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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 />

j::<br />

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o<br />

ao<br />

(o o<br />

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0)<br />

x<br />

3'<br />

0UIt'lTllil/AI Tllt 0LlCl([01R0 I<br />

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