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Power used to heat water and efficiency math problem

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data

    A household uses 100 gallons of hot water per day (1 gallon of water has a mass of 3.78 kg) and the water temperature must be raised from 10°C to 50°C

    a) find the average power needed to heat the water.

    b)Then assume you have solar collectors with 55% average efficiency at converting sunlight to thermal energy when tilted at your location's latitude. Using the table find the collector area needed to supply this power in Albuquerque in June.

    From looking at the text, the tilted altitude is 296.

    2. Relevant equations


    3. The attempt at a solution

    For part A I think I need to find the mass of 100 gallons of water? This would be 3780 kg. (make sure I am right on this)

    I would then use the formula I stated above. So

    Q= (3780) (4.184) (40) -is this correct for water?

    Then the answer would be 632620.8

    I am not sure if this is right.

    For part B I have no idea what formula I should be using
  2. jcsd
  3. Oct 15, 2013 #2


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    Homework Helper

    Your mass is 3780 kg per day, you need this in kg per second to get the units for power. So you have to divide your answer by 1 day converted to seconds.

    For part b, they are telling you that 55% of the total energy is what you calculated above.

    Also what table are they referring to?
  4. Oct 15, 2013 #3
    So to get it for a day I would need to know how many seconds are in a day? This would be 86400.
    That would then mean 3780 divided by 86400?

    The table is average monthly insolation (W/m^2) and temperature for selected US cities

    In part B how would I find the area?
  5. Oct 15, 2013 #4


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    Yes that is correct.

    The units of the insulation is W/m^2 and your energy is in W, so if you divide them, what units would you get?
  6. Oct 15, 2013 #5
    Would you end up with just m^2?

    I am confused with what I am trying to find in this problem still though
  7. Oct 15, 2013 #6


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    You would need to actually put the table, I assumed they were giving your the heat transferred per unit area.
  8. Oct 15, 2013 #7
    The specific heat of water of about 4.2 is in J/(g*K). Per gram, not per kilogram.
    So you are off by a factor of 1000, to start with.
    Last edited: Oct 15, 2013
  9. Oct 15, 2013 #8

    What the table gives is

    Albuquerque, NM for June:

    Horizontal- 338 W/m^2
    Titled at latitude- 296 W/m^2
    Average Temp (°C)- 23.4

    I am not sure what to do with this information to solve part B
  10. Oct 15, 2013 #9
    1. First find the correct value of the power.
    2. Then calculate how much power is necessary to collect from the sun if the conversion efficiency is 55%. In other words, how much power is necessary so that 55% of it is the answer in part (1).
    3. And third, you know that you 296 W for each m^2. How many m^2 you need to get the power in part (2).
  11. Oct 15, 2013 #10
    How would I go about calculating the power?
  12. Oct 16, 2013 #11
    Just the way you did. Find the energy and divide by time (one day in seconds, to get watts).
    I meant to use consistent units so you get the correct value.
    Either mass in grams and specific heat in J/(g*K) or mass in kg and specific heat in J/(kg*K).
    In the first case the value is 4.180, in the second 4180.
  13. Oct 16, 2013 #12
    I am confused on how to find the energy.

    I got 7322 watts from part A.

    I am not sure if this is right or relevant to part B? What numbers should I be using?
  14. Oct 16, 2013 #13
    Now I think you are 10 times higher. Did you use the mass as 3780 kg, as you wrote in the first post?
    100 gallons is about 378 kg. Not thousands.
    So the average power required is about 732 W.
    Now, the panel only converts 55% of the solar power.
    How much solar power is required so that 55% of it is 732 W?
    This is the first step for part B.
  15. Oct 16, 2013 #14
    I did (.04375)(4184)(40) and got 7322

    but if the answer is 732 would I do (.55) (732)?

    Am I looking for power input or output?
  16. Oct 16, 2013 #15
    What is .04375?:confused:

    Edit. Never mind. It's mass of water per second, right?
    It's just that you converted the gallons wrong. So that's why you get this higher values. I told you this already.

    No, you need to find of what number 732 is 55%.
    That number is obviously larger than 732, not smaller.
    Do you understand what 55% of something means?
  17. Oct 16, 2013 #16
    I got this for the mass?
  18. Oct 16, 2013 #17
    Sorry, I realized what you mean while you were posting yourself. Please see my edit above. (post 15)
  19. Oct 16, 2013 #18
    How would I convert the gallons correctly?

    So then would I be doing 732/.55 to get 1330?

    Sorry I am really confused
  20. Oct 16, 2013 #19
    1 gallon is about 3.78 liters. And 1 liter of water is about 1 kg.
    So 1 gallon of water has a mass of about 3.78 kg.
    Then 100 gallons have 378 kg. And not 3780 kg, as you used in your calculations.

    Yes, the solar power collected by the panels should be about 1330 W.
    Now there is just one more step. One square meter of panel collects 296 W.
    How many m^2 you need to collect 1330 W.
  21. Oct 16, 2013 #20

    So I am trying to find out how many m^2 are in 1330 W?

    Would I do 1330/296 to get 4.5 m^2?
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