Name Partner Thermal Physics Part I: Heat of Vaporization of Nitrogen Introduction: The heat of vaporization of a liquid, L v, is the energy required to vaporize (boil) a unit mass of substance. Thus if a mass, m, of liquid is vaporized, a quantity of heat, Q, must be added where, Q = L v m. (1) In this experiment you will use an electrical heater to add energy to a sample of liquid nitrogen. From measurements of the current, I, and the voltage, V, applied to the heater you can calculate the rate at which energy is added to the liquid. This rate is the electrical power, P = IV. (2) You will measure the rate at which the liquid s mass decreases, -dm/dt, and thus calculate the heat of vaporization from P Lv =. dm / dt A slight correction must be made to this expression because the liquid will be vaporizing without the addition of the thermal energy as a result of heat transfer from the surroundings. The resulting mass loss due to the surroundings must be measured and subtracted to find the mass loss due to the added heat. Thus, the equation used to calculate the heat of vaporization is L v P = ( dm / dt) ( dm / dt) In this equation the subscripts on and off refer to the electrical heater being turned on or off. Note that the negative sign in the numerator makes the heat of vaporization a positive quantity. Experimental Electrical Circuit: Assemble the electrical circuit shown below. Use digital multimeters to make both current and potential measurements. Have your instructor check the wiring before you proceed. on off. (3) (4) Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 1 of 6
Vaporizing Nitrogen: General Instructions: You will now fill a styrofoam cup with liquid nitrogen and lower the heating element into the liquid. By measuring the rate of mass loss you can find the rate of evaporation due to the surrounding energy inputs. You will then turn on the heater, recording current and voltage so you can calculate the power, P, the rate at which electrical energy is added, as well as measuring the mass loss. Finally, you will then turn the heater off and once again measure the mass loss with no added electrical energy. Specific Procedures: 1. Place the two nested styrofoam cups on the electronic balance and press the TARE button to zero the balance. 2. Remove the nested cups and set them on the lab bench. Slowly add liquid nitrogen from the dewar (the insulated thermos bottle). Be Careful! Liquid nitrogen will boil vigorously when first added to the cup. Splashes can burn your skin. Fill the cup near, but not to the top. 3. Place the nested cups on the electronic balance, and slowly lower the heating coil into the liquid. Again, be careful of splashing liquid. 4. When boiling has subsided, slowly add more liquid nitrogen until the cup is almost full. 5. Beginning at time, t = 0, record mass as a function of time for 240 seconds. Take readings every 30 seconds. 6. Turn on the digital voltmeter, digital ammeter, and power supply. Slowly raise the voltage of the power supply until the current flowing through the heating element is almost, but not quite, two amperes. Record the voltage and current readings and continue to monitor the two meters to ensure that they do not change significantly during the experiment. 7. Continue recording mass as a function of time for an additional 240 seconds, taking readings every 30 seconds. 8. Turn the heater off and continue to take readings of the mass as a function of time for an additional 240 seconds. Data Analysis: To analyze your results using Graphical Analysis, your data must be entered into the computer as three separate data series, corresponding to the three time intervals (power off, power on, and power off again) during which you took data. The computer can then calculate the slopes of each series individually on a single graph. Procedure: 1. Since the electronic balances measure to 0.1 grams you will need to set the computer to accept data good to one decimal place. So before entering you data, choose option R from the main menu. To set to one decimal place, delete the current setting and enter 1D. Then return to the main menu. 2. Choose option I to enter your data. 3. Remembering that you will be plotting mass vertically and time horizontally, choose appropriate labels for the axes of your graph. Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 2 of 6
4. Enter your data for the first 240 seconds, then press the ESC key to return to the main menu. 5. Choose option I. From the submenu which appears, choose option S-Add data as the new data series. 6. Enter your data for the 240 seconds during which the power was on. This will be data series 2. 7. Return to the main menu. 8. Again choose option I, then option S. 9. Enter your data for the last 240 seconds. 10. Return to the main menu and choose option P to plot your graph. 11. The differences in the slope of the graph for the three data series will probably show up better if you do not start the vertical axis at the origin. Therefore, choose automatic scaling for the horizontal (time) axis, but choose manual scaling for the vertical axis at the closest multiple of ten that is still less than your smallest mass value. For example, if your lowest mass value was 55.4 grams, start you vertical axis at 50. For pixel size, simply go with the computer s choice. 12. For your graph style choose options P, R, and S. Leave the grid off your graph as it can make the statistics hard to read in a case like this when the computer will be displaying several lines of information. 13. If your graph is satisfactory, print out a copy and proceed with the rest of your calculations. Consult your instructor if you have any problems. 14. Find the average of the slopes for data series 1 and 3. The result is the rate of mass loss without the added heat input (heater off). 15. The slope for data series 2 is the rate of mass loss with the heater on. 16. Use Equation (4) to find the heat of vaporization for liquid nitrogen. Include approximate uncertainties. Compare with the accepted value, 199.7 J/g. Part II: Measuring Work Transfer Introduction: In the eighteenth century, before the Law of Conservation of Energy was established, heat was thought to be a substance called caloric, and was, therefore, measured in calories. In about 1800, Benjamin Thompson (Count Rumford) performed a series of experiments that showed heat was not a substance, but was instead a form of energy. Today we know that the internal energy of a substance can be changed either by adding heat to the substance or by doing mechanical work on it. You will do work on a metal cylinder by rubbing a cord against it. The apparatus to be used consists of a copper cylinder that can be rotated on a shaft by a hand crank. A cord, which is wrapped several times around the cylinder, supports a block of mass M. As the crank is turned, the cord slips on the cylinder so that the block does not move. Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 3 of 6
How much work is done? If, as the crank turned through N complete revolutions, the cord had not slipped, then the block would have risen a vertical distance h = N (2BR), where R is the radius of the cylinder. The potential energy of the block would have been raised by an amount (5) E = M g h = 2π M g N R. In the experiment however, however, the cord does slip, the block does not rise, and the potential energy is not increased. The work done, 2B M g N R, calculated above, must have increased the internal energy, U, of the copper cylinder. The change in internal energy, )U, causes a temperature rise that depends on the mass, m, and specific heat, c, of the cylinder: (6) U = mc T = 2π M g N R. By measuring the radius of the cylinder, R, the mass, M, of the hanging block and counting the number of turns, N, you can find the work that you did. By measuring the mass, m, and temperature rise, )T, you can then calculate the specific heat, c, of the copper. Procedure: 1. Measure the mass, m, and diameter, 2R, of the copper cylinder and record the mass, M, of the hanging block. 2. Cool the cylinder to about 2 o C below room temperature by immersing it in cold water. Dry the cylinder completely before proceeding. 3. Screw the cylinder on the insulating shaft and wind the cord clockwise (as viewed from the handle) on the cylinder until the cord just reaches the support pin. 4. Connect the block to the hook on the other end of the cord. 5. Place a drop of two of glycerol on the tip of the thermometer and carefully insert it into the end of the cylinder through the plastic cover. Be extremely careful with the thermometer. It is fragile and expensive. 6. Turn the crank slowly until the block is raised a few centimeters off the floor. Make sure the crank can be turned while the cord slips on the cylinder, so the block rises no further. Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 4 of 6
7. Record the initial cylinder temperature. 8. Turn the crank rapidly through about 100 revolutions, recording the exact number. 9. Stop turning and record the highest temperature the cylinder reaches. 10. Calculate the specific heat of the cylinder using Equation (6). Questions: 1. How does the value of the specific heat of coppper compare with the value in the C.R.C. Handbook of Chemistry and Physics? 2. If the block continued to rise as you turned the crank, would the value of the specific heat you calculated be too high or too low? 3. Some heat is carried away from the copper cylinder by convection currents in the air. How will this affect your value of the specific heat? 4. Would our results be improved if you turned the crank 10 revolutions instead of 100? 1000 revolutions instead of 100? Explain. 5. Suppose 1 g of water had adhered to the surface after cooling it. How would that have affected your result? 6. In the United States the energy content of food is computed in Calories. Assuming your body to be 50% efficient in converting food energy into mechanical work, how many turns of the crank would be necessary to work off an average breakfast of 600 Calories? Part III: Measuring Heat Transfer Introduction: When experiments are performed on various substances, it is found that different amounts of heat are required to produce the same change in temperature of the same mass for different substances. For example, if one kilogram each of lead, iron, aluminum and water are heated to 100 o C and dropped separately into equal masses of cold water, the aluminum will warm the cold water twice as much as the iron, the iron about three times as much as the lead and the hot water about thirty times as much as the lead. It is found experimentally that the quantity of heat transferred into or out of a body, Q, while it is undergoing a temperature change )T = T 2 - T 1, is directly proportional to the mass, m, of the body and the change in temperature and depends on the particular substance under investigation. That is, Q m and Q T2 T1, or, Q = mc T, (7) where c is the specific heat of the substance. It is the amount of heat required to change the temperature of a unit mass of the substance by one Kelvin. The specific heat of water is (8) Calorimetric Measurements: cw = 4188 J / kg * K. Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 5 of 6
Calorimetric means literally the measuring of calories, it is the science of measuring quantities of heat transferred. The method that is most often used in determining specific heats is known as the method of mixtures. If two bodies, originally at two different temperatures, are placed in thermal contact an exchange of heat takes place exclusively between the two bodies, then the heat given up by one body will equal the heat absorbed by the other. The experiment consists of dropping a known mass of some specimen heated to a known high temperature into a measured mass of water at a known low temperature and measuring the resulting temperature. The heat transferred to the water and its containing vessel can be easily computed and equated to the heat transferred from the hot metal. From the results the unknown specific heat of the sample can be computed. Consider a sample, s. Let m s grams of the sample having specific heat c s be heated to a temperature T 1s. Drop this sample into a mass, m w, of water contained in a calorimeter of mass, m c. Originally the calorimeter (specific heat, c c ) and water are at a temperature T 1w. As the hot sample is dropped into the water the temperature of the sample will drop to T 2s while the temperature of the water and calorimeter will rise to T 2w. At equilibrium, (9) T = T w. 2s 2 If no heat is gained or lost to the surroundings, then m c ( T T ) = m c ( T T ) + m c ( T T ), s s 1s 2s w w 2w 1w c c 2w 1w at equilibrium. This equation enables us to determine the specific heat of the sample. Procedure: 1. Fill the electric steam generator to about 1/3 to ½ full of water and then heat it to the boiling point. 2. Weigh the (dry) mass and suspend it in the steam generator, monitoring the temperature of the thermometer which is also suspended in the steam generator. Take care not to touch the body of the hot steam generator and be careful of escaping hot steam when you open the top of the generator. Also, be extremely careful in handling the thermometers. 3. Determine the mass of the cup and stirrer. The calorimeter cup and stirrer are made of aluminum. 4. Put about 100 grams of water in the calorimeter cup and acurately determine the mass of the water. Best results will be obtained when the initial water temperature is a few degrees below room temperature and the final temperature is a few degrees above room temperature; in this way radiation losses to and gains from the room are approximately negated. 5. Measure the initial temperature of the water. 6. Quickly, but carefully, transfer the hot mass from the steam generator to the calorimeter, cover the calorimeter and stir the water, noting the temperature on the thermometer. 7. Record the highest temperature the water reaches. 8. Calculate the specific heat of the sample using Equation (10). Follow this procedure to determine the specific heat of three different samples. Check your results against those listed in the C.R.C. Handbook of Chemistry and Physics. Questions: (10) Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 6 of 6
1. What is the specific heat per mole for each of the three samples which you studied? This value should be almost the same for all three. Compare with the known value. 2. Why is the aluminum sample physically so much larger than the other samples used in this experiment? 3. Is it desirable for the coolant in your car s radiator to have a large or a small heat capacity? Explain. Physics Laboratory Manual. Sept.2003 Thermal Physics - Page 7 of 6