GLOBAL KNOWLEDGE ARK

Materials Science · Resources, reuse, recycling and lifecycle evidence

Resource extraction and material supply pathways

Useful background: Energy, stability and environmental context; Measuring and checking material claims. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

  • Trace material inputs beyond the final material mass.
  • Distinguish feed grade, element recovery, product purity and manufacturing yield.
  • Balance a target element through concentration, refining, shaping and an explicitly bounded return loop.
  • Compare matched outputs without equating gross feed mass, recycled content or assigned energy with environmental performance.
01

A finished kilogram has an upstream pathway

Extraction, concentration, refining, synthesis, transport and manufacture can require energy and additional materials. Different routes to a similar final composition can have different resource and environmental consequences. The product mass alone does not describe those pathways.

A copper-bearing rock is not a block of copper metal diluted with loose dirt. Copper can be chemically bound in mineral grains mixed with other minerals. Crushing and grinding can expose those grains; physical separation such as flotation can enrich a copper-bearing mineral fraction. This concentration step changes how much other material accompanies the copper. It does not by itself convert the mineral into pure metal.

In a sulfide-ore pathway, smelting involves chemical conversion and separation, followed by further refining. Electrorefining transfers copper from an anode through an electrolyte to a cathode by electrochemical reactions. Oxidized ores can follow a different pathway involving leaching. These are examples of different routes, not a claim that every metal or copper deposit uses one sequence. Extraction, separation and conversion must be identified before a final composition is compared.

Follow a named quantity through that route. Total dry ore, copper contained in that ore and refined copper are different quantities. A stream containing several kilograms of mineral matter may carry only a fraction of a kilogram of the target metal. Adding masses without their composition and moisture basis can therefore produce a plausible-looking but meaningless comparison.

The calculations below use a fictional target metal M. They are original conservation examples, not copper plant measurements or expected industrial yields. All accepted products are independently stipulated to meet the same pure-M product specification at the factory gate. Equal mass is a matched output here because that specification is fixed; it would not establish equal service for products of different quality or function.

02

Keep the fraction and its denominator together

Grade g is the mass of target M divided by the dry feed mass. For dry ore mass O with g = 0.0200, the contained target mass is gO. Concentration recovery R_c is the fraction of that contained target which reaches the concentrate: m_c = R_c gO. Its denominator is target mass entering the step, not total ore mass.

Concentrate grade c is a different fraction: target mass in the concentrate divided by total dry concentrate mass. Thus dry concentrate mass C = m_c/c. A concentrate can be 20% M while recovering 90% of the M originally in the ore. Those statements are compatible because they divide by different masses.

Refining recovery R_f = 0.950 means 95% of the target M entering refining reaches the new refined stock. It does not mean that the stock is 95% pure. Purity describes product composition and is independently specified here. Target-metal losses from refining are (1 − R_f)m_c; other elements, process additions and chemical products need separate accounts.

Supply g = 0.0200, R_c = 0.900 and R_f = 0.950. If new refined target mass N is required, work backward: N = R_f R_c gO, so O = N/(gR_cR_f). The denominator is dimensionless; kilograms of M divided by that product give kilograms of dry ore. For N = 1.25 kg, this gives 73.0994 kg ore. Using only O = N/g would omit both element losses.

Concentration residues are called tailings. In this invented dry separation, no solids are added and the two dry outputs are concentrate and tailings, so their masses sum to O. That is a deliberately limited total-dry-mass balance. A real refining or smelting plant also uses such inputs as oxygen, fluxes, fuels and water, and can produce gases, dust and slag. Conserving M alone neither balances those streams nor measures their consequences.

03

A return loop changes new input, not the meaning of throughput

Let accepted product mass be P = 1.00 kg. Shaping yield y = 0.800 is the accepted product mass divided by the gross pure-M mass passing through shaping. Consequently G = P/y = 1.25 kg passes through shaping and (1 − y)G = 0.250 kg becomes process scrap. Gross throughput counts material passing through that operation, including material that has passed through it before.

Now assume steady operation with no changing stock. Of the generated process scrap, a fraction r = 0.750 successfully returns as clean, specification-matched shaping feed. Returned target mass is R = r(1 − y)G = 0.1875 kg. The successful return is stipulated to have no further target-metal loss; collection alone would not establish it. Processing that return can still use energy and other materials.

New refined feed N plus internal return R must supply G, so N = G − R = P[1 − r(1 − y)]/y = 1.0625 kg. The unrecovered shaping scrap is (1 − r)(1 − y)G = 0.0625 kg. Looking across the whole shaping-and-return boundary gives the same result directly: N = P + 0.0625 kg. Do not add internal return again as a new external input.

For this steady operating account, the returned material can come from preceding operation of the same process. No product is credited with an unobserved future end-of-life return. Nor is this 75% fraction a claim about postconsumer recycled content: it refers only to the generated shaping scrap in the stated boundary. Scrap produced during manufacture, material recovered after product use and same-plant process return can have different reporting boundaries.

With the same upstream grade and recoveries, O = 1.0625/(0.0200 × 0.900 × 0.950) = 62.1345 kg dry ore. Returning 75% of the scrap reduces new metal and ore demand by 15% relative to no return, because scrap was only 20% of gross shaping input. The gross shaping throughput remains 1.25 kg per kilogram of product. At perfect successful return, r = 1, new refined M still equals the 1 kg leaving as product; the return loop does not create that metal.

One kilogram of matched pure-M product. Exact fictional fractions define a steady account; displayed masses are rounded. Both routes use identical grade, step recoveries and shaping yield.
Quantity and basisNo process return75% of shaping scrap returns
Gross shaping input G (kg M)1.25001.2500
Internal successful return R (kg M)00.1875
New refined feed N (kg M)1.25001.0625
Unrecovered shaping scrap (kg M)0.25000.0625
Dry ore O (kg total dry matter)73.099462.1345
M lost to tailings (kg M)0.1461990.124269
M lost at refining (kg M)0.06578950.0559211
Follow target M through the return loopOriginal fictional steady pathway for one kilogram of accepted pure-M product. Dry ore is 62.1345 kilograms containing 1.242690 kilograms M. A concentration step recovering 90 percent of target M produces 5.592105 kilograms dry concentrate containing 1.118421 kilograms M; 56.5424 kilograms dry tailings contain 0.124269 kilograms M. Refining recovers 95 percent of incoming target M, supplying 1.0625 kilograms new refined M and leaving 0.0559211 kilograms M outside that stock. New stock mixes with 0.1875 kilograms successfully returned M to supply 1.25 kilograms gross shaping throughput. Shaping at 80 percent product yield supplies 1.00 kilogram product and 0.2500 kilograms generated scrap. 75 percent of this scrap returns without further target loss; 0.0625 kilograms M leaves the loop unrecovered. Solid arrows show the forward path, dashed arrows successful return and dotted arrows M not reaching the next main output. Arrow widths, distances and box sizes are schematic, not proportional to mass. Only the concentration dry solids and target-M accounts close here; other refining inputs and chemical output streams are not quantified. All values are invented, not plant observations.Follow target M through the return loopFictional steady account; P = 1.00 kg accepted pure MArrows and box sizes are schematic, not mass-scaled.Dry ore: 62.1345 kgContained M: 1.242690 kgConcentrationR_c = 0.900; grade c = 0.200Dry concentrate: 5.592105 kgContained M: 1.118421 kgRefiningR_f = 0.950 of incoming MNew refined M: 1.0625 kgN: external input to the loopGross shaping: G = 1.25 kg MProduct yield y = 0.800Accepted productP = 1.00 kg pure MGenerated scrap0.2500 kg MTailings56.5424 kg dry0.124269 kg MOutside new stock0.0559211 kg MR = 0.1875 kg M75% of generated scrapUnrecovered: 0.0625 kg MForwardReturnOther M outputInternal return is counted once; no stock change is assumed.
Original fictional steady pathway for one kilogram of accepted pure-M product. Dry ore is 62.1345 kilograms containing 1.242690 kilograms M. A concentration step recovering 90 percent of target M produces 5.592105 kilograms dry concentrate containing 1.118421 kilograms M; 56.5424 kilograms dry tailings contain 0.124269 kilograms M. Refining recovers 95 percent of incoming target M, supplying 1.0625 kilograms new refined M and leaving 0.0559211 kilograms M outside that stock. New stock mixes with 0.1875 kilograms successfully returned M to supply 1.25 kilograms gross shaping throughput. Shaping at 80 percent product yield supplies 1.00 kilogram product and 0.2500 kilograms generated scrap. 75 percent of this scrap returns without further target loss; 0.0625 kilograms M leaves the loop unrecovered. Solid arrows show the forward path, dashed arrows successful return and dotted arrows M not reaching the next main output. Arrow widths, distances and box sizes are schematic, not proportional to mass. Only the concentration dry solids and target-M accounts close here; other refining inputs and chemical output streams are not quantified. All values are invented, not plant observations.
04

Compare a purchased secondary pathway on the same output

Consider an alternative source outside this internal loop: purchased recovered feed with target-M mass fraction q = 0.800 on a dry basis. Its independently supplied target recovery through sorting and refining is R_s = 0.900. The resulting pure-M stock is stipulated to meet the same product specification. The same shaping yield and successful return apply, so it must supply the same N = 1.0625 kg of new refined M.

Required purchased dry mass is S = N/(qR_s) = 1.475694 kg. It contains qS = 1.180556 kg M, of which 0.118056 kg does not reach refined stock. Its other 0.295139 kg is non-M feed matter. That last quantity is not target-metal loss, and its destination cannot be inferred from the target-M balance.

The primary pathway requires 62.1345 kg dry ore and this alternative requires 1.475694 kg purchased dry feed. This compares two named external feed masses for matched output; it is not a complete comparison of resource extraction or environmental impact. Purchased feed has a collection and prior-use history, and reprocessing has its own streams and energy requirements. Its much higher starting M fraction explains much of the gross-mass difference.

A lower purchased-feed recovery increases required feed even if its grade is unchanged. A higher concentrate grade can reduce transported concentrate mass without reducing ore demand if concentration recovery stays fixed. These are separate levers. In an actual process, grade, recovery, energy and acceptable composition may be coupled, so changing one parameter independently is a conditional calculation that needs evidence before being used as a forecast.

Successful return and new upstream demandOriginal conditional guide, not an observed recycling curve. Successful return r is the percentage of generated shaping scrap that reaches the same process as accepted pure-M feed, without additional target loss. For fixed one-kilogram product, shaping yield 0.800, ore grade 0.0200, concentration recovery 0.900 and refining recovery 0.950, dry ore demand is O = [1−0.2 r]/[0.8×0.0200×0.900×0.950] kilograms when r is written as a fraction. The horizontal linear axis runs 0–100 percent and the vertical linear ore-mass axis runs 0–80 kilograms. A circle marks no return at 73.0994 kilograms ore; a diamond marks 75 percent successful scrap return at 62.1345 kilograms; a triangle marks 100 percent at 58.4795 kilograms. The 75 percent case saves 15 percent of ore relative to no return, not 75 percent. Gross shaping throughput stays 1.25 kilograms M for every point; new refined M declines from 1.25 to 1.00 kilograms. This fixed-parameter relation does not establish real recovery feasibility, energy demand, certified recycled content or environmental benefit.Successful return and new upstream demandFixed output, grade, step recoveries and shaping yieldDry ore demand (kg per 1.00 kg accepted product)020406080025507510073.099462.134558.4795Successful return (% of generated shaping scrap)At r = 75%: 15% less ore than at r = 0Gross shaping input G = 1.25 kg M throughout.At r = 100%, new M still replaces 1.00 kg product.A mass result alone is not an environmental ranking.
Original conditional guide, not an observed recycling curve. Successful return r is the percentage of generated shaping scrap that reaches the same process as accepted pure-M feed, without additional target loss. For fixed one-kilogram product, shaping yield 0.800, ore grade 0.0200, concentration recovery 0.900 and refining recovery 0.950, dry ore demand is O = [1−0.2 r]/[0.8×0.0200×0.900×0.950] kilograms when r is written as a fraction. The horizontal linear axis runs 0–100 percent and the vertical linear ore-mass axis runs 0–80 kilograms. A circle marks no return at 73.0994 kilograms ore; a diamond marks 75 percent successful scrap return at 62.1345 kilograms; a triangle marks 100 percent at 58.4795 kilograms. The 75 percent case saves 15 percent of ore relative to no return, not 75 percent. Gross shaping throughput stays 1.25 kilograms M for every point; new refined M declines from 1.25 to 1.00 kilograms. This fixed-parameter relation does not establish real recovery feasibility, energy demand, certified recycled content or environmental benefit.
05

Physical balances and assigned burdens answer different questions

An output that is sold or used as a coproduct is not automatically a discarded loss. First record what physically leaves each process and which product specifications it meets. Only then ask how shared upstream requirements are to be associated with the products. Conservation alone does not select that accounting rule.

For a separate fictional joint process, suppose the measured delivered-energy input is 100 MJ and the outputs are 1.00 kg M and 0.250 kg saleable Q, with no other products in this accounting example. A declared allocation proportional to output mass assigns 100 × 1.00/(1.00 + 0.250) = 80.0 MJ to M. If the two outputs instead receive equal shares of total sales revenue, a declared revenue-based allocation assigns 50.0 MJ to M. In either account the joint physical process still used 100 MJ; assigned energy is not a separate meter reading.

These assignments are examples of stated choices, not instructions that either is universally preferred. They describe sharing an existing inventory. Claiming that Q avoids production elsewhere asks a further question about which alternative production actually changes; it needs an explicit comparison and evidence, not an automatic negative entry. Functional units and lifecycle boundaries develops the broader comparison task.

Even a complete delivered-energy inventory would not by itself measure climate effects, water scarcity, land disturbance, toxicity or supply reliability. Which electricity and fuels supplied the stages, where water was used and how residues were managed matter to those additional questions. A low ore mass, high return fraction or low allocated energy can be useful evidence without being a universal environmental score.

06

State the system and data context

Identify feedstock, geography, energy mix, process yield and relevant coproduct treatment. Do not turn one regional or historical dataset into a universal property of the material. Compare pathways delivering the same specified function.

For the numerical pathway, draw the boundary from the stated dry feed to the factory-gate product, including the stated shaping return. Ore-development overburden, upstream energy supply, transport, service and end-of-life are not quantified. Record those exclusions instead of silently treating them as zero. If stock is accumulating, add its change to input-minus-output; the steady loop equation would no longer be the complete account.

An evidence record should pair a representative mass measurement with an assay for each relevant stream, state whether masses are dry or include moisture, and relate inputs and outputs over a matched period. Residue assays test the target-metal losses; product analysis tests specification; inventories and successful return records test the loop boundary. A high weighed collection rate alone does not demonstrate a high successful metal return.

Use changed observations to discriminate explanations. If concentrate dry mass halves while its target mass and final output remain fixed, enrichment has changed; ore demand need not have fallen. If weighed return rises but new refined demand is unchanged, check stock changes, rejected return, changed product yield and the accounting period before claiming a conservation violation. A specific balance points to missing evidence; it does not uniquely identify the industrial cause.

Worked examples

Work through the case

Two routes produce one kilogram of the same alloy but have different scrap yields. Are equal output masses enough to compare resource input?

  1. Trace input mass and usable output.
  2. Include recoverable and unrecovered streams.
  3. Retain the boundary and recovery assumptions.

No. Upstream requirements and losses can differ.

Work backward from a matched product

With no process return, produce 1.00 kg pure-M product at shaping yield 0.800. Ore grade is 0.0200 M, concentration recovery 0.900, concentrate grade 0.200 and refining recovery 0.950. Find dry ore and concentrate requirements and check the target-M balance.

  1. Shaping needs G = 1.00/0.800 = 1.25 kg pure M. With no return, this is also new refined demand N.
  2. Before refining, the concentrate must carry m_c = 1.25/0.950 = 1.315789 kg M. Before concentration, ore must carry 1.315789/0.900 = 1.461988 kg M.
  3. Divide by the appropriate grades: O = 1.461988/0.0200 = 73.0994 kg dry ore; C = 1.315789/0.200 = 6.578947 kg dry concentrate. Dry tailings mass is O − C = 66.5205 kg under the specified no-added-solids concentration boundary.
  4. The target losses are 1.461988 − 1.315789 = 0.146199 kg in tailings, 1.315789 − 1.25 = 0.0657895 kg at refining and 1.25 − 1.00 = 0.250 kg at shaping. Their sum plus product is 1.461988 kg, the target mass entering in ore.

The route needs about 73.0994 kg dry ore and 6.578947 kg dry concentrate per 1.00 kg accepted product. Grade controls stream mass; step recovery controls how much M survives.

A large scrap-return fraction can give a smaller input saving

Keep the preceding product and process parameters, but successfully return 75.0% of generated shaping scrap without further M loss. How much new refined M and ore are required? Is the ore saving 75%?

  1. Gross shaping input is still 1.25 kg and generated scrap is 0.250 kg. Successful return is 0.750 × 0.250 = 0.1875 kg.
  2. New refined M is 1.25 − 0.1875 = 1.0625 kg. Equivalently, it replaces the 1.00 kg product and 0.0625 kg unrecovered shaping scrap leaving the loop.
  3. Upstream ore is 1.0625/(0.0200 × 0.900 × 0.950) = 62.1345 kg. Concentrate target mass is 1.118421 kg M and dry concentrate is 5.592105 kg. Tailings contain 0.124269 kg M; refining loses 0.0559211 kg M.
  4. Compare unrounded ore demands: (73.099415 − 62.134503)/73.099415 = 0.150. The 75% return acted on the 20% scrap share of gross throughput. Global target closure is 1.242690 = 1.000000 + 0.124269 + 0.055921 + 0.062500 kg, within rounding.

New refined demand is 1.0625 kg and dry ore demand about 62.1345 kg. The model saving is 15.0%, while gross shaping input remains unchanged.

More concentrated does not necessarily mean less ore

For the 75%-return route, a separate process change raises concentrate grade from 0.200 to 0.400. Concentration recovery, refining recovery and output are independently unchanged. Which masses change?

  1. The downstream product and return account still require N = 1.0625 kg new refined M. Refining therefore still needs 1.0625/0.950 = 1.118421 kg M in concentrate.
  2. The concentrate total dry mass changes from 1.118421/0.200 = 5.592105 kg to 1.118421/0.400 = 2.796053 kg. The target mass has not doubled.
  3. Concentration recovery and ore grade are fixed, so ore is still 1.118421/(0.900 × 0.0200) = 62.1345 kg. Under the same no-added-solids separation assumption, dry tailings become 62.134503 − 2.796053 = 59.338450 kg rather than 56.542398 kg.
  4. The lower concentrate transport mass could matter to a further transport inventory, but no energy coefficient or distance is supplied. The independent-recovery assumption must be checked for a real process change.

Concentrate dry mass halves; ore demand and contained M in concentrate stay fixed. More non-M matter is separated into tailings in this conditional account.

Separate composition from recovery in a purchased feed

The same 1 kg product and shaping loop require 1.0625 kg new refined M. Purchased dry feed is 80% M and its M recovery is 90%. If recovery falls to 80% while all downstream conditions and product quality remain fixed, what changes?

  1. Initially S = 1.0625/(0.800 × 0.900) = 1.475694 kg. Feed contains 1.180556 kg M, so 0.118056 kg M does not reach refined stock.
  2. At lower recovery S = 1.0625/(0.800 × 0.800) = 1.660156 kg. Feed contains 1.328125 kg M, of which 0.265625 kg does not reach refined stock.
  3. Required feed rises by a factor 0.900/0.800 = 1.125, or 12.5%. A ten-percentage-point decrease in recovery is not a ten-percent increase in required input.
  4. The product and shaping return remain the same by construction. Whether the additional feed, residues or processing energy make this route preferable to a different route needs a broader matched comparison.

Purchased feed rises from 1.475694 to 1.660156 kg, a 12.5% increase. Target loss rises from 0.118056 to 0.265625 kg; no environmental ranking follows from these masses alone.

Try the reasoning

  1. Is recycled feedstock free of processing requirements?

    Show the answer and reasoning

    No.

    Collection, sorting and reprocessing still require resources.

  2. Why record data year?

    Show the answer and reasoning

    Processes and supply conditions can change.

    The comparison is context-dependent.

  3. A 10.0 kg dry feed contains 5.0% M. Concentration recovers 80% of that M into a 40%-M concentrate. Find the concentrate dry mass and M lost to tailings.

    Show the answer and reasoning

    1.00 kg dry concentrate; 0.100 kg M in tailings.

    Feed contains 0.500 kg M. Recovered target is 0.800 × 0.500 = 0.400 kg, which occupies 0.400/0.400 = 1.00 kg concentrate. The unrecovered 0.100 kg M is distinct from the 9.00 kg total dry tailings under a no-added-solids separation.

  4. In the 75%-return primary route, ore grade falls from 2.00% to 1.00% M. Independently keep every recovery, shaping parameter and output fixed. What dry ore mass is required?

    Show the answer and reasoning

    124.269 kg dry ore, twice the original 62.1345 kg.

    O is inversely proportional to g when the other stated quantities are fixed. This is a conditional consequence, not evidence that a real lower-grade deposit has the same recovery, energy or processing route.

  5. Produce 2.00 kg product at y = 0.800, but successfully return only 50.0% of shaping scrap. Keep g = 0.0200, R_c = 0.900 and R_f = 0.950. Find new refined M and dry ore.

    Show the answer and reasoning

    2.250 kg new refined M and 131.579 kg dry ore.

    G = 2.00/0.800 = 2.50 kg. Generated scrap is 0.500 kg and return is 0.250 kg, leaving N = 2.250 kg. Divide N by 0.0200 × 0.900 × 0.950. Doubling the old 75%-return result would incorrectly preserve a condition that changed.

  6. Can perfect successful process-scrap return, r = 1, make the new refined input zero for the 1.00 kg product? Calculate its dry ore requirement in the same model.

    Show the answer and reasoning

    No. New refined M remains 1.00 kg; dry ore is 58.4795 kg.

    At steady state the product leaves the boundary. Returning all shaping scrap eliminates that loss but cannot replace the product metal: N = P. Upstream concentration and refining losses still require more contained M in ore than reaches the product.

  7. A supplier states a 95% target-metal refining recovery. Does that establish that its output is 95% pure or meets the product specification?

    Show the answer and reasoning

    Neither conclusion follows.

    Recovery divides target mass leaving successfully by target mass entering. Purity divides target mass in an output by total output mass. Product composition and acceptance evidence must be supplied separately.

  8. In the 75%-return example, what fraction of gross shaping input is internal return? Does it establish 75% postconsumer recycled content?

    Show the answer and reasoning

    15.0% of gross shaping input; no postconsumer-content conclusion.

    R/G = 0.1875/1.25 = 0.150. The stated 75% refers to generated process scrap, and no recovered used-product input has been specified. Possible future collection is not a present successful return.

  9. A joint process uses 100 MJ and supplies 1.00 kg M plus 0.250 kg Q. A mass allocation assigns 80 MJ to M, while an equal-revenue allocation assigns 50 MJ. Did switching the rule save 30 MJ at the plant?

    Show the answer and reasoning

    No. Physical input stays 100 MJ; the amount assigned to M changes by 30 MJ.

    The assignments answer a burden-sharing question under two declared rules. Neither is a separate meter reading, and neither proves avoided production elsewhere. Preserve the rule and the joint physical inventory.

  10. A shaping boundary receives 1.20 kg new M during an interval and releases 1.00 kg product plus 0.050 kg unrecovered scrap. No other target-M flow crosses it. What stock change closes the account? Can the steady N = P + loss equation be used unchanged?

    Show the answer and reasoning

    M stock increases by 0.150 kg; the steady equation is incomplete for this interval.

    Input minus output is 1.20 − 1.00 − 0.050 = 0.150 kg. Internal circulation does not cross this boundary. Accumulation must be included instead of treating the difference as an unexplained recovery improvement.

  11. Collected scrap mass rises, but new refined M per accepted product does not fall. Name two distinct explanations and the evidence that separates them.

    Show the answer and reasoning

    For example, the extra collection may be rejected or lost during reprocessing, or stored rather than returned during the matched interval.

    Weigh and assay successfully accepted return, measure reprocessing residues and reconcile stock changes over the same interval. Changed shaping yield or product specification is another possibility. A collection record alone does not identify successful return or its cause.

What to carry forward

  • Trace the supply route and losses.
  • Keep geography, time and boundaries visible.
  • Grade, recovery, purity and shaping yield divide different quantities; keep each denominator explicit.
  • Balance external target input against product, unrecovered losses and stock change; count an internal loop once.
  • A smaller feed mass or assigned burden is a bounded result, not a complete environmental ranking.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. Global Material Flows and Resource ProductivityUNEP / International Resource PanelOfficial report abstract and metadata, 2016wedocs.unep.orgRead source
  2. Copper Statistics and Information · OverviewU.S. Geological Survey / National Minerals Information CenterOfficial mineral pathway overviewusgs.govRead source
  3. T. G. Goonan / U.S. Geological Survey3 sources
    1. Flows of Selected Materials Associated with World Copper SmeltingT. G. Goonan / U.S. Geological SurveyOpen-File Report 2004–1395; selected indexed text · PDFpubs.usgs.govRead source
    2. Material flows generated by pyromet copper smeltingT. G. Goonan / U.S. Geological SurveyOfficial authored summary, Mining Engineering, 2005usgs.govRead source
    3. Copper Recycling in the United States in 2004T. G. Goonan / U.S. Geological SurveyCircular 1196-X, 2009; selected indexed text · PDFpubs.usgs.govRead source
  4. Critical minerals in subduction-related magmatic-hydrothermal systems of the United States · Copper–molybdenum processing passageU.S. Geological SurveyScientific Investigations Report 2023–5082; selected indexed textpubs.usgs.govRead source