Melting point of ice

The general temperature scale now in use is the Celsius scale, based nominally on the melting point of ice at 0°C and the hoiling point of water at atmospheric pressure at 100°C. (By strict definition, the triple point of ice is 0.01°C at a pressure of 6.1 mhar.) On the Celsius scale, absolute zero is -2 73.15°C. 

The liquid is the more dense phase (Figure 9.7b). The liquid-solid line is inclined to the left, toward the y-axis. An increase in pressure favors the formation of liquid that is, the melting point is decreased by raising the pressure. Water is one of the few substances that behave this way ice is less dense than liquid water. The effect is exaggerated for emphasis in Figure 9.7b. Actually, an increase in pressure of 134 atm is required to lower the melting point of ice by 1°C. 

The freezing points of electrolyte solutions, like their vapor pressures, are lower than those of nonelectrolytes at the same concentration. Sodium chloride and calcium chloride are used to lower the melting point of ice on highways their aqueous solutions can have freezing points as low as —21 and — 55°C, respectively. 

Temperature and pressure are the two variables that affect phase equilibria in a one-component system. The phase diagram in Figure 15.1 shows the equilibria between the solid, liquid, and vapour states of water where all three phases are in equilibrium at the triple point, 0.06 N/m2 and 273.3 K. The sublimation curve indicates the vapour pressure of ice, the vaporisation curve the vapour pressure of liquid water, and the fusion curve the effect of pressure on the melting point of ice. The fusion curve for ice is unusual in that, in most one component systems, increased pressure increases the melting point, whilst the opposite occurs here. 

Focus Question Does the melting point of Ice Increase, decrease, or remain constant when the pressure Is increased ... 

As expected, this switchover occurs at the normal melting point of ice. 

Estimate the lowering of the melting point of ice by the weight of a 180-lb skater. Assume that his entire weight is on one skate and that the skate makes an area of contact of 100 mm2 with the ice. 

If a mixture of ice and water at 1 atm pressure and 0°C is placed in an insulated container and all of the air is pumped away and the container sealed, what will happen As was shown in the previous section, at pressures lower than 1 atm, the melting point of ice is above 0°C. Water will, therefore, be solid at 0°C and reduced pressure. However, when some liquid water freezes, its latent heat is released and the temperature of the system is slightly increased. Equilibrium is reestablished at the higher temperature and reduced pressure. The pressure in the system is the vapor pressure of both liquid and solid water slightly above 0 K. The new equilibrium point of the system is called the triple point of water and is at 0.0098°C and 611 Pa. Three phases—solid, liquid, and gas—coexist at the triple point, and the chemical potential of water in each of the phases must be equal ... 

Increase of pressure or temperature will shift the equilibrium from left to right. In other words, melting point of ice is lowered by an increase of pressure or temperature. 

This curve shows the effect of pressure on the melting point of ice or freezing point of water. As the curve OC slopes towards the pressure axis, the melting point is lowered as the pressure is increased. This fact can also be predicted by Le-Chatelier s principle, as ice melts with decrease in volume. 

The triple point, posseses fixed values for pressure and temperature, i.e., 4.58 mm. of Hg and +0.0075°C, respectively. (It is clear that O is not the actual melting point of ice, i.e., 0°C. Its value has been increased due to the fact that 0°C is die normal melting point of ice at 760 mm of Hg and decrease of pressure will increase the melting point of ice. Since a decrease of pressure by 1 atmospheric pressure or 760 mm of Hg increases the melting point by 0.008°C, therefore, a decrease of pressure to 4.58 mm wilf raise the melting point to +0.0075 C.)... 

That such ought to be the case was first realised by James Thomson 1 who, in 1849, showed that from theoretical considerations a connection must exist between the melting-point of a solid and the pressure. The following year this was experimentally demonstrated, by his brother W. Thomson (Lord Kelvin),2 who found that under a pressure of 8-1 atmospheres the melting-point of ice was lowered by 0-059° C., equivalent to a fall of 0-0078° per atmosphere. In the table on p. 251 are given the more accurate determinations of Tammann,3 the third colunm giving the results calculated in terms of atmospheres.4... 

Ice exerts a definite vapour pressure which, at 0° C., is identical with that of water, so that at this point the three phases—solid, liquid, vapour—can co-exist. This is not the real triple point, because pressure lowers the melting-point of ice, and by definition 0° C. is the melting-point of ice under a pressure of one atmosphere. The true triple point therefore lies at +0 0076° C., and m the absence of air, OB, which represents the vapour pressure of ice at various temperatures, is termed the sublimation curve (fig. 43). 

Thus, the melting point of ice decreases by 0.0075 if pressure is increased. 

It has just been argued that the conductivities of simple ionic liquids, on the one hand, and liquid sihca and water, on the other, are vastly different because a fused salt is an unassociated liquid (it consists of individual particles) whereas both molten silica and water are associated liquids with network structures. What is the situation with regard to the viscosities of fused salts, water, and fused silica Experiments indicate that whereas water and fused NaCl have similar viscosities not far above the melting points of ice and solid salt, respectively, fused silica is a highly viscous liquid (Table 5.46). Here then is an interesting problem. 

Case I a temperature at which the vapor pressure of the solid is greater than that of the liquid. At this temperature the solid requires a higher pressure of vapor than the liquid to be in equilibrium with the vapor. Thus, as vapor is released from the solid to try to achieve equilibrium, the liquid absorbs vapor in an attempt to reduce the vapor pressure to its equilibrium value. The net effect is a conversion from solid to liquid through the vapor phase. In fact, no solid can exist under these conditions. The amount of solid steadily decreases and the volume of liquid increases. Finally, there is only liquid in the right compartment, which comes to equilibrium with the water vapor, and no further changes occur in the system. The temperature for Case 1 must be above the melting point of ice, since only the liquid state can exist. 

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