Two density readings that bracket the answer
A Grade 9 student is asked to find the density of an irregular piece of limestone she picked up from the playground. She records the following. Mass on an electronic balance: 84.6 g (the balance reads to 0.1 g). Water in a 100 cm³ measuring cylinder before the stone is added: 50.0 cm³ (smallest division 1 cm³, read to the nearest half division). Water level after sliding the dry stone in: 78.0 cm³. She calculates 84.6 ÷ 28.0 = 3.02 g/cm³. Her textbook gives 2.6–2.8 g/cm³ for limestone. She dries the surface of the same stone, lowers it into a fresh 50.0 cm³ of water, and this time reads 84.0 cm³, giving 2.49 g/cm³. She notices that in the first trial the stone released small bubbles that clung to it, and that afterwards it felt heavier. A classmate tells her to simply write 2.7 g/cm³ because "that is the correct answer".
1. Calculate the density from each trial. Using the two competing errors - water soaking into the pores and air bubbles clinging to the surface - explain why one result lies above the reference range and the other below it.
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Trial 1: volume = 78.0 − 50.0 = 28.0 cm³, so density = 84.6 ÷ 28.0 = 3.02 g/cm³.
Trial 2: volume = 84.0 − 50.0 = 34.0 cm³, so density = 84.6 ÷ 34.0 = 2.49 g/cm³.
The two errors push the measured volume in opposite directions, so one error cannot explain both results.
- Trial 1 used a dry, porous stone. Some water soaked into open pores instead of being displaced, so the level rose by less than the stone's true outer volume. Volume too small → density too large (3.02 is above 2.8).
- Trial 2 used the same stone already saturated, so it could absorb little more water, but bubbles stayed trapped against the rough surface. Bubbles displace water without adding mass, so the level rose by more than the stone's volume. Volume too large → density too small (2.49 is below 2.6).
A true value of 2.7 g/cm³ would need a volume of about 31.3 cm³, which sits neatly between 28.0 and 34.0. The fact that both readings bracket the reference range is the real evidence for this explanation.
2. Evaluate the classmate's advice. Describe a modified procedure that would give a value the student can defend.
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Writing 2.7 g/cm³ because the book says so is fabricating a result. The marks in a practical assessment come from the method, the recording and the treatment of error; a copied value contradicts her own measurements and cannot be defended if she is asked how she obtained it.
A defensible procedure would deal with both errors at once:
- Saturation: soak the stone in water for about ten minutes until bubbles stop, blot the surface, then use the saturated mass with the saturated volume, so no water is absorbed during the measurement.
- Bubbles: use boiled-and-cooled (deaerated) water, tilt the cylinder and slide the stone in slowly, then tap the cylinder gently to dislodge bubbles before reading.
- Resolution: use a 25 cm³ or 50 cm³ cylinder instead of 100 cm³ so the same volume change spans more scale divisions.
- Repetition: take three readings and average, discarding any obvious outlier with a stated reason.
Alternatively, coat the stone thinly in wax (weighing the wax separately so it can be subtracted) or collect the displaced water in an overflow can and weigh it - both avoid reading the cylinder at all.
3. Each reading of the cylinder is uncertain by ±0.5 cm³. Estimate the uncertainty in the final density and decide whether reading error alone could explain the gap between 3.02 and 2.49.
Read the worked answer
The volume is a difference of two readings, so the uncertainties combine: worst case 0.5 + 0.5 = ±1.0 cm³ (adding in quadrature gives ±0.7 cm³; the worst case is the safer estimate at this level).
- Trial 1: ±1.0 in 28.0 cm³ is about 3.6%, giving 3.02 ± 0.11 g/cm³.
- Trial 2: ±1.0 in 34.0 cm³ is about 2.9%, giving 2.49 ± 0.07 g/cm³.
- The mass uncertainty, ±0.1 g in 84.6 g, is about 0.1% and can be ignored.
The gap between the two results is 3.02 − 2.49 = 0.53 g/cm³, roughly five times the combined reading uncertainty. Random reading error therefore cannot account for it: there must be a systematic error, and the porosity and bubble explanation supplies exactly that. The correct conclusion is not "the readings are unreliable" but "the readings are precise enough to reveal a systematic fault in the method", which is why the method, not the arithmetic, must be changed.