A jar filled partially with water has an object floating in it. The jar is open. Now, if the jar is closed and a vacuum pump is used to pump the air out, what will happen to the floating object? will it rise up more or sink or ...? and why?
Buoyancy depends only on density. If the object is more dense
than water, it will sink to the bottom; if it is less dense, it will
float to the top; if it has the same density as water, it will float
within the water, moving randomly around due to Brownian motion.
Sucking the air out of the jar will create a vacuum that will
eventually be replaced with water vapor as equilibrium is established.
Sucking the air out of the jar does not change the density of either
the water or the object, so the object will stay floating as
before. However, applying the vacuum will actually cause the
water to "boil" as the water vapor fills the vacuum, and this boiling
will certainly jostle the object around.
If you want to get hypertechnical, the energy needed to vaporize the
water (that fills the vacuum left behind after the air is sucked out)
will come from the liquid water, and therefore the liquid water will
cool somewhat. In general, the density of water increases as it
cools (hence thermoclines in lakes). Since the density of the
object would not change, the object would then be less dense than the
water and would therefore float to the top. In a standard-sized
mason jar, however, I don't think you'd notice this effect since the
water temperature would equilibrate with the air temperature on the
outside of the jar, which remains constant. Maybe you'd notice if
you had very little water in the jar. Might be worth doing an
Posted by Snydley
on 2005-10-18 21:22:35