That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
I believe it would be an unbalanced force. Because the forces are unbalanced, one side is stronger and, therefore, the object will move.
Hey again!
Ok..
Now... The melting Point of this solid is 90°C.
Meaning That as soon as it gets to this temp... It STARTS Melting.
So at that temp... It still has some solid parts in it.
You can say its a Solid Liquid Mixture.
Additional Heat being applied at that point is not raising the temperature;rather its used in breaking the bonds in the solid. This is the Fusion stage.
After Fusion...It'd then Be a Pure Liquid with no solids in it.
So
Q'=MC∆0----- This is the heat needed to take the solid's temp from 30°c - 90°c
Q"=ml ----- This is the heat used in breaking the bonds holding the solids in the solid-liquid phase.
So
Q= Q' + Q"
Q= mc∆0 + ml
∆0 = 90°c - 30°c = 60°c
Q= 2.5(390)(60) + (2.5)(4000)
Q=6.9 x 10⁴Joules
the answer wil be 2.36+3.38+0.355+1.06=7.155
Quantitative because it tells you how many wheels