glass in losing 36.2° of temperature is used in melting the ice and raising the resulting water to 43.8°. Heat given out by the water ries. 18,100 calo = Heat given out by the beaker calories. = 500 X 36.2 x 1 Total heat given out 18,524 calories. Heat taken up by ice in melting = 150 × heat of fusion calories. = = = 100 × 36.2 x .117 = = 18,524. 11,954. = 79.7. 150 X 43.8 x 1 = 424 = 150 F The heat of fusion of ice, as found by careful experiments, is 80 calories. This means that it takes as much heat to melt one gram of ice as it would to raise one gram of water from 0° to 80°. €6570 307. Heat of Vaporization. When heat is applied to a beaker of water, its temperature will rise until the boiling point is reached. After this no further increase in the temperature will take place, however rapid the boiling. By continuing the experiment, the water can all be changed into vapor. The number of heat units required to vaporize a unit mass of liquid without changing its temperature is its heat of vaporization. When the vapor changes to a liquid, the same amount of heat will be given out. Experiment has shown that the heat of vaporization of water is 537 calories. This means that the heat required to boil away 1 g. of water is 5.37 times as much as is required to raise its temperature from zero to 100°. = 308. Curve of Heat Effects. An effective way of showing the relation between the heat units applied and the Ice at 0° effects produced is indicated in Fig. 257. This shows graphically the relation between the heat units and the changes produced when heat is applied to 1 g. of ice at -18° and continued until the latter is changed into steam at 120°. Horizontal distances represent calories or heat units (H. U.), and vertical distances represent changes of temperature. Since the specific heat of ice is 0.502, Mts = 1 X 18 X 0.502 = about 9 calories required to raise the ice to zero. 80 calories will be required to melt it; 1 X 100 × 1 = 100 Steam at 120° Heating Melting 80 H.U. 9 B.U. Water at 0° = FIG. 257 = calories to raise the water to 100°; 537 calories to vaporize it; and, since the specific heat of steam is 0.48, Mts 1X 20 X 0.48 9.6 calories to raise the steam to 120°. The total heat applied will be the sum of these amounts, or 735.6 calories. 9.6 H.U. = A study of this curve will be of great assistance in making clear the relation between heat units, specific heat, heat of fusion, and heat of vaporization, and will help in the solution of problems that include these quantities. Questions 1. Suppose you were making a number of determinations in specific heat, would there be any advantage in using the same calorimeter for them all? 2. What do t and t' mean in the equation Mts M't's'? 3. What effect does the high specific heat of water have upon the rate at which the temperature of a lake will be changed? 4° C., what will be the result 4. If the temperature of a room is of bringing pans of water into it? 5. Will it take longer to melt a piece of ice or to raise the resulting water to the boiling point, if the heat is uniformly applied? 6. What do horizontal parts of the curve of heat effects, Fig. 257, mean? Problems 1. How many calories are required to heat 200 g. of water from 15° C. to the boiling point? 2. How many B.T. U. are required to heat 5 lb. of water from 50° F. to the boiling point? 3. A piece of nickel at 100° C. was dropped into an equal weight of water at 0° C. and the resulting temperature was 9.8° C. Find the specific heat of the nickel. 4. What is the water equivalent of a glass beaker weighing 80 g.? 5. What is the water equivalent of a copper calorimeter of the same weight? ļ 6. What is the water equivalent of an aluminum calorimeter of the same weight? 7. A beaker whose water equivalent is 14 g. has in it 120 g. of water at 14° C. What is the resulting temperature when 56 g. of water at 90° C. is poured into it? 8. Into the same beaker, containing 50 g. of water at 96° C., 200 g. of mercury at 20° C. is poured. Find the resulting temperature. 9. An aluminum beaker weighing 10 g. contained 250 g. of water at 12° C. What was the resulting temperature when 250 g. of water at 92° C. was poured into it? 10. What mass of water at 90° C. will just melt 2 kg. of ice at 0°? 11. It takes 10 minutes to melt a piece of ice. How long will it take to raise the resulting water to the boiling point over the same source of heat? 12. How many calories will it take to melt 120 g. of ice and raise the resulting water to 16° C.? 13. How many B. T. U. will it take to melt 6 lb. of ice and raise the resulting water to 48° F.? 14. The normal temperature of the body is 98.4° F. How many calories are required to raise a drink of 300 g. of ice water to that temperature? 15. It takes 8 minutes to raise the temperature of a certain quantity of water from the freezing to the boiling point. How long will it take to boil the water away? 16. How much steam at 100° C. must be mixed with 1 litre of water at 18° C. to raise its temperature to 90° C.? 17. How many calories would be required to melt 40 g. of ice at zero, raise the resulting water to 100° C., and vaporize it? 18. How many grams of ice would be required to change the temperature of 1200 g. of water from 60° F. to 40° F.? 19. A piece of aluminum weighing 27 g., after being heated to 100° C., is placed in 50 g. of water in a copper beaker weighing 35 g. What was the resulting temperature, if the temperature of the beaker and water was zero? V. HEAT AND WORK 309. General Law. We have seen that both friction and collision give rise to heat. The work required to overcome the friction in a machine is, from the mechanical standpoint, lost work; in reality it is work transformed into heat. The relation between mechanical work and heat was investigated by Joule, who established the following principle: The disappearance of a certain amount of mechanical energy produces an equivalent amount of heat. The converse of this law is equally true: The disappearance of a certain amount of heat produces an equivalent amount of mechanical energy. 310. The Mechanical Equivalent of Heat. The number of units of work required to produce one heat unit is called the mechanical equivalent of heat. Joule's experiments determined that the number of foot pounds of work neces sary to heat 1 lb. of water 1° F. is 772, or to heat 1 lb. of water 1° C. is 1390. This is called Joule's equivalent. More recent determinations by Rowland give 778 and 1400 instead. To heat 1 kg. of water 1° C. requires 427 kilogrammeters of work. axle of a wheel, with paddles at the other end of the axle. This was arranged so that, on letting the weight fall, the paddles were caused to rotate in a known quantity of water in a vessel. The weight multiplied by the distance through which it falls gives the mechanical work, and the mass of water multiplied by the change of temperature gives the heat units. Joule's experiment showed that a 10-lb. weight falling through 77.2 ft. would raise the temperature of 1 lb. of water through 1°F. FIG. 259 -Demonstration. -Place 311. The Pressure of Water Vapor.a quantity of water in the flask A (Fig. 259). Raise the water to the boiling point, and after it has boiled a short time remove the flame |