18 minutes 75 V 15 A 20 minutes 75 V 15 A Mass of empty inner cup m c 5484 g

18 minutes 75 v 15 a 20 minutes 75 v 15 a mass of

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18 minutes 7.5 V 1.5 A 20 minutes 7.5 V 1.5 A Mass of empty inner cup m c = 54.84 g Mass of inner cup containing water m c +m w = 296.5 g
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Effective mass of the water from step in the procedure m’ w = 244.16 g Voltage- 7.5 V Current- 1.5 A Power P=IV= 11.25 W (J/s) Time t= 1200 seconds Total electrical energy provided to the system: E=Pt= 13,500 J Initial temperature of water: T i = 20 Final temperature of water: T f = 32 Total heat gained by the system Q = m w c w ( T f T i )+ m c c c ( T f T i ) : 3071 cal Electrical equivalent of heat Je=E/Q: 4.39 J/cal Now that we found the electrical equivalent of heat, we had to compare it to the accepted value of 4.184 J/cal. Percent Error: 4.184 4.39 4.184 100 =4.9% In this experiment, we incurred a very low error, which means that for the most part, we did the experiment correctly and we were accurate with our data. Discussion 1. Although the mechanical equivalent of heat lab was not performed, we can still say that the results in that experiment are relatively close with the electrical equivalent of heat that we got in this lab. This is so, because no matter what type of energy it is, if the energy is
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being converted into another form of energy, then due to the conservation of energy law, the electrical and mechanical equivalent of heat would be the same. 2. After comparing our value of electrical equivalent of heat to the accepted value of 4.184, we came to realization that our result was larger than the accepted value. One source of error could have been from the reading of the voltmeter and ammeter. We assumed that
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  • Spring '16
  • Heat

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