Question

Part B MISSED THIS? Read Section 7.2 (Pages 264 - 266) A particular frost-free refrigerator uses about 780 kWh of electrical energy per year. Express this amount of energy in kilojoules (kJ). 281x106 kJ Submit Previous Answers Correct The energy in kilojoules is equal to the energy in joules divia Significant Figures Feedback: Your answer 2808.10.2 differently or used a different number of significant figures the Part C Express this amount of energy in Calories (Cal). 90 AXPO ? Submit Request Answer

Answer 1

I kwh is the Consumed when amount of electrical energy a 1000 watt 13 used for hour. an Now I watt 13 sec and hour - 3600&ca So, lkwh = 1000 watter = 1000 x 3600 Joule 3.6 x 10 Joules. 780 kwh = 780 X3.6 x 106 Joule - 2808 x 100 Joule - 2.81 x 10 Joule - 2.81 x 106 kJ Now, - - Ical = 4.184 Joules. 1 Joule = t. Cal 4.184 780 kwh = 2.81 x 10 = 2.81 x 10 2 Joule x 1 Cal. = 0.6716 X 10 - 6.71 x 108 Cal Cal