Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition

14-1: Heat as Energy Transfer
14-3 and 14-4: Specific Heat; Calorimetry
14-5: Latent Heat
14-6 to 14-8: Conduction, Convection, Radiation

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 4

An average active person consumes about 2500 Cal a day.

  1. What is this in joules?
  2. What is this in kilowatt-hours?
  3. If your power company charges about 10 ¢ per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day?
  1. 1.0×107 J1.0 \times 10^7 \textrm{ J}
  2. 2.9 kW hr2.9 \textrm{ kW hr}
  3. 29 cents. No, 29 cents would not buy a day’s food.29 \textrm{ cents. No, 29 cents would not buy a day's food.}
Giancoli 7th Edition, Chapter 14, Problem 4 solution video poster

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This is Giancoli Answers with Mr. Dychko. The energy the average person consumes in a day in joules is 2,500 calories times 4,186 joules per calorie. And this gives about 1.00 times 10 to the 7 joules. Converting that into kilowatt hours and multiplied by 1 watt second per joule, and so the joules cancel. And then what second is just another way of writing joules. And then multiplied by 1 kilowatt per 1,000 watts and then times by 1 hour for every 3,600 seconds, and we're left with kilowatt hours. And this is work out to about 2.9 kilowatt hours. And then if your cost per kilowatt hour is 10 cents, this works out to a total of 29 cents. And it would not be possible to buy a day's food with 29 cents in a, you know, Western country like the Canada or United States.

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