1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.
  22. Sep 19, 2010 #21
    u are probably the smartest most helpful online free tutor. every website i go people give me crappy answers to try and seem smart but they never help.

    anyways. here it goes.
    From a hot reservoir at a temperature of T1, Carnot engine A takes an input heat of 5550 J, delivers 1500 J of work, and rejects heat to a cold reservoir that has a temperature of 495 K. This cold reservoir at 495 K also serves as the hot reservoir for engine B, which uses the rejected heat of the first engine as input heat. Engine B also delivers 1500 J of work, while rejecting heat to an even colder reservoir that has a temperature of T2. Find the temperatures

    (a) T1 and
    (b) T2.
  23. Sep 19, 2010 #22
    I'm sorry I can't help you for this.

    Im also studying Physics at this moment and the Carnot engine is not in my field of studies. So I suppose your first line is wrong then :)
  24. Sep 19, 2010 #23
    okay that's fine. you were still helpful. and if you want to redeem urself here is a slightly easier one:
    The drawing shows two thermally insulated tanks. drawing is located at:
    They are connected by a valve that is initially closed. Each tank contains neon gas at the pressure, temperature, and volume indicated in the drawing. When the valve is opened, the contents of the two tanks mix, and the pressure becomes constant throughout.

    (a) What is the final temperature? Ignore any change in temperature of the tanks themselves.(Hint: The heat gained by the gas in one tank is equal to that lost by the other.)
    (b) What is the final pressure?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook