Saturday, February 25, 2012

Regelation : effect of pressure on melting point







Ice crystal.

Change of volume on solidification

  When water freezes to form ice expansion occurs and the ice takes up a bigger volume than the water. For this reason, water pipes are liable to brust during very cold weather, although the leaks do not occur until a thaw sets in.
  Sometimes the expansion of a substance on solidification serves a useful purpose. Molten type-metal, for example, expands very slightly when it solidifies, and so takes up a sharp impression of the mould.
Paraffin wax experiment.




  Most substances, however, contract in volume when they solidify. Paraffin wax is a typical example. When some molten paraffin wax is allowed to become solid in a test-tube the shrinkage in volume is shown by a deep cleft in the surface ( as you can see in the image ).
  

Effect of pressure on melting point. Regelation

  If a substance expands on solidifying, then the application of pressure lowers the melting point. Conversely, substances which contract in volume on solidifying have their melting points raised by pressure. Thus, the freezing point of water is lowered by just over 0.007 °C per atmosphere increase in pressure , while the freezing point of paraffin increases by about 0.04 °C per atmosphere.
Regelation experiment.

 

Experiment to illustrate the effect of pressure on the melting point

  The experiment illustrated in the image shows the effect of pressure on the melting point of ice in a rather striking manner.
  1- A block of ice rests on two supports, and a thin copper wire with heavy weights at each end is hung over it.

  2- After an hour, more or less, depending on the size of the block, the wire cuts right through it and falls to the floor, leaving the ice still in a solid block. This phenomenon is called regelation = refreezing.
  Several factors are involved here. The pressure of the wire lowers the melting point of the ice in contact with it, and so the ice melts and flows above the wire. The latent heat required for the melting comes, in the first instance, from the copper wire. As soon as the water passes above the wire it is no longer under pressure and therefore freezes. In so doing it gives out latent heat, and this heat is conducted down through the wire to provide heat for further melting of the ice beneath..
  It must be realized that rapid conduction of heat down through the copper is an important factor in the process. An iron wire of smaller thermal conductivity cuts through much more slowly. A thin string of very low conductivity will not pass at all.
  

Regelation is a factor in the making of snowballs. Compression of the snow by hand causes slight melting of the ice crystals, and when the pressure is removed refreezing occurs and binds the snow together. In very cold weather the pressure exerted is insufficient to melt the snow, and so it fails to bind.

  

Why ice is slippery

 
  The ease with which a skater glides over the ice depends on the formation of a thin film of water between the blade of the skate and the ice. At one time this was generally believed to be caused by melting under pressure. However, a simple calculation shows that the pressure exerted by a skater is about 10 atmospheres and this would lower the melting point by less than 0.1 °C. Yet skating is possible even when the temperature is several degrees below zero.
  It was first pointed out by F. P. Bowden, and later by J. Fremlin that the water film in this case is more likely to be brought about by the work done against friction. This work becomes transformed into internal energy and the ice melts in consequence.
  Worked example
  Dry steam is passed into a well-lagged copper can of mass 250 g containing 400 g of water and 50 g of ice at 0 °C.
  The mixture is well stirred and the steam supply cut off when the temperature of the can and its contents reaches 20 °C. Neglecting heat losses, find the mass of steam condensed.
  Specific heat capacities:

water, 4.2 J/g °C; copper,0.4 J/g °C.

Specific latent heats: steam, 2260 J/g; ice, 336 J/g.
Solution
  Using the principle of conservation of energy, we may say,
  heat given out by steam = heat received by ice, water and can.

  Let mass of steam condensed = m grams.

Heat in joules given out by:

Steam condensing to water at 100 °C = m × 2260

Condensed steam cooling from 100 °C to 20 °C = m × 4.2 × 80


    Total = m × 2596

  Heat in joules received by:

  Ice melting to water at 0 °C = 50 × 336

  Melted ice warming from 0 °C to 20 °C = 50 × 4.2 × 20

  Calorimeter warming from 0 °C to 20 °C = 250 × 0.4 × 20

    Total = 56000 J

  Hence,    m × 2596 = 56000

or    Mass condensed, m = 21.8 g.
I hope this post was interesting.