Problem and solution Blundell and Blundell Concept of Thermal Physics Chapter 28 Phase transitions Masatsugu Sei Suzuki, SUNY at Binghamton (Date: November 10, 2019) ______________________________________________________________________________ 28-1 ((Solution)) 4 10 101 . 1 s kg/m 3 , 4 10 065 . 1 l kg/m 3 , 4 10 90827 . 0 1 s s V m 3 /kg, 4 10 938971 . 0 1 l l V m 3 /kg, 5 . 24 f L kJ/kg (latent heat of fusion for lead) 600 273 327 M T K 7 4 3 10 33 . 1 10 0307 . 0 10 5 . 24 600 1 1 s l f s l s l V V L T V V S S dT dP Pa/K =131.3 atm/K where 1 atm = 1.01315 x 10 5 Pa. When 99 P atm, 75 . 0 3 . 131 99 3 . 131 dP T C which means that the melting temperature increases as
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For a very heavy skater (100 kg), only making the contact with ice over an area 10 cm x 1 mm =
10-4 m2.
410
8.9100
P = 9.8 x 106 Pa = 96.7 atm
Then
Fig. TP (triple point): T = 0 C (273.16 K) and P = 0.00637 atm. The stating point : T = -5 C
and P = 1 atm.
We now suppose that the temperature is -5 C. At 1 atmosphere, the ice is in the state on the P-T
plane. There is no water present. When the ice skater puts pressure on the ice, the state moves
TP
Log P atm
T C 05
1 atm
106.7 atm
661 atm
Ice
Water
dP dT 132.2 atm K
along the constant temperature line ( ). In theory, as soon as the phase boundary is
reached at the state , some ice melts so that the edge of the skate sinks in fractionally, with the
load now spread over a large area stabilizing the pressure. The state thus remains fixed at , with
the liberated water acting as a lubricant.
The question must be asked: Is 106.7 atm or so sufficient to achieve this goal? From
dT
dP -132.2 atm/K.
So a pressure increase of (132.2 atm/K) x 5 K = 661 atm would be required to go to from to . Thus this explanation for the success of the skater appears inadequate.