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QUESTIONS AND PROBLEMS
1. A barrelful of tepid water, when poured into a snowdrift, melts much more snow than a cupful of boiling water does. Which has the greater quantity of heat?
2. Why is a liter of hot water a better foot warmer than an equal volume of any substance in the preceding table?
3. The specific heat of water is much greater than that of any other liquid or of any solid. Explain how this accounts for the fact that an island in mid-ocean undergoes less extremes of temperature than an inland region.
4. How many calories are required to heat a laundry iron weighing 3 kg. from 20° C. to 130° C.?
5. How many B. T. U. are required to warm a 6-pound laundry iron from 75° F. to 250° F. ?
6. If 100 g. of mercury at 95° C. are mixed with 100 g. of water at 15° C., and if the resulting temperature is 17.6° C., what is the specific heat of mercury?
7. If 200 g. of water at 80° C. are mixed with 100 g. of water at 10° C., what will be the temperature of the mixture? (Let x equal the final temperature; then 100 (x −10) calories are gained by the cold water, while 200 (80 — x) calories are lost by the hot water.)
8. What temperature will result if 400 g. of aluminium at 100° C. are placed in 500 g. of water at 20° C.?
9. Eight pounds of water were placed in a copper kettle weighing 2.5 lb. How many B. T. U. are required to heat the water and the kettle from 70° F. to 212° F.? If 4.3 cu. ft. of gas was used to do this, and if each cubic foot of gas on being burned yields 625 B. T. U., what is the efficiency of the heating apparatus?
10. If a solid steel projectile were shot with a velocity of 1000 m. (3048 ft.) per second against an impenetrable steel target, and all the heat generated were to go toward raising the temperature of the projectile, what would be the amount of the increase?
CHANGE OF STATE
192. Heat of fusion. If on a cold day in winter a quantity of snow is brought in from out of doors, where the temperature is below 0° C. and placed over a source of heat, a thermometer plunged into the snow will be found to rise slowly until the temperature reaches 0° C., when it will become stationary and remain so during all the time the snow is melting, provided only that the contents of the vessel are continuously and vigorously stirred. As soon as the snow is all melted, the temperature will begin to rise again.
Since the temperature of ice at 0° C. is the same as the temperature of water at 0° C., it is evident from this experiment that when ice is being changed to water, the entrance of heat energy into it does not produce any change in the average kinetic energy of its molecules. This energy must therefore all be expended in pulling apart the molecules of the crystals of which the ice is composed, and thus reducing it to a form in which the molecules are held together less intimately, that is, to the liquid form. In other words, the energy which existed in the flame as the kinetic energy of molecular motion has been transformed, upon passage into the melting solid, into the potential energy of molecules which have been pulled apart against the force of their mutual attraction. The number
*This subject should be preceded by a laboratory exercise on the curve of cooling through the point of fusion, and followed by a determination of the heat of fusion of ice. See, for example, Experiments 21 and 22 of the authors' Manual.
of calories of heat energy required to melt one gram of any substance without producing any change in its temperature is called the heat of fusion of that substance.
193. Numerical value of heat of fusion of ice. Since it is found to require about 80 times as long for a given flame to melt a quantity of snow as to raise the melted snow through 1o C., we conclude that it requires about 80 calories of heat to melt 1 g. of snow or ice. This constant is, however, much more accurately determined by the method of mixtures. Thus, suppose that a piece of ice weighing 131 g. is dropped into 500 g. of water at 40° C., and suppose that after the ice is all melted the temperature of the mixture is found to be 15° C. The number of calories which have come out of the water is 500 × (40 – 15) = 12,500. But 131 × 15 = 1965 calories of this heat must have been used in raising the ice from 0° C. to 15° C. after the ice, by melting, became water at 0°. The remainder of the heat, namely, 12,500 – 1965 = 10,535, must have been used in melting the 131 g. of ice. Hence the number of calories required to melt 1 of ice is 10535 g. 80.4. To state the problem algebraically, let = the heat of fusion x of ice. Then we have
131 x + 1965 = 12,500; that is, x = 80.4.
According to the most careful determinations the heat of fusion of ice is 80.0 calories.
194. Energy transformation in fusion. The heat energy that goes into a body to change it from the solid state to the liquid state no longer exists as heat within the liquid. It has ceased to exist as heat energy at all, having been transformed into molecular potential energy; that is, the heat which disappears represents the work that was done in effecting the change of state, and it is, therefore, the exact equivalent of the potential energy gained by the rearranged molecules. This is strictly in accord with the law of conservation of energy.
195. Heat given out when water freezes. Let snow and salt be added to a beaker of water until the temperature of the liquid mixture is as low as -10° C. or - 12° C. Then let a test tube containing a thermometer and a quantity of pure water be thrust into the cold solution. If the thermometer is kept very quiet, the temperature of the water in the test tube will fall four or five or even ten degrees below 0° C. without producing solidification. But as soon as the thermometer is stirred, or a small crystal of ice is dropped into the neck of the tube, the ice crystals will form with great suddenness, and at the same time the thermometer will rise to 0° C., where it will remain until all the water is frozen.
The experiment shows in a very striking way that the process of freezing is a heat-evolving process. This was to have been expected from the principle of the conservation of energy; for since it takes 80 calories of heat energy to turn a gram of ice at 0° C. into water at 0° C., this amount of energy must reappear when the water turns back to ice.
196. Use made of energy transformations in melting and freezing. A refrigerator (Fig. 170) is a box constructed with double walls so as to make it difficult for heat to pass in from the outside. Ice is kept in the upper part of one compartment so as to cool the air at the top, which, because of its greater density when cool, settles and causes a circulation as indicated by the arrows. To melt cach gram of ice 80 calories must be taken from the air and food within the refrigerator. If the ice did not melt, it would be worthless for use in refrigerators.
FIG. 170. A refrigerator
The heat given off by the freezing of water is often turned to practical account; for example, tubs of water are sometimes placed in vegetable cellars to prevent the vegetables from freezing. The effectiveness of this procedure is due to the fact that the temperature at which the vegetables freeze is slightly
lower than 0° C. As the temperature of the cellar falls the water therefore begins to freeze first, and in so doing evolves enough heat to prevent the temperature of the room from falling as far below 0° C. as it otherwise would.
It is partly because of the heat evolved by the freezing of large bodies of water that the temperature never falls so low in the vicinity of large lakes as it does in inland localities.
197. Melting points of crystalline substances. If a piece of ice is placed in a vessel of boiling water for an instant and then removed and wiped, it will not be found to be in the slightest degree warmer than a piece of ice which has not been. exposed to the heat of the warm water. The melting point of ice is therefore a perfectly fixed, definite temperature, above which the ice can never be raised so long as it remains ice, no matter how fast heat is applied to it. All crystalline substances are found to behave exactly like ice in this respect, each substance of this class having its characteristic melting point. The following table gives the melting points of some of the commoner crystalline substances:
We may summarize the experiments upon melting points of crystalline substances in the two following laws:
1. The temperatures of solidification and fusion are the same. 2. The temperature of the melting or solidifying substance remains constant from the moment at which melting or solidification begins until the process is completed.
198. Fusion of noncrystalline, or amorphous, substances. Let the end of a glass rod be held in a Bunsen flame. Instead of changing suddenly from the solid to the liquid state, it will gradually grow softer