Regelation
The melting point of the ice under the bar will be decreased. The decrease is given by the
Clausius-Clapyron equation.
 12
T (v  v)
 dP 
T  P
where
v  v
so
Solid - liquid ,

 
 12
 dT 12 T (v  v)
2mg
2mgT (v  v)
P 
T 
,
so
bc
bc 12
The ice under the bar melts, and slips out from under the bar. It is no longer under pressure, so it
re-freezes. When it freezes, it gives off heat which the ice under the bar needs to melt. Thus, the
bar slowly sinks into the ice. The pressure of the bar allows the ice under the bar to melt, and the
water slips around to the top of the bar where it re-freezes. The latent heat flows through the bar,
and the rate of heat flow controls the rate of descent through the ice. The rate of heat flow is
given by,
1 dQ
dT
dQ
dT
,
or
 
  A
A dt
dy
dt
dy
The heat flow is related to the rate at which the ice melts by,
1
dQ
dm
dm
1 dQ
  12

, or
.
dt
dt
dt  12 dt
dm
dV

, and this is
dt
dt
dy
1 dm
1 dQ


so
dt bc dt bc  12 dt
The volume rate of melting is related to the mass rate of melting by,
related to the rate of descent of the bar by,
dm
dy
 bc ,
dt
dt
dy
 A dT
 A T


.
dt
bc  12 dy
bc  12 y
dy
 A 1 2mgT (v  v)
 bc 1 2mgT (v  v)  2mgT (v  v)



2
dt
bc  12 y
bc 12
bc  12 y
bc 12
y
bc  12
Then,

y
is tricky because the heat flows through the steel bar and a thin layer of water. Conduction
through two separate materials must be considered. A value of

must be found for the
y
combination. The conductivity of steel is much larger than that of water, but the water layer is
much thinner than the steel bar.
The rate of heat flow must be the same for both materials since all the heat that goes through the
steel goes through the water.
1 dQ
T
T
1 dQ ys
1 dQ yw
 s s  w w , so
Ts  
and Tw  
A dt
ys
yw
A dt s
A dt w
1 dQ
T
1 dQ

For the combination,
, or
 
  T where T  Ts  Tw .
A dt
y
A dt
y
1 dQ
  1 dQ ys 1 dQ yw 
,
   

A dt
y  A dt s
A dt w 
In this case,

y

1
y s
s

y w
w
,
so
so 1 
y ys yw
  ys yw 

 , and


.


s
w
y  s
w 
dy
 2mgT (v  v)

becomes,
2
dt y
bc  12
dy
2mgT (v  v)
.

dt  ys yw 
2

bc  12

w 
 s
For this problem, m = 10 kg, a = b = 1 mm, c = 10 cm.
s = 0.15 cal·K-1·cm-2
w = 0.00133 cal·K-1·cm-2
yw = 0.001 cm = 0.01 mm
dy
 0.0014 cm/s , or 1000 s for 1.4 cm (1.98 in/hr)
dt
2