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Help with Heat

  • Thread starter frozen7
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  • #1
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A leaf of area 40 cm2 and mass 4.5 x 10-4 kg directly faces the sun on a clear day. The leaf has an emissivity of 0.85 and the specific heat of 0.8 kcal/kg.K.
a) Estimate the rate of rise of the leaf’s temperature. (ans: 2.3 oC/s)
b) Calculate the temperature of leaf would reach if it lost all its heat by radiation (the surroundings are at 20 oC). (ans: 84 oC)


How to do this question??

Any hints?
 

Answers and Replies

  • #2
mukundpa
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What is the rate at which it receives heat? Intensity of sun's radiation at that place is not given.
 
  • #3
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That`s all the information i have. Does it mean more information is needed to solve this question?
 
  • #4
Tom Mattson
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Yes, but it's the sort of information that you can look up. You might want to to a Google search on "power output sun" or something like that. If you know the total solar power output [itex]P[/itex] then you can find the solar intensity [itex]I[/itex] at the Earth by:

[tex]I=\frac{P}{4\pi r^2}[/tex],

where [itex]r[/itex] is the Sun-Earth distance.
 
  • #5
Integral
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The energy from the sun is constant everywhere as long as you are considering an area oriented normal to the suns rays. That is what I would interpert the pharse "faces the sun" to mean. IIRC the solar contstant is ~2cal/m^2
 
  • #6
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How if i have another extra value of 1000W/m^2?

I optain the value of another temperature after using this value in radiation formula. However,how to find the rate of rise of temperature?
 
  • #7
mukundpa
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If I is the energy incident per unit time per unit area(intensity) then rate of heat absorbed

dU/dt = I A = mc dT/dt

m is the mass and c is specific heat of the material. with proper units you will get dT/dt.

Absorbity is not mentioned hear but I think it is equal to emissivity of the surface.

For the second part you can use stefan's law of radiation, the radiant heat must be equal to heat absorbed.
 
  • #8
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mukundpa said:
dU/dt = I A = mc dT/dt
Why dU/dT = IA? But not supposed to be IeA The given e of leaf is 0.85. How come substitute e = 1?
 
  • #9
mukundpa
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mukundpa said:
Absorbity is not mentioned here but I think it is equal to emissivity of the surface.
I think you have missed it.
 
  • #10
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So,

IeA = solar energy
1000(0.85)(20cm^2) = [tex]\sigma[/tex] (T^4) (Surface area of Sun)
T = xxx

Is it?
If so, how to find the rate of rise of temperature then?
 
  • #11
mukundpa
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not energy, power. incident energy per unit time.


IeA = m c dT/dt Where m is the mass, c is specific heat and T is temperature.
 
  • #12
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erm.....why IeA = m c dT/dt ??
 
  • #13
mukundpa
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I (Intensity of any wave or radiation) is the amount of energy received per unit area per tine unit time, normal to the surface. Multiplied by area will give the amount of energy received per unit time. Multiplied by e will give energy absorbed per unit time.

Specific heat (c) is the amount of heat required to raise the temperature of unit mass of the substance by 1 degree. To make per unit time we have to divide it by time and hence dT/dt will give the rate of increase in the temperature.
O.K.?
 
  • #14
164
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OK..Thanks for it.
 

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