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Solar Energy problem

  1. Dec 3, 2007 #1
    1. The problem statement, all variables and given/known data
    In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is 14.0% (that is, 86% of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 250 L of water in the tank from 21°C to 46°C in 1.6 h when the intensity of incident sunlight is 650 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.

    2. Relevant equations
    equations that i used or tried to use were the energy equation where
    and Q=cmT
    and W=pdv

    3. The attempt at a solution
    i dont know what the pressure of the object is....unless my method is wrong
    i also tried the equation with 0.14E because its 14% of the energy
  2. jcsd
  3. Dec 5, 2007 #2

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    What has this got to do with pressure of the object?

    Energy falling per unit area per unit time is given. You can calculate the energy reqd to heat up the water, if you know ms(t2-t1). The time is given. So, you should be able to find the area.
  4. Dec 5, 2007 #3
    well i redid the question where i used the equation
    where m is the density times the volume
    then with Q i found Power which is Q/t
    then with power i plugged it into the intensity equation of
    I=P/A (p is power not pressure and A is area) and solved for A, but for somereason im not getting the right answer....is it a problem with conversion or is the process im using wrong....
  5. Dec 5, 2007 #4

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    Show me the steps. Maybe we can work it out.
  6. Dec 5, 2007 #5
    ooo nvm i got it....i forgot about the 14% of energy being used....
    the method i used was right its just at the end when u solve for intesity
    I=P/A its actual 0.14I=P/A
    thanks for the help
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