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AT THE<br />
BLACKBOARD I<br />
The lriclion and pre$$tlre ol slralinU<br />
Also glaciers, pressure cookers, and the Carnot theorem<br />
by Alexey Chernoutsan<br />
F YOU WERE TO ASK A TEN-<br />
lear-old child why ice skates<br />
glide so easily on the ice, the an-<br />
I I*"r you'll probably get is the<br />
simple and obvious one: "The skates<br />
rub against the ice and they make a<br />
thin watery film, and this helps the<br />
skates slide along the ice." A student<br />
more experienced in physics,<br />
however, would find this answer too<br />
simple and rather boring. "No, no,<br />
no," the budding physicist will say,<br />
"it's not a matte'r of friction but of<br />
the pressure of the skates on the ice.<br />
As the pressure increases, the melting<br />
point of ice becomes less than<br />
0oC, so the ice melts under the<br />
skates." There is some merit in this<br />
explanation-the melting point of<br />
ice actually does decrease with an<br />
increase in external pressure. However,<br />
physics is a quantitative science.<br />
If we want to find out whether<br />
this physical phenomenon has any<br />
bearing on ice skating, we need to<br />
produce the appropriate numerical<br />
calculations.<br />
First of all, what is the melting<br />
point of ice, and why is it interesting?<br />
As you may know, when the<br />
temperature increases to this point,<br />
it can't be raised any further, and<br />
any additional heat goes into melting<br />
the ice. If no heat is provided<br />
from outside, the ice and water coexist<br />
in thermal equiiibrium. Thus, the<br />
meltingpoint is the equilibrium temperature<br />
of water and ice at a given<br />
pressure. For example, it's equal to<br />
0'C when the pressure is 1 atm.<br />
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