Speed of metal melting through ice

1. Nov 29, 2011

grepecs

1. The problem statement, all variables and given/known data

Question no. 4 in this document (there's a helpful picture, too):

2. Relevant 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.

3. 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?

2. Nov 30, 2011

grepecs

No one?

Perhaps I should state my answer to b) explicitly: the speed with which the bar sinks is

$$v=\frac{∆z}{∆t}=\kappa\frac{∆\tau}{2mga}.$$

Is this correct?

3. Nov 30, 2011

grepecs

I'd really need some help.

Substituting $$∆\tau$$ in the last expression for

$$\frac{\delta p ∆v\tau}{l}$$

(a rearrangement of the Clausius-Clapeyron equation, and using the fact that δT and $$∆\tau$$ are equal), I get

$$v=\frac{\kappa bc\delta p \tau}{2mgal}(\frac{1}{\rho_i}-\frac{1}{\rho_w})$$

It kind of resembles the correct answer, but as I said, the term 2mg is in the wrong place, and I don't know what to do with δp.