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Thermodynaics, solving for minimum power to heat water

  1. Sep 19, 2010 #1
    An electric hot water heater takes in cold water at 15.8°C and delivers hot water. The hot water has a constant temperature of 45.6°C, when the "hot" faucet is left open all the time and the volume flow rate is 5.0 multiplied by 10-6 m3/s. What is the minimum power rating of the hot water heater?

    I really need help. I don't know how to connect Power and Thermal Energy. I know the individual formulas but can't connect the two. This is for WebAssign and is due within the hour. Please provide quick solution. Thank you very much.
  2. jcsd
  3. Sep 19, 2010 #2


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    Ok well let's start simple, what equation connects a temperature difference and heat energy?
  4. Sep 19, 2010 #3
    Power is just the rate at which energy is converted, remember that.
  5. Sep 19, 2010 #4
  6. Sep 19, 2010 #5


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    And we know that mass = density*volume = ρV and then divide by time, what do we get?
  7. Sep 19, 2010 #6
    wait i'm getting lost. sorry. so do i do Q=[cpv(deltaT)]/t
  8. Sep 19, 2010 #7
    so for my answer it would be Power= 4186(1.94)(5E-6)(45.6-15.8) where c=4186 and the density of water is 1.94
  9. Sep 19, 2010 #8
    is this correct?
  10. Sep 19, 2010 #9
    oh no wait. i got the density wrong. i have no idea what the density is since the temperature is different
  11. Sep 19, 2010 #10
    The density of water is 1.94 for feet^3, while your question is in m^3
  12. Sep 19, 2010 #11
    0.591312 m^3?
  13. Sep 19, 2010 #12
    The average density of water should be 1g/ml, or 1g/cm^3,

    whilst 1.94 is for lb/ft^3 if I recall right.
  14. Sep 19, 2010 #13
    yeah. 1.94 is ft^3

    so in the equation i would actually plug in .001kg/m^3 since were using kg not g
  15. Sep 19, 2010 #14
    Your unit conversion is wrong, if you were to use m^3 you should get something else.
  16. Sep 19, 2010 #15
    Power= 4186(1000)(5E-6)(45.6-15.8)
    so the conversion is 1000kg/m^3
    is the above formula correct?
  17. Sep 19, 2010 #16
    It should be right, yes.
  18. Sep 19, 2010 #17
    thank you so much. i actually learned something for once.
  19. Sep 19, 2010 #18
    My pleasure, copitlory8
  20. Sep 19, 2010 #19
    would it be pushing it if i asked you another physics question?
  21. Sep 19, 2010 #20
    That'd be fine, as long as I can answer it.
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