Homework Statement
Question no. 4 in
this document (there's a helpful picture, too):
A steel bar of rectangular cross section (height a and width b) is placed on a
block of ice (width c) with its ends extending a trifle as shown in the figure. A weight of mass m is hung from each end of the bar. The entire system is at T = 0° C. As a result of the pressure exerted by the bar, the ice melts beneath the bar and refreezes above the
bar. Heat is therefore liberated above the bar, conducted through the metal, and then absorbed by the ice beneath the bar. (We assume that this is the most important way in which heat reaches the ice immediately beneath the bar in order to melt it.) Find an approximate expression for the speed with which the bar sinks through the ice. Take the latent heat of fusion per gram of ice to be l, and the densities of ice and water to be ρi and ρw, respectively.
(a) Let’s say the bar sinks a distance ∆z in time ∆t. Calculate ∆U, the amount of energy
required for this to happen.
(b) The energy calculated in part (a) must pass through the bar via the process called
thermal conduction, described by the following equation
F=-\kappa\frac{\delta \tau}{\delta z}\approx -\kappa\frac{∆\tau}{a}
Here F is the heat flux, i.e. total energy crossing unit area per unit time, constant κ is
the coefficient of thermal conductivity, and ∆τ is the difference between the temperatures
under and above the bar. Note that F is proportional to the negative of the gradient of
temperature (as expected, since heat flows from high to low τ ). Use Eq. (1) and your result from part (a) to calculate the speed of the bar, v = dz/dt, as a function of ∆τ and other given quantities.
(c) Finally, eliminate ∆τ from your result in part (b) to obtain the final expression for v:
v=\frac{2mg\kappa\tau}{abcl^2\rho_i}(\frac{1}{\rho_i}-\frac{1}{\rho_w})
Homework Equations
The Clausius-Clapeyron equation:
\frac{\delta p}{\delta \tau}=\frac{l}{\tau ∆v},
where v is the volume per unit mass, i.e., the inverse of the density.
The Attempt at a Solution
The answer to question a) is simply the change in potential energy of the bar divided by time, which is
\frac{∆U}{∆t}=-2mg\frac{∆z}{∆t}.
Since the rate of energy transport through the steel bar is
\frac{F}{bc}=\frac{Q}{\delta t},
where bc is the surface area of the bar, I think I should be able to set
\frac{Q}{\delta t}=\frac{∆U}{∆t}
so that I now have
2mg\frac{∆z}{∆t}=\kappa\frac{∆\tau}{a}
From here on I'm pretty lost, though. It seems to me that since ∆z/∆t is the speed, the factor 2mg will end up in the denominator instead of in the numerator. Also, I don't really know what to do with the Clausius-Clapeyron equation. I can see that the term ∆v will eventually give me the fractions of the densities that you can see in the final expression, but it also contains a pressure term that I'm not quite sure I understand. Would that be the difference in pressure between the ice and the water? If so, should I perhaps restate it in terms of energy and volume, i.e., ∆U/∆v?