The Plastic Cube and Four Liquids
A student is given an irregular plastic cube, a balance, and a 100 mL graduated cylinder. The cube's mass is 45.0 g. To find its volume, she adds 40.0 mL of water to the cylinder, tips the cylinder, and slides the cube under the surface. The water level rises to 90.0 mL. She also has four liquids in beakers: rubbing alcohol (0.79 g/cm³), vegetable oil (0.92 g/cm³), water (1.00 g/cm³), and corn syrup (1.38 g/cm³). The cube does not dissolve and does not soak up liquid. Based on the word "plastic," she predicts the cube will float in all four liquids.
1. Calculate the cube's density, then predict whether it floats or sinks in each of the four liquids. Explain each prediction.
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Volume = 90.0 mL − 40.0 mL = 50.0 cm³. Density = 45.0 g ÷ 50.0 cm³ = 0.90 g/cm³.
The rule: an object floats when its density is less than the liquid's and sinks when it is greater.
- Rubbing alcohol (0.79): 0.90 > 0.79 → the cube sinks.
- Vegetable oil (0.92): 0.90 < 0.92 → the cube floats, but barely. Only about 0.90/0.92 ≈ 98% of the cube is underwater, so it looks almost sunk, with just a sliver above the surface.
- Water (1.00): floats, with about 90% submerged.
- Corn syrup (1.38): floats high, with only about 65% submerged.
Her prediction is wrong in alcohol and misleading in oil: "plastic" is not a property that decides floating - density compared with the liquid does.
2. She cuts the cube into two equal pieces. She says, "Now each piece is lighter, so each piece will float in rubbing alcohol." Evaluate her reasoning.
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Her reasoning is incorrect. Cutting the cube in half halves the mass and halves the volume: each piece is 22.5 g and 25.0 cm³.
Density = 22.5 g ÷ 25.0 cm³ = 0.90 g/cm³ - exactly the same as before. Density is a property of the material, not of how big the sample is.
Since 0.90 g/cm³ is still greater than alcohol's 0.79 g/cm³, both halves sink in rubbing alcohol, just like the whole cube. The pieces are lighter, but they are also smaller, so the comparison that actually matters - density of cube versus density of liquid - has not changed.
3. The student wants the cube to hang suspended in the middle of a liquid instead of floating or sinking. Describe how she could make that happen and explain why it works.
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She needs a liquid whose density is exactly 0.90 g/cm³ - the same as the cube. Then the buoyant force pushing up equals the cube's weight pulling down, so the cube neither rises nor sinks; it stays where it is placed.
She cannot get there with any one of the four liquids (0.79, 0.92, 1.00, 1.38). She can get there by mixing: combine rubbing alcohol (0.79) with water (1.00), which bracket 0.90, adding one to the other and testing with the cube until it stays suspended. Salt would only push water's density higher, away from 0.90.
Caveat worth stating: the match must be made at a steady temperature, because warming a liquid usually lowers its density and would make the cube drift upward.