Then set the voltage to 10v finally set the data to

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Then set the voltage to 10V. Finally, set the data to show the area under the curve of power and this will give the resistor energy. Now, take the mass of the water and then hook up the system (shown in Appendix A Figure 1) and turn on the power. After the temperature change raises about 2 degrees, turn off the power. The temperature will eventually level off and that will be the change in temperature. And the area under the curve up to the second before the power supply is turned off. This is the energy of the resistor. Data Analysis: Table 1: Date acquired from experiment and determined values after certain calculations Mass of Foam Cup 9.7g Mass of Aluminum Cup 33.1g Mass of Foam Cup with Water 157.1g Mass of Water 147.4g Water Temp Min 21.2°C Water Temp Max 25.2°C Change in Temp 4.0°C Electrical Energy (E) 2384J Thermal Energy (Q) 589.6 cal Accepted Value (J) 4.184 J/cal Experimental Value (J) 4.043 J/cal Percent Error 3.37% Questions: 1) Was the energy (Q) gained by the water greater than the energy (E) dissipated by the resistor? Explain why it was greater or less. Our electrical energy result divided by our thermal energy result yielded a value smaller than that which is accepted (4.043J/cal<4.184J/cal). Therefore, the energy gained by the water was greater than the energy dissipated by the resistor (2384J>985.3J). This is because the process is not 100% efficient. Not all of the energy from the resistor is going to be directly transferred to heating the water. Environmental effects and simple human error prevent a perfect system such as this.
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