Keep the Heat. Procedure Determine the mass of BB's in the cup and record their temperature to the nearest 0.1 o C.

Size: px

Start display at page:

Download "Keep the Heat. Procedure Determine the mass of BB's in the cup and record their temperature to the nearest 0.1 o C."

Cordelia Iris Wiggins
5 years ago
Views:

Keep the Heat PART I: MECHANICAL EQUIVALENT OF HEAT When an object falls from a height, it converts Gravitational Potential Energy into Kinetic Energy.

1 Keep the Heat PART I: MECHANICAL EQUIVALENT OF HEAT When an object falls from a height, it converts Gravitational Potential Energy into Kinetic Energy. When it hits something which stops its fall, all of its kinetic energy is converted into heat. Gravitational potential energy of the object at the top of its fall is the work the gravitational field can do on it. Since the distance an object falls and its mass can be measured, the Gravitational Potential Energy and the heat which is produced from the conversion of Work to G.P.E. to Kinetic Energy to Heat is easily determined. Procedure Determine the mass of BB's in the cup and record their temperature to the nearest 0.1 o C. Final Temperature Initial Temperature 23.8 o C 22.0 o C Mass of BB's: g kg Change in Temperature 1.8 o C The length of the tube is equal to the distance the BB's fall, which is 1.00 meter. Place the BB's inside the tube and securely screw the top on the tube. Invert the tube 200 times. Do this quickly so that the shot falls freely and does not rub the sides of the tube during its fall. Quickly open the tube and pour the BB's into the cup and record the temperature of the BB's. Calculate the work performed lifting the BB s during the experiment. [ Work (Joules) ] = [ mass (kg) ] x [ g (9.80 m/s 2 ) ] x [ distance (m) ] 503 j = ( kg) x (9.80 m/s 2 ) x (200.0 m) The specific heat of the BB's is (390 J/kg o C). Calculate the heat energy gained by the BB's. [ Heat (Joules)] = [ specific heat (J/kg o C) ] x [ mass (kg) ] x [ T ( o C) ] 180 j = ( kg) x (390 j/kg o C) x (1.8 o C) Analysis How close do the two values for the heat gained by the BB's and the work done on the BB s agree? Do your values support the law of conservation of energy? The energy measured from the change in temperature of the BB s is approximately half of the work done on the BB s. This does not support the Law of Conservation of Energy since the heat energy out is not equal to the work done on the BB s. Identify some of the sources of error in this experiment. How might you change the procedure or apparatus to reduce the experimental error? The Law of Conservation must be obeyed! If the change in temperature of the tube and the air inside the tube were accounted for, the two values would be closer in value. Keep the Heat 1

2 If an equal mass of lead pellets (specific heat = 190 J/kg o C) was substituted for the BB's, would you expect the change in temperature for the lead to be greater than, less than, or equal to the change in temperature you obtained for the BB's? Explain your reasoning. Since lead has a smaller specific heat than the BB s, the change in temperature would be greater for the lead. Materials with a lower specific heat require less energy to change their temperature. PART II: WORK AND EFFICIENCY As you saw in the previous experiment, the GRAVITATIONAL POTENTIAL ENERGY an object possesses is equal to the work done by the force of gravity when an object falls from a particular point to its lowest position. The object obtained this energy because you lifted it to a particular height. Lifting an object involves work. For an object like a rubber ball, the difference in the height of the ball when released and the height of its bounce is approximately equal to the energy required to compress the ball and overcome the frictional forces of air. This amount of energy is converted to heat. The maximum KINETIC ENERGY for an object occurs when the object is traveling at its maximum velocity. For a falling object this is usually at the time of impact. Thus the kinetic energy of an object (K.E. = ½ m v 2 ) is dependent upon the mass of the object and its velocity. Notice that for any given object, doubling the velocity increases the kinetic energy of the object by 4 times. PROCEDURE Obtain a high bounce ball and a two-meter stick. Determine the mass of your ball. Drop the ball from a series of heights (2.0 m, 1.5 m, 1.0 m, 0.5 m) and record the heights of the first bounce. For each height determine: 1. The gravitational potential energy of the ball when released; 2. The gravitational potential energy at the top of the first bounce; 3. The energy lost in joules (difference in G.P.E. from release point to first bounce); 4. The percent efficiency = G.P.E. at bounce height G.P.E. at release height X 100 Mass: g = kg Weight: N Bounce Height Height Released Trial 1 Trial 2 Trial 3 Average Keep the Heat 2

Height Dropped G.P.E. Ave. Bounce Height G.P.E. Energy Lost % Efficiency 2.0 0.980 1.55 0.760 0.220 78% 1.5 0.735 1.25 0.613 0.122 83% 1.0 0.490 0.80 0.392 0.098 80% 0.5 0.245 0.38 0.186 0.

3 Height Dropped G.P.E. Ave. Bounce Height G.P.E. Energy Lost % Efficiency % % % % Compare the loss of energy of the ball at each of its heights with the total gravitational potential energy of the ball before it is released. Is the percentage of energy lost during the bounce consistent or is a greater percentage of energy lost at greater heights? The amount of energy lost is smaller at successively shorter heights. However, the percentage of energy lost is approximately the same (within experimental error) at each of the heights. How do you account for this lost energy? Heat! As the ball falls, its gravitational potential energy is converted to kinetic energy. At the point of impact, the kinetic energy is converted to elastic potential energy (the ball is compressed upon impact with the floor). Some of the kinetic energy is converted to heat through the friction of the ball with the floor. The stored elastic potential energy is then converted to kinetic energy as the ball expands to its original shape, propelling the ball upward. PART III: GO EAT WAX! In this experiment you will determine the Caloric content of wax. Wax is also called paraffin, which is derived from kerosene, a distilled component of petroleum. Determine the mass of the candle. Add 50.0 ml of water to the soup can and record the temperature of the water. Place the soup can on the stand so that it is supported about 2 cm above the candle. Ignite the candle and heat the water for about 10 minutes. Stir occasionally. Record the final temperature of the water and the final mass of the candle. Initial Temperature of Water 22.0 o C Change in Temperature Final Temperature of Water 58.3 o C 36.3 o C Initial Mass of Candle g Change in Mass Final Mass of Candle g 0.24 g Determine the heat absorbed by the water j = ( kg) x (4190 j/kg o C) x 36.3 o C Keep the Heat 3

Determine the heat released (per gram) by the candle. 7600 j / 0.24 g = 31,700 j/g The known caloric content of paraffin (also called heat of combustion) is 42,000 J/g.

4 Determine the heat released (per gram) by the candle j / 0.24 g = 31,700 j/g The known caloric content of paraffin (also called heat of combustion) is 42,000 J/g. How does this value compare with your experimental determination? Where is the greatest source of error in this experiment? The experimental value is about 25% lower than the known value of paraffin. Some of the energy released by the candle was not absorbed by the water (went into heating the surrounding air) and some of the heat was used to heat up the can. The Law of Conservation must be obeyed!! Use the data from the label to determine the energy (Joules) in one gram of Cheeto? Is this more or less than the energy obtained from burning wax? 150 Calories / 28 g = 150,000 calories / 28 g = 5360 calories / g = 22,400 joules / g This is about 2/3 of the energy released by the paraffin. Yes, eating wax does yeild a greater amount of energy, but Cheetos are much more delicious. POST-LAB QUESTIONS 1. COLD WATER DIET The label on a bottle of Poland Springs water indicates that water contains 0 Calories. However, drinking cold water requires energy! For example, assume you drink 4.00 liters of ice cold water on a hot summer day. The water is 5.0 o C, and to excrete the water your body must warm the water to body temperature, or 37.0 o C. How many Calories are required for your body to process 4 liters of ice cold water? Can you justify eating the Butterfinger candy bar? (4.00 kg) x (4190 j/kg o C) x (32.0 o C) = 536,000 j = 128,000 calories = 128 Calories It takes about a half of a candy bar (about 250 Calories per bar) just to heat up the water! Just stay close to a bathroom! Keep the Heat 4

5 2. I was walking by Henry Barnard School the other day and observed some kids playing jump rope, which got me thinking. Suppose you decide to jump rope with the children to burn off some junk food. What is your weight? (in Newtons) kg = 980 N If every time you jump the rope you jump 30.0 cm off the ground. How much work (in Joules) do you perform jumping once? (Work = Force x distance) 980 N x m = 294 joules per jump 4.18 Joules is equivalent to 1 calorie. How many calories is equivalent to jumping rope one time? How many Calories? 294 j = 70.0 calories = Calories per jump How many times would you have to jump the rope to burn off the chemical energy supplied by a miniature Butterfinger candy bar of 175 Calories? 175 Calories / Calories per jump = 2500 jumps! If you jump once every 2.0 seconds, how many seconds would you have to jump rope? What is your power output in Watts for this period? 2500 jumps x 2.0 seconds per jump = 5000 seconds (almost 1 ½ hours!) 294 joules / 2.0 seconds = 147 Watts Keep the Heat 5

WORK, POWER, & ENERGY

WORK, POWER, & ENERGY In physics, work is done when a force acting on an object causes it to move a distance. There are several good examples of work which can be observed everyday - a person pushing a