Mass — Question 4

Question 4

Mass removed is not necessarily mass delivered

Powder is poured from a source vial into a receiving cup. The two vessels are weighed before and after transfer. Source and receiver may use different balances. The source vial with contents reads $32.684\ \mathrm{g}$ before and $28.417\ \mathrm{g}$ after transfer. The receiving cup with contents reads $55.120\ \mathrm{g}$ before and $59.362\ \mathrm{g}$ after transfer. For each balance, a reading is $R=M+b+e$, where $M$ is the vessel-plus-contents mass, $b$ is an unknown offset that is constant between that vessel’s two weighings, and each reading error satisfies $|e|\leq0.002\ \mathrm{g}$. The offsets of the two balances need not be equal. Every combination of the allowed reading errors is possible.

Assume neither vessel changes mass. Apart from the powder transfer, there is no evaporation, adsorption, or other mass exchange. Ignore buoyancy and all effects not in the stated error bounds. Mass is additive in this model.

1. Find central estimates for the mass removed from the source, the mass delivered to the receiver, and the mass not delivered. Which difference measures the quantity actually collected?

2. Derive a hard interval for each estimate. Can zero undelivered mass be reconciled with the readings and error model?

3. Why does an unknown constant offset not spoil either vessel’s difference? Would that argument still work for a changing offset?

4. Does the observed discrepancy prove that powder spilled onto the bench, or that mass was destroyed? State what it establishes and suggest one measurement that could locate the missing powder.

Question 4 - Solution

Mass removed is not necessarily mass delivered

(a) Use the source for removal and the receiver for delivery.

$$m_{{\mathrm{removed}},0}=32.684-28.417=\boxed{4.267\ \mathrm{g}},$$ $$m_{{\mathrm{delivered}},0}=59.362-55.120=\boxed{4.242\ \mathrm{g}}.$$ The receiver’s increase measures the collected powder. The central undelivered amount is their difference: $\boxed{m_{{\mathrm{lost}},0}=4.267-4.242=0.025\ \mathrm{g}}$. “Lost” here means absent from the receiver, not destroyed.

(b) Bound differences by the extreme allowed errors.

Each two-reading difference has error $e_1-e_2$, between $-0.004$ and $+0.004\ \mathrm{g}$. Hence $$\boxed{m_{\mathrm{removed}}\in[4.263,4.271]\ \mathrm{g}},\qquad \boxed{m_{\mathrm{delivered}}\in[4.238,4.246]\ \mathrm{g}}.$$ For the loss, use the smallest removal with the largest delivery, and conversely: $$\boxed{m_{\mathrm{lost}}\in[4.263-4.246,\,4.271-4.238]\ \mathrm{g} =[0.017,0.033]\ \mathrm{g}}.$$ The interval excludes zero. Within all supplied assumptions, at least 17 mg left the source without entering the receiver. Hard bounds are combined by allowed extremes, not by a statistical rule.

(c) Common offsets cancel, but offset changes do not.

$$\Delta R=\Delta M+\Delta b+\Delta e,\qquad\Delta b=0\ \Longrightarrow\ |\Delta R-\Delta M|\le0.004\ \mathrm g.$$

For either vessel, $R_2-R_1=M_2-M_1+e_2-e_1$: the same $b$ subtracts away. Thus the two balances may have different constant offsets. If the offset changes, the difference also includes $b_2-b_1$, which requires its own correction or bound.

(d) A mass discrepancy does not identify a destination.

The data locate the loss relative to the two chosen vessel boundaries. Powder could remain on a funnel or transfer tool, become airborne, or spill elsewhere. Weighing a transfer funnel before and after use, with an error bound small enough to resolve tens of milligrams, could test one destination. The data establish neither destruction of mass nor a unique location for it.