Phy 213: General Physics III Page 1 of 1Instructor: Tony ZableExperiment: Joule’s Law of HeatingPRELIMINARY QUESTIONS1) A thermally isolated container holds 50 mL water. A 5  resistor, connected to a 10 V power supply, is placed into the water and the circuit is turned on for 5 minutes. During this time heat generated by the resistor is absorbed by the water, which heats up from an initial temperature of 20 oC to a final temperature of 48.7oC.a) What is the rate of energy (i.e. power) dissipated through the resistor, in Joules per second?b) How much energy is dissipated during the 5 minute interval?c) The specific heat capacity of water, c, is 1.000 calories/goC . How many calories of heat are absorbed by the water during this experiment?d) If no energy is lost to the container or outside environment, the energy dissipated (lost) by the circuit is equal to the energy gained by the water. Both of these values were calculated above. Based on the results of your calculations, what is the relationship betweenelectrical energy in Joules and heat energy.Phy 213: General Physics III Page 2 of 2Instructor: Tony ZableINTRODUCTIONWhen an electric current flows through a resistance energy is expended (i.e. work is required to drive charge through the resistor). Potential difference is defined as the work per unit charge required to move a charge from one point to another. Symbolically, V = W/q (1)where V is potential difference (also known as the voltage or electromotive force), W the work done, and q the quantity of charge transported. The unit of potential difference is the volt, which is the potential difference such that 1 joule of work is used to move 1 coulomb of charge (or 1 volt = 1 J/C).Quantity of charge, q, is the product of the intensity of the current (i) and the time (t) the current flows. For a steady current,q = it (2)The energy expended in an electric circuit can be written as,W = Vq = Vit = i2Rt (in joules) (3)Recall from last term that the heat (Q) absorbed by a substance is related to the mass of the substance (m), the specific heat capacity of the substance (c), and the change in temperature of the substance (T) according to the relation,Q = cmT (in calories) (4)where Q is in calories, c is in cal/goC, m is in g, and T is in oC. When immersed in a volume of water, the energy expended in a circuit, which is released as heat, is related to the heat absorbed by the liquid. For water, recall the specific heat capacity is 1.00 cal/g oC. The relationship between heat absorbed by the water and the energy expelled by the circuit isW = JQ (5)where J is called Joule’s constant. Experiments have determined the value of J to be approximately 4.18 J/cal. Rearranging equation (5), Q = W/J = i2Rt/J (6)Combining equations (4) and (6) then rearranging terms,cmT = i2Rt/J (7)or T = (R.t/c.m.J).I2(8)In this experiment, we are going to use this relationship to determine Joule’s constant.in calories. This value is known as Joule’s constant.Phy 213: General Physics III Page 3 of 3Instructor: Tony ZableOBJECTIVES To study the heating effect due to an electric current  To determine Joule’s constant (J) using an electrocalorimeterMATERIALSLogger Pro wiresLabPro Interface clips to hold wiresVernier Temperature Probe adjustableVernier Current and Voltage Probe electric calorimeteradjustable 5-volt DC power supply+ -IR e s i st o rB lac kR e dCu rr e n t p r o b eV o l t a g e p r o b eFigure 1EXPERIMENTAL SETUP1. Prepare the computer for data collection by opening “Exp 25” from the Physics with Computers experiment files of Logger Pro. The Meter window should display potential andcurrent readings.2. Connect a Voltage Probe to Ch1 on the LabPro. Connect Current Probe to Ch2. 3. With the power supply turned off, connect the power supply, electric calorimeter, wires, and clips as shown in Figure 1. Take care that the positive lead from the power supply andthe red terminal from the Current & Voltage Probe are connected as shown in Figure 1. Note: Attach the red connectors electrically closer to the positive side of the power supply.4. Select Zero from the Experiment dropdown box. A dialog box will appear. Select “Zero Potential” then “Zero Current”. This sets the zero for both probes with no current flowing and with no voltage applied.Phy 213: General Physics III Page 4 of 4Instructor: Tony Zable5. Have your instructor check the arrangement of the wires before proceeding. Turn the control on the DC power supply to 0 V and then turn on the power supply. Slowly increasethe voltage to 2.5 V. Record the current and potential values. Calculate the resistance of the electric calorimeter using the relationshipR = V/i6. Connect the temperature probe to Ch 3 and update the sensor settings for Ch3 in the Set-Up menu. Add the temperature reading to the meter window.PROCEDURE1. Record the value of the resistor in the data table.2. Place a known mass (volume) of water into the electric calorimeter. Use the thermometer to measure the initial temperature of the water.3. With the circuit disconnected, set the power supply to 5 V and the rheostat to near its lowest setting.4. Connect the circuit and let run for 5 min. During the run, record the current and potentialdifference across the calorimeter (these values should stay fairly constant!). Stir occasionally to insure good heat transfer between the water and the coil.5. At the end of the 5 minutes, disconnect the circuit and record the final temperature of thewater. Be sure to wait until the water reaches its peak temperature.6. Replace the water with fresh water. Change the setting of the rheostat and repeat steps 2-5.7. Repeat process for a total of 5 trials.DATA TABLESet-up  V = _____V i = _____A R = Trial # Mass of water(g)Current(A)Initial Temp.(oC)Final Temp.(oC)T12345Phy 213: General Physics III Page 5 of 5Instructor: Tony ZableANALYSIS1. Using Graphical Analysis, create a plot of T vs i2 for your data points. Is the graph linear? 2. Generate a best-fit line for your graph. The slope of the line is ________±_______.3. Calculate Joule’s Constant, J, from the slope of the graph.4. Compare your value for J with the accepted value. What is the % error? 5. What is the significance of the value of