Experiment P39: Electrical Equivalent of Heat (Power Amplifier, Temperature Sensor)
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1 PASCO scientific Vol. 2 Physics Lab Manual: P39-1 Experiment P39: (Power Amplifier, Temperature Sensor) Concept Time SW Interface Macintosh file Windows file energy 45 m 700 P39 Elect. Equivalent of Heat P39_ELEC.SWS EQUIPMENT NEEDED Interface (2) Patch Cords Power Amplifier Resistor for heating water (10 Ω, 1W) Temperature Sensor Styrofoam cup with lid Balance (for measuring mass) water PURPOSE The purpose of this laboratory is to heat water and to show that the energy dissipated by a heating resistor in the water is equal to the energy absorbed by the water. This concept is referred to as joule heating. From the law of conservation of energy the electrical equivalent of heat can be found. This is the number of joules of electrical energy that are equivalent to one calorie of thermal energy. THEORY The water is heated by submerging a heating resistor in water and running a current through the resistor. The Joule heat from the resistor is transferred to the water and the resulting temperature rise is recorded by the Temperature Sensor. Using the law of conservation of energy, if there are no energy losses to the surroundings, all the energy given off by the resistor should be absorbed by the water. The energy, E, dissipated by the resistor is E = Pt where t is the time during which the current flows through the resistor and P is the power given by P = IV where I is the current through the resistor and V is the voltage across the resistor. The energy gained by the water is given by Q = mc T where m is the mass of the water, c is the specific heat of water (1 cal/g C), and T is the change in temperature of the water. dg 1996, PASCO scientific P39-1
2 P39-2: Physics Lab Manual Vol. 2 PASCO scientific PROCEDURE In this activity, the Power Amplifier supplies electrical energy to a heating resistor at a set voltage. The energy dissipated by the resistor warms a measured quantity of water. The Temperature Sensor measures the change in temperature of the water. The Power Amplifier measures the current through the heating resistor. The program records the voltage and current supplied by the Power Amplifier to the heating resistor as well as and the change in temperature of the water. The program calculates the electrical energy by integrating the electrical power (voltage x current) over time. You will calculate the thermal energy gained by the water based on the known mass of water and its measured temperature change. The electrical energy (in joules) is set equal to the energy gained by the water (in calories) to determine the electrical equivalent of heat. PART I: Computer Setup 1. Connect the interface to the computer, turn on the interface, and turn on the computer. 2. Connect the Temperature Sensor DIN plug to Analog Channel A on the interface, and the Power Amplifier DIN plug to Analog Channel B. 3. Open the document titled as shown: Macintosh P39 Elect. Equivalent of Heat Windows P39_ELEC.SWS The document opens with a Signal Generator window, a Digits display of Temperature, and a Graph display with plots of Temperature vs. Time and Power Output vs. Time. The Power Output is a calculation based on the voltage and current measurements for the heating resistor. The area under the Power Output vs. Time plot will be the electrical energy. P , PASCO scientific dg
3 PASCO scientific Vol. 2 Physics Lab Manual: P39-3 The Signal Generator is set to output 10 volt DC. It is on Auto so it will automatically start when you click REC or MON and automatically stop when you click STOP or PAUSE. Note: For quick reference, see the Experiment Notes window. To bring a display to the top, click on its window or select the name of the display from the list at the end of the Display menu. Change the Experiment Setup window by clicking on the Zoom box or the Restore button in the upper right hand corner of that window. PART II: Sensor Calibration and Equipment Setup You do not need to calibrate the Temperature Sensor or the Power Amplifier. The Temperature Sensor produces a voltage that is proportional to temperature (10 mv = 1.0 Celsius). The default calibration is C = V and C = V. 1. If you have a lid that will fit over the top of the cup, make one hole in the lid for the Temperature Sensor, and a second hole in the lid for the heating resistor. Measure the mass of the Styrofoam cup and lid. Record the mass in the Data section. NOTE: Use water that is about three degrees Celsius below room temperature (about 18 or 19 C) when data collection begins. Take data until the temperature of the water is about three degrees above room temperature (approximately 25 C). This minimizes the effect of the surroundings because the water gains heat from its surroundings for half the activity and loses heat to its surroundings for the other half of the activity. 2. Put about 200 ml of water in the cup and weigh the cup, lid, and water. Measure and record the total mass. Subtract the mass of the cup and lid from the total mass of the cup with water to find the mass of the water. Record the water s mass in the Data section. 3. Put the heating resistor through its hole in the lid. Submerge the resistor in the water. Connect the banana plugs of the heating resistor into the output jacks of the Power Amplifier. 4. Put the Temperature Sensor through its hole in the lid of the cup. dg 1996, PASCO scientific P39-3
4 P39-4: Physics Lab Manual Vol. 2 PASCO scientific TO INTERFACE LID TO INTERFACE CUP TEMPERATURE SENSOR POWER AMPLIFIER HEATING RESISTOR CAUTION: Be sure the resistor is submerged in water when the current is flowing through it. Otherwise it can burn up! PART III: Data Recording 1. Turn on the switch on the back of the Power Amplifier. Click the REC button ( ) to begin collection data. Note the beginning temperature in the Digits display. IMPORTANT: While the data is being taken, gently agitate the water. This will ensure uniform heating. 2. When the temperature reaches three degrees above room temperature, turn off the Power Amplifier switch on the back panel, but continue to agitate the water and collect data. The temperature will continue to rise as the last thermal energy from the resistor is slowly dissipated. 3. When the water temperature begins to drop, click the STOP button ( ) to stop collecting data. Run #1 will appear in the Data list in the Experiment Setup window. P , PASCO scientific dg
5 PASCO scientific Vol. 2 Physics Lab Manual: P39-5 ANALYZING THE DATA 1. Click the graph to make it active. Click the Statistics button ( ) to open the Statistics area on the right side of the graph. Click the Autoscale button ( ) to rescale the graph to fit the data. 2. Click the Statistics Menu button ( ) in the Statistics area for the plot of Temperature vs. Time. Select All of the Above from the Statistics menu. 3. Record the minimum and maximum temperatures (values of y) displayed in the Statistics area. Calculate and record the change in temperature of the water. 4. Click the Statistics Menu button in the Statistics area for the plot of Power Output vs. Time. Select Integration from the Statistics menu. Record the integration value as the electrical energy (joules) used by the heating resistor. 5. Calculate (in calories) the energy (Q) absorbed by the water using Q = mc T, where m is the mass of the water, c is the specific heat of water (1cal/g C), and T is the change in temperature of the water. Record this value in the Data section. 6. By the law of conservation of energy, the electrical energy used by the resistor should equal the thermal energy gained by the water, neglecting losses to the surroundings. Solve for the number of joules per calorie: Electrical Equivalent ( J Electrical Energy ) = cal Thermal Energy Calculate the percent difference between this experimental value and the accepted value (4.184 J/cal). dg 1996, PASCO scientific P39-5
6 P39-6: Physics Lab Manual Vol. 2 PASCO scientific DATA Mass of Foam Cup Mass of Foam Cup with Water Mass of Water = grams = grams = grams Water Temperatures (C) Tempmin Tempmax Temp Electrical Energy (J) Thermal Energy (calories) Accepted Value = joules = calories = J/cal = J/cal Percent difference = % QUESTIONS 1. Was the energy (Q) gained by the water greater or less than the energy (E) dissipated by the resistor? Explain why it was greater or less. 2. The heating resistor is rated at 10 Ω and 1 watt. By how much was its power rating exceeded? (Remember, P = V 2.) Why didn t the resistor burn up? R 3. What are some factors that could account for the percent difference between the experimental and accepted values? P , PASCO scientific dg
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