Question

A 0.130-kg cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.20 cm . The level of the glycerine is well below the top of the cylinder.

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

Express your answer with the appropriate units.

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?

Answer 1

A 0.130-kg cube of ice

inside radius 3.20 cm

First question: what volume of the cube is immersed in the glycerol?
the weight of the displaced liquid equals the weight of the cube (or the mass of the displaced liquid equals the mass of the cube). And
mass = density*volume
m = ρ*V
0.130 kg = 1.26 kg/L * V
V = 0.130kg/1.26 kg/L = 0.10317 L (= dm^3 = displaced liquid)
Since the initial volume vi of the glycerol without the cube stays the same after the cube is added, the new volume (up to the surface only) in the cylinder after addition of the cube is the initial volume + the immersed volume
i.e., V = Vi + 0.10317 dm^3.

Now let the cube melt: 0.130 kg ice give 0.130 dm^3 water.
The volume in the cylinder is now vi + 0.130 dm^3.

The volume difference before, and after the melting, is then 0.130 - 0.10317 = 0.02683dm^3

now find the height of 0.02683 dm^3 in the cylinder:

V = pir^2*h

h = V/(pir^2) with r in dm

h = 0.02683/(pi*0.32^2)

h = 0.0834 dm = 0.834 cm

> the liquid level is 0.834 cm higher after the melting.