Question
An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of running it for one year if electricity costs 12.0 cents/KWh12.0\textrm{ cents/KW}\cdot\textrm{h}? See Figure 20.41.
<b>Figure 20.41</b> An electric on-demand water heater.
Figure 20.41 An electric on-demand water heater.
Question by OpenStax is licensed under CC BY 4.0
Final Answer

$438 / year

Solution video

OpenStax College Physics, Chapter 20, Problem 50 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. An electric water heater with a power consumption of 5.00 kilowatts operating for 2.00 hours per day at an electricity cost of 12.0 cents per kilowatt-hour would use 438 dollars to operate it per year in electricity so we will show how to figure this out. So we converted this rate for the power into dollars per kilowatt-hour by multiplying by 1 dollar for every 100 cents and we have cost per year then is the power of 5.00 kilowatts multiplied by 2.00 hours everyday so this gives us kilowatt-hours per day and then we need to get rid of this day unit and have it per year so we multiply by 365.25 days per year and now we have kilowatt-hours per year but we want dollars per year so we multiply by 0.120 dollars for every kilowatt-hour and then the kilowatts and the hours cancel and we are left with dollars per year and that is 438 dollars of electricity used per year.

Comments

why 365 and a quarter days? why not just 365...

Hi Hanna, thank you for the question. 365 days would be totally fine, but I chose to account for leap years which have 366 days and occur once every four years giving an average of 3×365+3664=365.25\dfrac{3 \times 365 + 366}{4} = 365.25 days per year.
All the best,
Shaun