A 0.190-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is

Question

A 0.190-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.90 cm . The level of the glycerine is well below the top of the cylinder.

Part A

If the ice completely melts, by what distance does the height of liquid in the cylinder change?

Express your answer with the appropriate units.

Part B

Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted?

Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted?

Explanation / Answer

Part A)   Here, weight of the displaced liquid equals the weight of the cube

=> mass = density*volume

=>   0.190 kg = 1.26 kg/L * V

=>   V = 0.1508 L

=> the new volume in the cylinder after addition of the cube is the initial volume + the immersed volume
=>    V = Vi + 0.1508

=> The volume difference before, and after the melting = 0.190 - 0.1508 = 0.0392 L

=>   h = V/(pir2)

=>   h   =   39.2/(3.14 * 3.902)

           =   0.8207 cm

=>   distance does the height of liquid in the cylinder change =   0.8207 cm

Part B)     B)The level of liquid in the cylinder rises.

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