That’s not valid. The glaciers have a surrounding volume of water infinitely greater than a bathtub. You would have to ensure the relative container size (bathtub, ocean) are relative to the size of the block of ice (10lb, glacier). The pressure differential is nowhere near the same. Water has weight, so the further you go down, the higher the pressure. This creates a push upward for buoyant objects to the least path of resistance.
It’s about the buoyancy support of the surrounding water. A 10lb block of ice would float in the ocean. It would not in a bath tub because of volume bouyancy support not being equivalent...the pressure differential being orders of magnitude apart.
Yes, it is. The explanation I gave is valid for any size block of ice from an ice cube to a giant glacier as long as the top and bottom of the block are flat and the sides are vertical.
Let h be the height of the block.
Let A be the area of the top or bottom.
Let d be the depth of the water from the bottom of the block to the surface.
Let r1 = density of fresh water
Let r2 = density of seawater = 1.029 x r1
Let r0 = density of ice = 0.9167 x r1
(all densities given as weight per unit volume)
Volume of ice block = h x A
Weight of ice block = h x A x r0
Pressure at bottom of ice block = d x r2
Force on the bottom of the ice block = d x r2 x A
The ice block will float when the force from water pressure on the bottom is greater than the weight of the block:
d r2 A > h A r0
This is the same as:
d r2 > h r0
or:
d/h > r0/r2
But r0/r2 = 0.9167/1.022 = about 89%
So the block will float only when:
d/h > 89%
A 10lb block of ice would float in the ocean.
If it was in a very calm part of the ocean where the water depth was less than 89% of the height of the block, it would not float. The size of the body of water is irrelevant.
The volume of extra water is equal to about 2 1/2 percent of the volume of the ice block.
Weight of block = h x A x r0
Volume of fresh water from melted block = (h x A x r0)/r1
Volume of block below the water line = d x A
Volume of extra water produced when an ice block resting on the bottom melts:
h x A x r0/r1 - d x A
= ((0.9167 x h) - d) x A
Volume of extra water produced when a floating ice block melts:
= h x A x (r0/r1 - r0/r2) = h x A x (0.9167 - 0.8909) = h x A x 0.0258
So it might be possible to trick someone by having some ice floating in a glass filled with very salty water. When the ice melts the glass should overflow. (Because of surface tension you would have to make sure that there is a lot of ice.)