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: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook