separates them, and exposes all six faces of every cube.

After preparation, each route receives a surface film. Its mass per unit exposed area is $$\sigma=0.0200\ \mathrm{mg\,cm^{-2}}.$$ No film was present before division. Treat the coating mass as $m_{\mathrm{film}}=\sigma A$, use the uncoated face areas, and ignore all other changes. Values are exact in this model. A cube has volume $a^3$ and area $6a^2$.

1. Find the common uncoated mass, the edge of each small cube, and total exposed area for each route.

2. Find the acquired film mass for each route and express it as parts per million of the uncoated mass, where $1\ \mathrm{ppm}=10^{-6}$.

3. For division into $n$ equal cubes, derive the ratio of total area to the original area. What does this imply about preserving total bulk mass while changing susceptibility to a surface film?

4. If a reference artifact acquires such a film, has the kilogram changed? Explain what must be controlled when using physical artifacts to transfer a mass calibration.

Question 8 - Solution

The same bulk mass can acquire different surface mass

(a) Equal total volume does not mean equal exposed area.

The uncoated mass is $8.00(10.0)^3=\boxed{8000\ \mathrm{g}=8.00\ \mathrm{kg}}$. Each small cube has volume $1000/1000=1\ \mathrm{cm^3}$ and edge $\boxed{1.00\ \mathrm{cm}}$. The areas are $$A_A=6(10.0)^2=\boxed{600\ \mathrm{cm^2}},\qquad A_B=1000\,[6(1.00)^2]=\boxed{6000\ \mathrm{cm^2}}.$$

(b) Apply the same film mass per unit area.

$$10^6\frac{m_{{\mathrm{film}},A}}{M}=10^6\frac{12}{8\times10^6}=1.50,\qquad10^6\frac{m_{{\mathrm{film}},B}}{M}=15.0\quad\text{(ppm)}.$$

$$m_{{\mathrm{film}},A}=0.0200(600)=\boxed{12.0\ \mathrm{mg}},\qquad m_{{\mathrm{film}},B}=0.0200(6000)=\boxed{120.0\ \mathrm{mg}}.$$ The common uncoated mass is $8.00\times10^6\ \mathrm{mg}$. Thus the fractional gains are $\boxed{1.50\ \mathrm{ppm}}$ and $\boxed{15.0\ \mathrm{ppm}}$, respectively. The extra mass comes from deposited material, not from division itself.

(c) Derive the area scaling from volume preservation.

If each small cube has edge $b$, then $nb^3=a^3$, so $b=a/n^{1/3}$. Therefore $$\frac{A_n}{A_1}=\frac{n(6b^2)}{6a^2}=\boxed{n^{1/3}}.$$ This assumes division into equal cubes is geometrically possible and every face is exposed. At $n=1000$, the ratio is 10. With fixed $\sigma$, the coating mass increases by the same factor even though uncoated mass is unchanged.

(d) The artifact changes; the unit does not.

A coating adds to the artifact’s mass and may invalidate its previous assigned value. The SI kilogram is not redefined by that change. Surface condition, cleaning history, storage, and exposure must be controlled or assessed when an artifact transfers a calibration. Stable bulk material alone does not establish a stable total artifact mass.