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Rooftop solar hot water system

  1. Oct 19, 2014 #1
    1. The problem statement, all variables and given/known data
    You would like to put a solar hot water system on your roof but your not sure if it's feasible. A reference book on solar energy shows that the ground level solar intensity in your city is 800W/m^2 for at least five hours a day throughout most of the year. Assuming that the completely black collector plate loses energy only by radiation and that the air temperature is 20 degrees Celsius. What is the equilibrium temperature of a collector plate directly facing the sun? Note that while the plate has two sides only the side facing the sun will radiate as the other side is well insulated.

    2. Relevant equations

    Q/delta t = e x sigma x A x T ^4

    Sigma= stefan-Boltzmann constant = 5.67 x 10^-8

    3. The attempt at a solution
    My exams in a month and this is one of my study questions, the answer in the back of the book is 110 degrees Celsius.

    I tried using the radiation equation provided to solve for T.

    I subbed in 800/(5x60x60) = 0.04444 as Q/delta t.
    Seeing as the solar intensity was given in W/m^2, I figured that they had already divided Q by A which left (e x sigma x T^4) on the RHS
    I divided the 0.044444 by sigma and then assuming e =1 since the plate is black I was left with:
    783852.63 = T^4
    T = 29 K
    Which is so far off. I feel like I'm making the same stupid mistake over and over but I can't figure out what it is?
  2. jcsd
  3. Oct 19, 2014 #2


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    Wrong comment. Contents deleted. Sorry
    Last edited: Oct 19, 2014
  4. Oct 19, 2014 #3


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    Staff: Mentor

    Hi Meera.sheeda, Welcome to Physics Forums.

    I suspect that the details of the hours of sun yielding 800 W/m^2 is unnecessary information here. The question specifically asks "What is the equilibrium temperature of a collector plate directly facing the sun?" So the sun must be up and the plate directly facing it for your calculation.

    The plate is therefore absorbing heat energy at the specified rate. It is also losing heat at a rate that depends upon the ambient temperature. Take a look at the Hyperphysics exposition on the Stefan-Boltzmann law.
  5. Oct 19, 2014 #4

    Thank you so much for your answer but this wasn't right either :/
  6. Oct 19, 2014 #5


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    Yes, I have just deleted it. Sorry.
  7. Oct 19, 2014 #6

    Thank you!!!! I did have to use the 800W/m^2 but your link showed me how to include the surrounding temp in my calculations and I ended up getting the same answer as in the back of the book! thank you heaps for all your help :)
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