Compare to Q=m,Cw T energy absorbed by water, where the specific heat of water Cw=4186 J kgK Procedure 1) Go to the following link http://seilias.gr/go-lab/html5/joulelaw.plain.html 2 ) Set the current to 1 A and resistor to 2 . The temperature will always begin at 20 C. Close the switch and press the play. 3) The temperature will rise until it reaches 95 C, at which point the timer will automatically stop. Record the total time and change in temperature in the table below. The mass of the water is 80g. Calculate E and Q, and find the percent error between the two. Use 2 significant figures for all calculations of E.Q. averages, and percent errors Table 1 [[A] 1 AT[ c] R Q t [s] E [J] Q[J] 2 4 5 8 10 E avg [ J] Qavg] % 8 E 4) Set the current to 3A and repeat. Each time you reset you will need to change that current to 3A. Table 2 I[A] 3 AT[ C] R $2] t[s] E [J] Q 2 4 6 8 Credits: Joshua SondereggerAnalysis 1) How does the energy gained by the water compare to the energy given off by the resistor in the water (was it higher/lower)? What possible reasons could account for this if it was a physical experiment? 2) When the resistance of the resistor in the water was changed, does the current also change? What enables the current to do this? Credits: Joshua Sonderegger Theom: (For additional insight you may refer to Experiment 27 of your Lab Manual) Resistors lower the amount of electrical energy in a circuit by converting electrical energy into thermal energy. This change in energy is displayed as a voltage difference. The amount of power a resistor element is given by: P=IV=IzR p= t Therefore E :12 Rtenergy generated by the resistor in the water. Credits: Joshua Scnderegger