Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Speed of metal melting through ice

  1. Nov 29, 2011 #1
    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:

    [tex]\frac{\delta p}{\delta \tau}=\frac{l}{\tau ∆v},[/tex]

    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

    [tex]\frac{∆U}{∆t}=-2mg\frac{∆z}{∆t}.[/tex]

    Since the rate of energy transport through the steel bar is

    [tex]\frac{F}{bc}=\frac{Q}{\delta t},[/tex]

    where bc is the surface area of the bar, I think I should be able to set

    [tex]\frac{Q}{\delta t}=\frac{∆U}{∆t}[/tex]

    so that I now have

    [tex]2mg\frac{∆z}{∆t}=\kappa\frac{∆\tau}{a}[/tex]

    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. jcsd
  3. Nov 30, 2011 #2
    No one?

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

    [tex]v=\frac{∆z}{∆t}=\kappa\frac{∆\tau}{2mga}.[/tex]

    Is this correct?
     
  4. Nov 30, 2011 #3
    I'd really need some help.

    Substituting [tex]∆\tau[/tex] in the last expression for

    [tex]\frac{\delta p ∆v\tau}{l}[/tex]

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

    [tex]v=\frac{\kappa bc\delta p \tau}{2mgal}(\frac{1}{\rho_i}-\frac{1}{\rho_w})[/tex]

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook