The 'Copper' Wire and a Measuring Cylinder That Cannot Tell
A school science club is preparing an open-day display called 'Spot the Fake'. A local electrician donates a coil of bare wire and says it is copper; the club suspects it may be a cheaper alloy. Available equipment: a top-pan balance reading to 0.1 g, a 100 cm3 measuring cylinder graduated in 2 cm3 steps, a 30 cm ruler, thread and tap water. A 20.0 g length is tied with thread and lowered gently into water reading 42.0 cm3; the level rises to 44.5 cm3. A second 20.0 g sample from the same coil, wound more loosely, takes the level from 40.0 cm3 to 47.0 cm3, but the students can see tiny bubbles clinging to the metal. Densities: copper 8.9 g/cm3, brass 8.4 g/cm3, aluminium 2.7 g/cm3. The display card must be printed before lunch, and the club leader insists they should only make a claim that the measurements actually support.
1. Calculate the density from the first trial and state precisely what it does and does not rule out.
Read the worked answer
Calculation (trial 1)
- Volume displaced = 44.5 - 42.0 = 2.5 cm3
- Density = m/V = 20.0 / 2.5 = 8.0 g/cm3
What it seems to say: 8.0 is between brass (8.4) and copper (8.9), so a careless reading would confirm the electrician's claim.
What it actually supports: each reading can be misread by about half a graduation, i.e. ±0.5 cm3. Worst cases: volume 1.5 cm3 gives 13.3 g/cm3; volume 3.5 cm3 gives 5.7 g/cm3. The uncertainty is about 40% of the measurement, so the result cannot separate copper from brass at all, and cannot even exclude a mixture. The one thing it does rule out is solid aluminium, which would need 20.0/2.7 = 7.4 cm3 of displacement (a rise to about 49.4 cm3) - far outside the reading.
Judgement: a value with a ±2.6 g/cm3 spread is not evidence for a public claim. The trial looks precise because it is written to one decimal place, but the instrument, not the balance, sets the precision.
2. Decide which of the two trials the club should put on the display card, and justify the choice using the direction of the bubble error.
Read the worked answer
Trial 2: volume = 47.0 - 40.0 = 7.0 cm3, density = 20.0/7.0 = 2.9 g/cm3.
Trapped bubbles make the measured volume too large, so the calculated density is too small. The bubble error therefore pushes the answer for copper (8.9) downwards - but not nearly far enough. To get 8.9 g/cm3 the volume would have to be 20.0/8.9 = 2.2 cm3, not 7.0 cm3. The discrepancy is 4.8 cm3; a few clinging bubbles might account for 0.2-0.5 cm3. The claim of copper cannot be rescued by blaming the bubbles.
Trial 2 is also the more trustworthy measurement because the displacement (7.0 cm3) is three graduations rather than barely one, so a ±0.5 cm3 reading uncertainty is about 14% instead of 40%.
Judgement on trial 1: a 2.5 cm3 displacement is suspiciously small for the same 20.0 g. If the true volume is about 7 cm3, the level should have risen from 42.0 to about 49.0. The likely explanation is that the tightly wound coil trapped air or part of it sat above the water line, giving a volume that is too small and a density that is too high - an error in the opposite direction to trial 2.
Honest card: 'Our best measurement gives about 2.9-3.1 g/cm3, close to aluminium (2.7), not copper (8.9).'
3. The club has one afternoon left. Recommend the single change that would most improve their conclusion, and explain why it beats simply repeating the cylinder trial.
Read the worked answer
Recommendation: find the volume geometrically instead of by displacement.
Method with the equipment they already have:
- Straighten and measure a long length L of wire with the 30 cm ruler (e.g. 1.00 m, read to ±1 mm, about 0.1% uncertainty).
- Wind 30 turns tightly side by side on a pencil and measure the width of the coil with the ruler. Width 30 mm means diameter = 1.0 mm, so radius = 0.5 mm = 0.05 cm.
- Volume = pi x r^2 x L = 3.14 x (0.05)^2 x 100 = 0.785 cm3. Mass on the balance, then density = m/V.
Why this is better: every reading is large compared with the smallest division of the ruler, and there is no trapped air. Uncertainty falls from roughly ±30% to a few per cent - enough to separate 8.9 from 8.4 from 2.7.
Why repeating the cylinder trial does not help: repeating reduces random scatter but the graduation limit and the air-bubble fault are systematic. Ten repeats of trial 1 would still give about 8.0 g/cm3 with the same ±0.5 cm3 reading limit. A bigger sample (say 100 g of wire, about 35 cm3) or an overflow can would also work, but the geometric method is faster and uses only what is on the bench.