Temperature and States of Matter Help (page 3)
When matter is heated or cooled, it often does things other than simply expanding or contracting, or exerting increased or decreased pressure. Sometimes it undergoes a change of state . This happens when solid ice melts into liquid water or when water boils into vapor, for example.
Thawing And Freezing
Consider our old friend, water. Imagine that it is late winter in a place such as northern Wisconsin and that the temperature of the water ice on the lake is exactly 0°C. The ice is not safe to skate on, as it was in the middle of the winter, because the ice has become “soft.” It is more like slush than ice. It is partly solid and partly liquid. Nevertheless, the temperature of this soft ice is 0°C.
As the temperature continues to rise, the slush gets softer. It becomes proportionately more liquid water and less solid ice. However, its temperature remains at 0°C. Eventually, all the ice melts into liquid. This can take place with astonishing rapidity. You might leave for school one morning and see the lake nearly “socked in” with slush and return in the evening to find it almost entirely thawed. Now you can get the canoe out! But you won’t want to go swimming. The liquid water will stay at 0°C until all the ice is gone. Only then will the temperature begin to rise slowly.
Consider now what happens in the late autumn. The weather, and the water, is growing colder. The water finally drops to 0°C. The surface begins to freeze. The temperature of this new ice is 0°C. Freezing takes place until the whole lake surface is solid ice. The weather keeps growing colder (a lot colder if you live in northern Wisconsin). Once the surface is entirely solid ice, the temperature of the ice begins to fall below 0°C, although it remains at 0°C at the boundary just beneath the surface where solid ice meets liquid water. The layer of ice gets thicker. The ice near the surface can get much colder than 0°C. How much colder depends on various factors, such as the severity of the winter and the amount of snow that happens to fall on top of the ice and insulate it against the bitter chill of the air.
The temperature of water does not follow exactly along with the air temperature when heating or cooling takes place in the vicinity of 0°C. Instead, the water temperature follows a curve something like that shown in Fig. 11-4. In part a , the air temperature is getting warmer; in part b , it is getting colder. The water “stalls” as it thaws or freezes. Other substances exhibit this same property when they thaw or freeze.
Fig. 11-4 . Water as it thaws and freezes, (a) The environmental temperature is getting warmer, and the ice is thawing, (b) The environmental temperature is getting colder, and the liquid water is freezing.
Heat Of Fusion
It takes a certain amount of energy to change a sample of solid matter to its liquid state, assuming that the matter is of the sort that can exist in either of these two states. (Water, glass, most rocks, and most metals fill this bill, but wood does not.) In the case of ice formed from pure water, it takes 80 cal to convert 1 g of ice at 0°C to 1 g of pure liquid water at 0°C. This quantity varies for different substances and is called the heat of fusion for the substance.
In the reverse scenario, if 1 g of pure liquid water at 0°C freezes completely solid and becomes ice at 0°C, it gives up 80 cal of heat. The heat of fusion is thus expressed in calories per gram (cal/g). It also can be expressed in kilocalories per kilogram (kcal/kg) and will yield exactly the same numbers as the cal/g figures for all substances. When the substance is something other than water, then the freezing/melting point of that substance must be substituted for 0°C in the discussion.
Heat of fusion is sometimes expressed in calories per mole (cal/mol) rather than in calories per gram. However, unless it is specifically stated that the units are intended to be expressed in calories per mole, you should assume that they are expressed in calories per gram.
If the heat of fusion (in calories per gram) is symbolized h f , the heat added or given up by a sample of matter (in calories) is h , and the mass of the sample (in grams) is m , then the following formula holds:
h f = h / m
Heat of Fusion Practice Problem
Suppose that a certain substance melts and freezes at +400°C. Imagine a block of this material whose mass is 1.535 kg, and it is entirely solid at +400°C. It is subjected to heating, and it melts. Suppose that it takes 142,761 cal of energy to melt the substance entirely into liquid at +400°C. What is the heat of fusion for this material?
First, we must be sure we have our units in agreement. We are given the mass in kilograms; to convert it to grams, multiply by 1,000. Thus m = 1,535 g. We are given that h = 142,761 cal. Therefore, we can use the preceding formula directly:
h f = 142,761/1535 = 93.00 cal/g
This is rounded off to four significant figures because this is the extent of the accuracy of our input data.
Boiling And Condensing
Let’s return to the stove, where a kettle of water is heating up. The temperature of the water is exactly + 100°C, but it has not yet begun to boil. As heat is continually applied, boiling begins. The water becomes proportionately more vapor and less liquid. However, the temperature remains at + 100°C. Eventually, all the liquid has boiled away, and only vapor is left. Imagine that we have captured all this vapor in an enclosure, and in the process of the water’s boiling away, all the air has been driven out of the enclosure and replaced by water vapor. The stove burner, an electric type, keeps on heating the water even after all of it has boiled into vapor.
At the moment when the last of the liquid vanishes, the temperature of the vapor is + 100°C. Once all the liquid is gone, the vapor can become hotter than + 100°C. The ultimate extent to which the vapor can be heated depends on how powerful the burner is and on how well insulated the enclosure is.
Consider now what happens if we take the enclosure, along with the kettle, off the stove and put it into a refrigerator. The environment, and the water vapor, begins to grow colder. The vapor temperature eventually drops to +100°C. It begins to condense. The temperature of this liquid water is +100°C. Condensation takes place until all the vapor has condensed. (But hardly any of it will condense back in the kettle. What a mess!) We allow a bit of air into the chamber near the end of this experiment to maintain a reasonable pressure inside. The chamber keeps growing colder; once all the vapor has condensed, the temperature of the liquid begins to fall below +100°C.
As is the case with melting and freezing, the temperature of water does not follow exactly along with the air temperature when heating or cooling takes place in the vicinity of +100°C. Instead, the water temperature follows a curve something like that shown in Fig. 11-5. In part a , the air temperature is getting warmer; in part b , it is getting colder. The water temperature “stalls” as it boils or condenses. Other substances exhibit this same property when they boil or condense.
Heat Of Vaporization
It takes a certain amount of energy to change a sample of liquid to its gaseous state, assuming that the matter is of the sort that can exist in either of these two states. In the case of water, it takes 540 cal to convert 1 g of liquid at +100°C to 1 g of pure water vapor at +100°C. This quantity varies for different substances and is called the heat of vaporization for the substance.
In the reverse scenario, if 1 g of pure water vapor at +100°C condenses completely and becomes liquid water at +100°C, it gives up 540 cal of heat. The heat of vaporization is expressed in the same units as heat of fusion, that is, in calories per gram (cal/g). It also can be expressed in kilocalories per kilogram (kcal/kg) and will yield exactly the same numbers as the cal/g figures for all substances. When the substance is something other than water, then the boiling/condensation point of that substance must be substituted for +100°C.
Heat of vaporization, like heat of fusion, is sometimes expressed in calories per mole (cal/mol) rather than in cal/g. However, this is not the usual case.
If the heat of vaporization (in calories per gram) is symbolized h v , the heat added or given up by a sample of matter (in calories) is h , and the mass of the sample (in grams) is m , then the following formula holds:
h v = h / m
This is the same as the formula for heat of fusion, except that h v has been substituted for h f .
Heat of Vaporization Practice Problem
Suppose that a certain substance boils and condenses at +500°C. Imagine a beaker of this material whose mass is 67.5 g, and it is entirely liquid at +500°C. It heat of vaporization is specified as 845 cal/g. How much heat, in calories and in kilocalories, is required to completely boil away this liquid?
Our units are already in agreement: grams for m and calories per gram for h v . We must manipulate the preceding formula so that it expresses the heat h (in calories) in terms of the other given quantities. This can be done by multiplying both sides by m , giving us this formula:
h = h v m
Now it is simply a matter of plugging in the numbers:
h = 845 × 67.5
= 5.70 × 10 4 cal = 57.0 kcal
This has been rounded off to three significant figures, the extent of the accuracy of our input data.
Practice problems of these concepts can be found at: Temperature, Pressure, And Changes Of State Practice Problem
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