Experiment #4, Ohmic Heat

Experiment #4, Ohmic Heat 1 Purpose Physics 18 - Fall 013 - Experiment #4 1 1. To demonstrate the conversion of the electric energy into heat.. To demonstrate that the rate of heat generation in an electrical conductor is proportional to the product of the electric potential drop across it and the electric current flowing through it. 3. To demonstrate that liquid nitrogen vaporizes when an electric conductor connected to a DC power supply is placed in liquid nitrogen (LN ). In addition, to show that the rate at which the liquid nitrogen vaporizes is proportional to the power, VI, dissipated in the conductor. 4. To determine the value of the latent heat of vaporization of LN. Introduction Heat is generated when an electric current passes through a resistive conductor. Heat produced in this manner is referred to as ohmic heat. An electric current is the flow of free electrons through the conductor. These electrons suffer frequent collisions with the atoms (or lattice ions) in the conductor, continually transferring their kinetic energy to these atoms. This causes the atoms to vibrate with greater energy, raising the temperature of the conductor. In this way electrical energy (i.e., the kinetic energy of the free electrons) is converted into thermal energy the vibrations of the atoms. The work done W I, in units of joule (1 J = 1 kg. m /s ), by a battery or power supply in driving an electric current, I, through a resistor, R, in a time interval Δt can be shown to be: W I = VIΔt (joule) (1) where V (in volts) is the potential drop across the resistor. If I is measured in amperes (A), R in ohms ( ) and t in seconds then the work done has units of joule. Using Ohm's law (V=IR), we can express this result in terms of the voltage, V, i.e., W I = V Δt/R or the current, I, W I = I RΔt. Since this energy is converted into heat, the temperature of the conductor (and any parts of the experimental system with which the conductor is in thermal contact) will increase. This temperature changes the value of the resistance. The resistance may increase or decrease, depending on the material used for the resistor. Thus, if U is the thermal energy generated in the system over the time interval Δt for which the current flows, then: U = VIΔt (joule) () If the resistor is immersed in liquid nitrogen (LN ), the heat, U, generated in the resistor is rapidly transferred to the LN, causing some of it to be evaporated. The amount of heat energy Q required to evaporate a mass of LN is given by: Q = L v (joule) (3) where L v is the latent heat of vaporization of LN. If all the heat energy is transferred from the resistor to LN, then conservation of energy requires that Q = U = W I, and the amount of LN evaporated can be determined by equating Eqs. (3) and (1), i.e., L v = (VI)Δt or, (VI) LV (4) ( ) Δt

Physics 18 - Fall 013 - Experiment #4 Equation (4) shows that, for a current I flowing in a conductor, the rate (/Δt) at which the LN evaporates is proportional to VI, and accordingly a plot of VI vs. /Δt should produce a straight line whose intercept is equal to zero and whose slope is equal to L v. (Intercept may not be zero when doing the experiment.) 3 Experimental Apparatus and Procedure 3.1 Apparatus 1. DC power supply.. Agilent 34405A digital volt-ohmmeter (to be used for voltage and resistance measurements). 3. Fluke 75 digital multi-meter (to be used for all current measurements). 4. Carbon resistor. The circuit is shown in Fig. 1. The current I through the resistor is set by the power supply and when the switch is turned on a current will flow in the resistor R which is immersed in LN. The nominal value for the electrical resistance of the carbon resistor is Ω, so for a current of 1 amp the ohmic heat generated in the resistor will be joule/s. Carbon is a semimetal, and its resistance will increase when the temperature decreases. There is an energy gap between the valence and conduction bands and the probability of an electron going from valence to conduction is temperature dependent. When cooled, electrons are forced from the conduction band to the valence band and the resistance is increased. The experimental setup, with the resistor immersed in LN, is shown in Fig.. The LN is contained in a Styrofoam cup. To measure the mass of LN which is evaporated during the passage of an electric current in the resistor, the cup is mounted on the right-hand pan of a dual pan balance. Some LN will undergo evaporation even without current flow since heat will conduct through the cup, the cup is open at its top and the resistor leads will also conduct heat from the room into the LN. To counterbalance this, an identical Styrofoam cup and resistor arrangement is mounted on the lefthand balance pan. When the balance is brought into equilibrium by means of the slider weight, it should remain in equilibrium until you begin the current flow in the conductor on right side.

Physics 18 - Fall 013 - Experiment #4 3 3. Procedure 1. Place a pair of Styrofoam cups on each scale pan.. Suspend the resistors with their leads (Figure ) inside the cups. Be sure the resistors are supported so as not to touch the cup sidewalls or bottom. Please note that the resistor on the right pan of the scale forms part of the electrical circuit (Figure 1). 3. Turn on the Agilent 34405A digital volt-ohmmeter multimeter, and measure the resistance of the resistor in the circuit (on the right side). This is R a. Use the correct setting and range. 4. Pour liquid nitrogen (LN ) into each double Styrofoam cup on the scale until the liquid is one centimeter below the top of the cup. Measure the resistance of the resistor (on the right side) which is now immersed in LN. This is R b. 5. Turn on the Fluke 75 meter to the DC Amp current setting. Connect the red cable to the DC power supply, and turn it on. For safety, you should operate the power supply with only one hand. Adjust the power supply so it provides a current of 0.70 amps in the resistor. The Agilent 34405A should be set to measure voltage, and set to the correct range. Turn off the power supply. Set the timer to zero. 6. If necessary, add additional LN (one centimeter below the top of the cup). Carefully adjust the amount of LN until the scale is brought into balance. You may use the spoon to remove some of the LN from one cup to another. Observe that the scale remains in balance for a long period of time, indicating that the LN evaporates at the same rate from both sets of cups. (If the "dummy" resistor was not introduced into the second stacked cup, the scale would become unbalanced due to different rates of evaporation.) 7. Once the scale is balanced, move the top slider to the 1.00g position. The right side of the pan balance should go down. Turn on the DC power supply. Start the timer when the scale reaches the balance point. Observe that nitrogen gas is emitted from the right cup as the current flows in the resistor. Move the slider to the 5.00g position and record the time when the scale is again balanced, indicating that 4.00 grams of LN evaporated. Record the current and voltage at this time. Return the timer to zero. 8. Before turning off the power supply, reduce the voltage until you have a current of 0.65 amps. Turn off the power supply. 9. With the power supply off, enter data from your first run into Excel. If your calculation of VI t shows a large difference from 800.00J, return the current to 0.70 amps, and repeat your first measurement (step 5). 10. Before turning on the power supply for your next experimental run (0.65 amps), refill both double Styrofoam cups to the same original level (one centimeter below the top of the cup). Repeat steps 6-10 for current values of 0.65, 0.60, 0.55 and 0.50 amps. For each run, make sure both cups are always refilled with liquid nitrogen. 11. With the power supply off, disconnect the red cable from the power supply and use the ohmmeter setting on the Agilent 34405A to measure the resistance of the carbon resistor while it is still immersed in LN. This is R c. (If R c is not approximately the same value as R b, inform your TA.) Please note that the carbon resistor is at the lowest experimental temperature when immersed in liquid nitrogen and no current is flowing through it. Here, VI = 0.00 J/s.

4 Calculations, Analysis and Graph Physics 18 - Fall 013 - Experiment #4 4 1. All calculations should be done in the lab and for your lab report. Compute the value of V/I corresponding to each value of current I and voltage V. Compute the evaporation rate /Δt (grams/second). Compute VI in joule/second and VI T in joules.. Plot a graph (Excel in the lab and graph paper for your report) of VI vs /Δt. The graph should be a straight line. Compute the slope of this line. Use non-data points for this calculation on your graph. 3. Use your calculator s linear regression function to determine your slope, intercept and coefficient of determination, R, in your lab report. Be sure to enter only data that appears to be linear from your graph in part. Do not use data points that are far from your line of best fit. Include in your lab report the coefficient of determination R. The coefficient of determination reflects linearity. 4. Use your calculator s linear regression results to determine the final experimental value (L v,exp ) of the latent heat of vaporization of LN and compare it to the hand book value (L v = 00.00 joule/g). Remember, your Excel results in the lab are used for the Quick Sheet to confirm the results of your experiment. Your lab report should use results from your calculator. 5 Questions 1. Does the analysis of your data verify the validity of Eq. 4? If it does, explain how.. During the experiment, do the 5 values of V/I, the resistance R, remain constant? If they do not remain constant, why? When answering this question, you should equate the power term VI to reflect the temperature of the carbon resistor in each of the 5 experimental runs. 3. Compute the average value of the work W I ( W I = VI t ) done by the power supply. Include both the standard deviation (s) and the standard deviation of the mean (SDM) for W I. Use the 5 values of VI t from your data table. Why should the average value equal 800.00 J when 4.00 grams of liquid nitrogen is evaporated? You may use a calculation to answer this part of the question. 4. Use the following propagation of error expression to calculate the error L v in your experimental value of L v (L v,exp ). The physics lab Web site has an appendix on error analysis which gives details on this expression. As stated in 4.4, use your linear regression results for your experimental value (L v,exp ). SDMWI δm L v = L v,exp ( ) + ( ) WI δm where m is the error in reading the scale balance and M should be 4.00 grams. We use ( ) as δm two scale readings are used to determine the 4.00 grams. This term may too be written as: ( ). Including the uncertainty (error) L v yields: L v = L v,exp L v. 5. Is the difference between the accepted value of L v and your experimental value of L v within the limits of experimental error determined by ± L v,exp? If the difference falls within the limits of experimental error,

then the following should be true: Physics 18 - Fall 013 - Experiment #4 5 L v,accepted - L v,exp L v,exp i.e., that the positive absolute value ( ) of L v,accepted - L v,exp is less than or equal to L v,exp. Include your calculations when answering this question. 6. The accepted value of L v is 00.00 joule/g. What is this value in cal/g. Use 1cal = 4.184 J. 7. To answer the following questions, you will need to use equations and information from the experiments in The Specific Heat of Aluminum and The Latent Heat of Vaporization of Liquid Nitrogen. (a) (b) (c) A wire wound aluminum conductor (R = 50.00 Ω and m = 15.00 g) is initially at 0.0 C. The conductor, thermally insulated from its environment, is connected to a 10.00 V power supply for 3.00 minutes. What will be the final temperature of the aluminum? Assume the resistance remains constant. In this temperature range, c Al = 0.15 cal/g- C. With its thermal insulation removed and disconnected from the power supply, the resistor from part (a) is placed in LN. Calculate the amount of LN that evaporates while the resistor is cooling from room temperature to LN temperature. Assume room temperature is 0.0 C. Do not use experimental values when doing this calculation. Use accepted values only. While the uninsulated resistor is in liquid nitrogen, it is supplied with 10.00 V for 3.00 minutes. How much additional LN will evaporate? Assume the resistance remains constant. 6 Conclusions This section should have a clear statement of the results of the experiment and the extent to which the results are in agreement with the theory being tested. When the experiment results in a measurement of a constant, e.g., the latent heat of vaporization of LN, L v,exp, compare it with its established handbook values. Use percent error for this comparison. To make this comparison meaningful, you should include the impact of the experimental error on your results. This includes errors in plotting and reading linear graphs when determining their slope and intercept. In addition, please include a statement of what you have learned, a critique of the experiment, and any suggestions you have which you think could improve the experiment or the lab handout.

Ohmic Heat Data Table Physics 18 - Fall 013 - Experiment #4 6 Warning: Do Not Turn on the Power Supply until the Resistor is Immersed in Liquid Nitrogen. Measure the resistance of the carbon resistor before it is immersed in liquid nitrogen and with no current flowing through it. R a = Ω. Measure the resistance of the carbon resistor when it is immersed in liquid nitrogen and with no current flowing through it. R b = Ω. You are required to calculate VI t for each experimental run before you proceed to the next current. VI t should be approximately 800.00 J. Current I amp Volt V volt V/I R ohm Initial Slider Setting g Final Slider Setting g Final Time Δt s g /Δt g/s VI joule/s VI t (Work done) W I joule Once again, measure the resistance of the carbon resistor while it is still immersed in liquid nitrogen and with no current flowing through it. Power supply should be off. R c = Ω. Calculate the average value of work done. W I = VI t = J (Use above table data.) Estimated error in reading scale balance. Error M = m = g