Change of volume on solidification
Paraffin wax experiment.
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Effect of pressure on melting point. Regelation
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Experiment to illustrate the effect of pressure on the melting point
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.
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.
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.
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.
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.