A Floating Object and an Impossible Density
Students receive an irregular object with a small internal cavity and a pinhole opening. Its mass is 67.4 g. It floats in water. Using a 100-mL graduated cylinder, they record 50.0 mL of water, add the object, and read 58.5 mL. They calculate volume as 8.5 cm3 and density as 67.4/8.5 = 7.9 g/cm3, then claim the object is iron, whose density is about 7.87 g/cm3. The cylinder markings have an uncertainty of ±0.5 mL. When the object is placed in a transparent tank, about one-third of it sits above the waterline. A few air bubbles escape from the pinhole when it is first submerged. The teacher asks the class to reconcile the floating behavior with the calculated density and to design a better measurement.
1. Explain why the calculated 7.9 g/cm3 cannot be the average density of an object that floats freely with one-third of its volume above water. What does this imply about using 8.5 mL as the object's total volume?
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A freely floating object has an average density less than water, about 1.0 g/cm3. For a 67.4-g floating object, the buoyant force must equal its weight, so it must displace about 67.4 mL of water. The cylinder rose only 8.5 mL, which would displace only 8.5 g of water, far too little to support 67.4 g. Therefore the 8.5 mL cannot be the total outer volume of the floating object. Either the object touched the bottom or side of the cylinder, or water entered the cavity and the reading measured only part of the object's material volume, or the volume reading is wrong. The 7.9 g/cm3 calculation uses a volume that is not the whole-object volume, so it cannot establish that the object is iron. The conflicting evidence must be resolved, not averaged.
2. Describe a procedure to measure the object's total volume including the cavity, and explain how to reduce the effect of the pinhole and trapped air. Include an uncertainty estimate.
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Seal the pinhole with a thin waterproof coating or glue and allow it to cure, then use an overflow can filled exactly to the spout. Lower the sealed object gently with a thin wire until it is fully submerged, and collect the displaced water in a graduated cylinder large enough to hold it. Read the volume at eye level and repeat three times. Sealing prevents water from entering the cavity and prevents trapped air from escaping, so the displaced volume represents the outer volume. Estimate uncertainty from the cylinder markings, about ±0.5 mL per reading, plus the coating volume, perhaps ±0.2 mL; report the mean and range. If coating is not allowed, use fine dry sand displacement in a graduated container, but account for packing differences.
3. A sealed coating trial displaces 100.0 mL. Calculate the average density, decide whether the iron claim is supported, and explain how this result fits the floating observation.
Read the worked answer
Average density = mass/volume = 67.4 g / 100.0 cm3 = 0.674 g/cm3. This is less than 1.0 g/cm3, so the object should float, which it does. The result does not support the iron claim, because iron's density is about 7.87 g/cm3; the object is much less dense overall, consistent with a hollow or low-density body. The earlier 7.9 g/cm3 came from dividing mass by a volume that was not the whole-object volume, probably because water entered the cavity or because the object was not freely floating. The sealed trial gives the average density including the cavity, which is the relevant quantity for floating and sinking.