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Total power and specific heat

  1. May 6, 2007 #1
    If the total power per unit area from the Sun incident on a horizontal leaf is 9.00x10^2 W/m2, and we assume that 70.0% of this energy goes into heating the leaf, what would be the rate of temperature rise of the leaf? The specific heat of the leaf is 2.80 kJ/(kg°C), the leaf area is 7.00 x10^−3 m2, and its mass is 0.200 g.

    I don't know what equation to use.
    I tried using

    Power = kA (Temp/d)
    but did not get the right answer.

    The right answer is 7.88 degree celsius per sec.
  2. jcsd
  3. May 6, 2007 #2


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    Are you sure that's the right answer? It seems pretty big (or fast).
  4. May 6, 2007 #3

    yup, it says that is the correct answer
  5. May 6, 2007 #4


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    Well, I would use Q=mc(deltaT).

    The rate of change of thermal energy is the power.

    I don't quite get the same answer as the book though, I get 7.88x10^-4 oC/s. (which is much more reasonable, in my opinion)
  6. May 6, 2007 #5

    If I were to use the equation

    Q = (9.00 x 10^2 W/m2) * 0.7 (would this be right b/c it's 70% of energy?)
    m= 0.200 g
    c= 2.80 kJ/(kg C)
    (delta T) = what we are looking for

    Where woul the leaf area fit into the picture?
  7. May 7, 2007 #6
    can anyone help please?
  8. May 7, 2007 #7


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    Technically, since your working with power your equation should be;

    [tex]\frac{dQ}{dt} = mc\Delta T[/tex]

    Now, you've done everything correctly up to now. So, if the sun emits 9.00 x 10^2 watts per square meter; much power falls on a leaf with an area of 7.00 x10^−3 square meters?
  9. May 7, 2007 #8
    sorry i don't understand the hints.
    i did 7.00 x10^−3 / 9.00 x 10^2 watts for dQ/dt
    and plugged in m and c and solved for delta T.
    still not getting the right answer
  10. May 7, 2007 #9


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    Your very close here;
    But why are you dividing the area by the power per unit area when you want power? Lets have a look at the units;

    [tex]area = m^2[/tex]

    [tex]\text{power per unit area}=\frac{W}{m^2}[/tex]

    [tex]\text{power}= \frac{dQ}{dt}=W[/tex]
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