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Finding time to increase temp with power

  1. Jun 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A hot water heater is operated by using solar power. If the solar collector has an area of 6 m^2, and the power per unit area delivered by sunlight is 1000 W/m^2, how long in hours will it take to increase the temperature of 1 m^3 of water from 20C to 60C? (1 cal=4.186 J)



    2. Relevant equations
    I'm not really sure...
    I know that:
    power=(sigma)AeT^4
    sigma = Stefan -Boltzman constant = 5.6696x10^(-8) W/m^2*K^4
    A = surface area
    e = emissivity
    T = surface temp.
    but we weren't given emissivity so i dont think I use this...

    maybe mc(Tf-Ti)/P = (change in time)

    ?? I'm really clueless on what equation to use, please help!!
     
  2. jcsd
  3. Jun 22, 2008 #2

    Hootenanny

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    There is no need to invoke the Stefan-Boltzmann law here.

    What is the definition of power?
     
  4. Jun 22, 2008 #3
    P = force /area ??

    (the answer is stated as 7.8)
     
  5. Jun 22, 2008 #4

    Hootenanny

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    That's pressure, not power.
     
  6. Jun 22, 2008 #5
    oh yeah hahahha wow, whoops.

    P= Work/change in t
    or P = Q/ (delta T)
     
  7. Jun 22, 2008 #6

    Hootenanny

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    Correct. So you know that power provided and you can work out the work/energy required using one of your aforementioned equations.
     
  8. Jun 22, 2008 #7
    So do I use mc(Tf-Ti) to find Q?
    (1000 kg)(4186)(60-40)?
    I used 1000 kg for the mass because D = m/v
    we know the density of water is 10^3 = m/(1m^3)
    and solved for m ...= 1000
    but this is not giving me 7.8
     
  9. Jun 22, 2008 #8

    Hootenanny

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    You're good so far. What the next step?
     
  10. Jun 22, 2008 #9
    Ok so I did (1000)(4186)(60-40) = 83720000/(1000) = 83720
    ^ I rearranged the P = Q/ (delta t) solved for delta t = 83720 in secs

    then they want it in hours so dived by 3600s (83720)/3600 = 23.25 hrs

    the answer is 7.8 according to the book so im not sure where I am going wrong, please help!
     
  11. Jun 22, 2008 #10
    I figure I must be missing something since I am not using the area of 6 m^2
     
  12. Jun 22, 2008 #11

    Hootenanny

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    Indeed you are:
    This means that each square meter on the surface of the earth receives 1000W of energy from the sun.
     
  13. Jun 22, 2008 #12
    oh yeah i see that now
    soooo..
    P = 1000 W/m^2 and there are 6m^2 so it should be multiplied by 6
    1000)(4186)(60-40) = 83720000/(6000) = 13953.3333
    and then divide by 3600...3.875
    still not 7.8 =/
     
  14. Jun 22, 2008 #13
    I realized my mistake!
    It should be 60-20

    (1000)(4186)(60-20)/(6000)
    and then divided by 3600
    =7.75

    thank you so much for your help!!!!!!!!
     
  15. Jun 23, 2008 #14

    Hootenanny

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    A pleasure :smile:
     
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