nL;

How much will the temperature increase
when the pressure increases
to 1.01 atm?
Surprisingly, it is the Carnot
theorem that helps us calculate the
shift in the meltingpoint of ice. Yes!
The very theorem that deals with
the maximum efficiency of heat engines.
"What has this to do with the
melting point of ice?" you may ask.
The point is, Nicolas Camot proved
that the maximum efficiency of a
cyclic heat engine of any design does
not depend on what this engine uses
as its working substance-an ideal
gas, melting ice, or a soapy film.
Only one condition must be met:
the engine must acquire heat at a
temperature T, and release it at a
temperature T, (there must be no
heat exchange during the intermediate
steps). The efficiency of such an
ideal engine (known as a Carnot engine)
is attained by a slow, reversible
process and, regardless of the working
substance used, equals
W 7,.7"
n_ _= _;, (1)
qir
where I4l is the work performed by
the engine per cycle and Q, is the
amount of heat obtained at the temperature
7r.
Let's consider our imaginary
Carnot engine to be a vertical cylinder
with a piston (fig. 1a). Inside the
cylinder we put ice of massm under
PressureP, = 1 atm and temPerature
Figure 1
I atm
1.01 atm
b
TT
d
0'C (71 :273K\. Thepressureis stabilized
by a weight set on the piston.
To underscore the state of equilibrium
between the water and the ice,
the figure shows a small amount of
water at the corner of the cylinder.
Now let's describe, step by step/
what goes on in this kind of Carnot
engine during one complete cycle.
1. Let's set the cylinder on a thermal
reservoir at a constant temperature
7, and transmit heat energy Q,
= Lm to the system necessary to
melt all the ice (I is the latent heat
of fusion). As a result, the piston
sinks a little (fig. 1b), because the
volume of ice Vi: mlpi is larger
than that of water V* = m/p*. This
melting stage is represented by the
line 1-2 on the graph, where the coordinates
are pressure P and volume
v (fig.2l.
2. Now we take the cylinder away
from the thermal reservoir, isolate it
thermally (fig. 1c), and then increase
the pressure very slowly until it is
equal to P, + LP: 1.01 atm. (This
can be done by pouring sand slowiy
onto the piston.) This will result in
a decrease in the temperature down
toTr=Tr-LT, which is equal to the
melting point of ice at apressure of
1.01 atm.
3. Now we put the cylinder on a
thermal reservoir at a temperature ?,
and remove heat until the water
freezes again (fig. 1d). In figure 2 this
stage is shown by the line 3-4.
4. All that's left is to thermally
T
water
rt Aa
TA
ice
le,
r1*Ar I a
Tt- Yo,
^T
1 atm
1.01 a
Figure 2
isolate the cylinder and slowly remove
all the sand from the piston.
This brings us back to the initial
state.
Now let's do some calculations.
The work performed during the
cycle can be found from the graph
-
it's equal to the area outlined by the
cycle:
(*
-)
[Pi P* )
w = LP(vi - y,)= oPl "'- "'
l.
The amount of heat obtained from
the heater is
Qt = Lm'
Hence, from the Carnot theorem (1)
we get
^P(r- -'l
Ip, p* )
Lm
AT
or, for any arbttrary temperature 7t
:7,
' r)
^r = ^P!l L.
This is known as the Clapeyron-
Clausius equation. Substituting numerical
data in this equation gives
us A? = 9.2. l}-s K forM = 0.01 atm.
The effect is clearly very small. To
change the melting point by say, I K,
we need apressure of about 133 atm.
Now we can get back to our skating.
The pressure produced by an ice
skater can be estimated as P : mglS
600 N/2 cm2:30 atm. The corresponding
shift in the meltingpoint of
=
ice is about 0.3 K, which is surely too
little on a cold winter's day. So the
"l\aive" little kid was right after aII:
TT
28
JUIY/AUOlJST Igg4