• #1
Dear forum

I am working with thermal radiation. This is the specific formula:
P = σ ⋅ A ⋅ T4
P = emitted effect (W, J/s)
σ = Stefan-Boltzmann constant (5,67 ⋅ 10-8)
A = area of object (m2)
T = temperature of object (K)


How can I get to know the total emitted joules, within a certain range of temperature? Any formula out there? I think this is a tricky one because the more joules emitted, the less joules is being emitted (because temperature is lowered), if you know what I mean.

Best regards
 

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  • #2
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Is there some reason you think it isn't simply M*c*ΔT ?
 
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  • #3
Is there some reason you think it isn't simply M*c*ΔT ?

Where M is thermal emittance? And c is specific heat?

If M is thermal emittance, which thermal emittance? Since it is changing?

Thank you
 
  • #5
gleem
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From your initial equation you can write dP =4σA T3dt giving the power in the interval dt
 
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  • #6
PeterDonis
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How can I get to know the total emitted joules, within a certain range of temperature?

Under what conditions? Radiating into vacuum, with no absorption?

Also, why would you want it within a certain range of temperature?
 
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  • #7
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Under what conditions? Radiating into vacuum, with no absorption?

Why would you need vacuum for no absorption ? A body could be placed with another body in vacuum too - the OP's first post only mentions power emitted .

Also, why would you want it within a certain range of temperature?

I didn't understand what was wrong with asking this .

Thanks for your reply .
 
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  • #8
PeterDonis
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Why would you need vacuum for no absorption ?

If there are other objects present, they will also be emitting radiation, so the object the OP is asking about will be absorbing radiation as well as emitting it. That wil affect its temperature.

I didn't understand what was wrong with asking this .

I was just trying to understand the scenario the OP had in mind. Normally when people ask about total energy emitted by an object, they ask about energy emitted over some period of time, not over some range of temperature.
 
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  • #9
PeterDonis
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From your initial equation you can write dP =4σA T3dt giving the power in the interval dt

This gives the rate of change of power with respect to time, i.e., the change in power in the interval dt. It does not give the power itself.
 
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  • #10
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Post edited .
 
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  • #11
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Wouldn't it still remain m*c*ΔT if temperature of surroundings remain the same ?
 
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  • #12
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If there are other objects present, they will also be emitting radiation, so the object the OP is asking about will be absorbing radiation as well as emitting it. That wil affect its temperature.

But there could be other objects in vacuum too .
 
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  • #13
nasu
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I didn't understand what was wrong with asking this .
It's not wrong to ask but the question is not well defined.
You ask about the power radiated in an interval of temperatures. But you did not specify your system. Temperature of what?
You have in mind a body with a distribution of temperatures? Or a body whose temperature changes in time? Or maybe something else?

I am puzzled by how many people jump with answers before trying to figure out what the question is. :)
I am not referring to this thread in particular.
 
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  • #14
It's not wrong to ask but the question is not well defined.
You ask about the power radiated in an interval of temperatures. But you did not specify your system. Temperature of what?
You have in mind a body with a distribution of temperatures? Or a body whose temperature changes in time? Or maybe something else?

I am puzzled by how many people jump with answers before trying to figure out what the question is. :)
I am not referring to this thread in particular.

Sorry for being unclear, thought I should make this as general as possible but I'll be more specific. I want to calculate the total amount of joules being emitted from an object with a non-constant temperature. If you have a stove for example, you can calculate the total amount of joules being emitted every second (Watt, symbol P here) using this earlier mentioned formula: P = σ ⋅ A ⋅ T4 if you know the temperature and area (T and A, σ is a constant). Then you just multiply with during how many seconds and you have the total emitted joules during a certain time. But what if the temperature of the object that are emitting heat are not constant? The more joules emitted, the lower temperature, right? How can I calculate the total amount of joules being emitted during a certain time now?

Thank you
 
  • #15
PeterDonis
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what if the temperature of the object that are emitting heat are not constant?

Then you need to know how the temperature changes with time, so you can do an integral:

$$
E = \int \sigma A \left[ T(t) \right]^4 dt
$$

If ##T## is constant, then this just reduces to multiplying the power by the length of time the object radiates.
 
  • #16
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I am puzzled by how many people jump with answers before trying to figure out what the question is. :)
I am not referring to this thread in particular.

I still cannot understand why it would simply not be m*c*ΔT .
 
  • #17
PeterDonis
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there could be other objects in vacuum too

"Vacuum" means "no objects present", so a single object radiating into a vacuum means it's radiating into empty space with no other objects present.
 
  • #18
PeterDonis
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I still cannot understand why it would simply not be m*c*ΔT .

See post #14; the OP wasn't actually asking the question you are answering here.
 
  • #19
PeterDonis
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The more joules emitted, the lower temperature, right?

If there is no other energy source present, yes. But as your example of the stove shows, an object can emit heat while remaining at a constant temperature, if there is an energy source present (the stove is burning fuel to produce heat).
 
  • #20
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See post #14; the OP wasn't actually asking the question you are answering here.

No , my question was this - in any case , why would heat emitted by the body not be m*c*ΔT ?
 
  • #21
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"Vacuum" means "no objects present", so a single object radiating into a vacuum means it's radiating into empty space with no other objects present.

My bad .
 
  • #22
PeterDonis
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in any case , why would heat emitted by the body not be m*c*ΔT ?

It would be, but you would need to know ##\Delta T## in order to calculate it using this formula. The OP, as clarified in post #14, is considering a situation where you don't know ##\Delta T## in advance.
 
  • #23
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It would be, but you would need to know ΔT\Delta T in order to calculate it using this formula. The OP, as clarified in post #14, is considering a situation where you don't know ΔT\Delta T in advance.

Oh , ok , sorry for the confusion .
But do you have solution for this ? You mentioned it in your post #15 , but only for T as a function of t .
 
  • #24
No , my question was this - in any case , why would heat emitted by the body not be m*c*ΔT ?

It could be, where did you get that formula from? However, I think that the area of the object will matter, so I think that should be included.
 
  • #25
PeterDonis
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do you have solution for this ?

If you mean, for finding ##\Delta T## without having to solve the integral I gave in post #15, no, I don't have one. There isn't one. If you don't already know ##\Delta T## in advance, the only way to solve for it is to know ##T## as a function of ##t## and then solve the integral I gave. You also have to know the length of time you are interested in (since that's what gives you the limits of integration).
 

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