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Heating curve of pure substance

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The diagram below shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC tsteam at temperatures above 100 ºC, affects the temperature of the sample.

A: Rise in temperature as ice absorbs heat.B: Absorption of heat of fusion.C: Rise in temperature as liquid water absorbs heat.D: Water boils and absorbs heat of vaporization.E: Steam absorbs heat and thus increases its temperature.

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The specific heat capacity of water is 4.18 kJ kg -1 K -1. The specific heat capacityfor Ice and Steam are 2.09 kJ kg -1 K -1.

The heat of fusion of ice is 334 kJ kg -1, and the heat of vaporization of water is2260 kJ kg -1.

EXAMPLE Calculate the amount of heat required to completely convert 50 g of ice at -10 ºC

to steam at 120 ºC.Heat is taken up in five stages:

1. The heating of the ice2. The melting of the ice,

3. The heating of the water,4. The vaporization of the water and

5. The heating of the steam.

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P-T diagram for water

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The melting curve or fusion curve of ice/water is very special. It has a negative slopedue to the fact that when ice melts, the molar volume decreases. Ice actually melts at

lower temperature at higher pressure. Most North Americans skate, and the liquidformed between the skate and ice acts as a lubricant so that the skater moves gracefully

accross the ice. The skate apply a very high pressure on to the ice.

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The triple point of carbon dioxide occure at a pressure of 5.2 atm (3952 torr) and

216.6 K (-56.4 oC). At temperature of 197.5 K (-78.5 oC), the vapor pressure of

solid carbon dioxide is 1 atm (760 torr). At this pressure, the liquid phase is not

stable, the solid simply sublimates. Thus solid carbon dioxide is called dry ice ,

because it does not go through a liquid state in its phase transition at room

pressure.

The critical temperature for carbon dioxide is 31.1°C, and the critical pressure is

73 atm. Above the critical temeprature, the fluid is called super-critical fluid.

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ΔH sub > ΔH vap

Though both enthalpies involve the changing asubstance into its gaseous state, the change inenergy associated with sublimation is generally

greater than that of vaporization. This is because

of the initial state of the substances and theamount of initial energy that each substance has.Particles in a solid have less energy than those of a liquid, meaning it is takes more energy to excitea solid to its gaseous phase that it does to excite a

liquid to its gaseous phase.

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Hess's Law says that if a single process can be written as two or more steps, the enthalpy (heat) change for thatprocess is equal to the sum of the enthalpy changes of the individual steps. As long as the starting point and

end result are the same, the enthalpy change should be the same. Let's consider the case of water:

Sublimation of water: H2O(s) --> H2O(g)

The molar enthalpy change of this process is called the heat of sublimation, but we can imagine this changehappening in two steps:

Step 1: H2O(s) --> H2O(l)

Step 2: H2O(l) --> H2O(g)

The enthalpy change of Step 1 is the molar heat of fusion, ΔHfus. The enthalpy change of Step 2 is the molar heat of vaporization, ΔHvap. If we combine these two equations and cancel out anything that appears on both

sides of the equation (in this case liquid water), we're back to the sublimation equation:

Step 1 + Step 2 = Sublimation

Therefore the heat of sublimation, ΔHsub, must be equal to the sum of the heats of fusion and vaporiation:

ΔHfus + ΔHvap = ΔHsub

Unless ΔHfus is equal to zero (and it NEVER is), ΔHsub MUST be greater than ΔHvap.

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= BC/BA

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Useful links

http://webbook.nist.gov/chemistry/fluid/http://www.ecourses.ou.edu/cgi-bin/view_anime.cgi?file=th

http://www.kentchemistry.com/links/KentsDemos.htm