Thermal Radiation: Calc Total Emitted Joules in Certain Temp Range

[PytrTchaikovsky]

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

 

[Qwertywerty]

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

 

[PytrTchaikovsky]

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

 

[Qwertywerty]

Where M is thermal emittance?

M is mass of object ( Q = mcΔT ) .

 

[gleem]

From your initial equation you can write dP =4σA T³dt giving the power in the interval dt

 

[PeterDonis]

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?

 

[Qwertywerty]

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 .

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[Qwertywerty]

Post edited .

 

[Qwertywerty]

Wouldn't it still remain m*c*ΔT if temperature of surroundings remain the same ?

 

[Qwertywerty]

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 .

 

[nasu]

.
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.

 

[PytrTchaikovsky]

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

 

[PeterDonis]

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.

 

[Qwertywerty]

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 .

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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).

 

[Qwertywerty]

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 ?

 

[Qwertywerty]

"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 .

 

[PeterDonis]

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.

 

[Qwertywerty]

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 .

 

[PytrTchaikovsky]

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.

 

[PeterDonis]

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).

 

[PeterDonis]

where did you get that formula from?

That formula is a general relationship between the amount of heat emitted by an object and the change in its temperature. However, if you don't know either of those things in advance, you can't use that formula to solve anything. You need to do the integral I gave in post #15.

 

[Qwertywerty]

If you mean, for finding ΔT\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 ΔT\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).

I tried this - assuming time limits are known - on integrating dT and dt using the the two known relations P = ... and Q = ... , you get t as a complex function of T , terms of which include ln() and arctan .

Although you have said there is no solution , can this relation not be solved ?

*Please excuse me if this question sounds repetitive .

 

[PeterDonis]

on integrating dT and dt using the the two known relations P = ... and Q = ... , you get t as a complex function of T

You're going to have to show your work, this is too vague for me to understand what you're doing.

 

[Qwertywerty]

You're going to have to show your work, this is too vague for me to understand what you're doing.

dQ/dt = m*c*dT/dt ( dividing both sides by dt ) .
and
dQ/dt = σ*A*(T⁴ - T₀⁴) (where T₀ is temperature of surroundings ) .

∴m*c*dT/dt = σ*A*(T⁴ - T₀⁴) - then integrate it .

 

[PeterDonis]

dQ/dt = m*c*dT/dt ( dividing both sides by dt )

Differentiation is not an algebraic operation; you can't just "divide by dt", you have to have functions of $t$ on both sides that you are differentiating. In the original equation, you don't; you have $\Delta T$ on the RHS, not $T$. $\Delta T$ is not temperature as a function of time; it's the change in temperature between two states of the object that you happen to be interested in. That's not a function of time, and you can't differentiate it with respect to time.

 

[Qwertywerty]

Differentiation is not an algebraic operation; you can't just "divide by dt", you have to have functions of t on both sides that you are differentiating.

I have not done this - You can look at this in two ways - Q = m*c*ΔT and then I differentiate with respect to t ,
or , for small change in temperature , dQ = m*c*dT , and now I just simply divide both sides by dt .

In the original equation, you don't; you have ΔT\Delta T on the RHS, not TT. ΔT\Delta T is not temperature as a function of time; it's the change in temperature between two states of the object that you happen to be interested in. That's not a function of time, and you can't differentiate it with respect to time.

Yes , I made a mistake here . You could use this only for small temperature difference in T and T₀ . Sorry .

 

[Qwertywerty]

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.

A molecule of an object , loses heat on fall in temperature . So , greater number of molecules , greater heat lost .

Also , greater the loss in temperature , greater the loss of heat .

∴ heat lost depends on physical dimensions , density ,and temperature .

Physical dimensions and density together are replaced by m and ,
ΔQ ∝ m ;
ΔQ ∝ ΔT .

∴ΔQ = c*m*ΔT with c the constant of proportionality and equal to specific heat capacity .

 

[PeterDonis]

Q = m*c*ΔT and then I differentiate with respect to t

But, as you go on to note, you can't do this unless $\Delta T$ is infinitesimal, i.e., unless you really have $dQ$ and $dT$. You can't differentiate $\Delta T$ by $t$ because it isn't a function of $t$ ($T$ is, but $\Delta T$ is not).

for small change in temperature , dQ = m*c*dT , and now I just simply divide both sides by dt

This is the equivalent of differentiating by $t$, because $dQ$ and $dT$ are both infinitesimals (strictly speaking, you would take the limit as $dt \rightarrow 0$).

Also, this only works if both $m$ and $c$ are constant. Constant $m$ is no problem, but constant $c$ is; most substances have a heat capacity that changes with temperature. So the full formula would have to be

$$
\frac{dQ}{dt} = m c \frac{dT}{dt} + m T \frac{dc}{dT} \frac{dT}{dt} = m \frac{dT}{dt} \left( c + T \frac{dc}{dT} \right)
$$

Finally, as I said before, this formula doesn't help unless you already know $T$ as a function of $t$ (and $c$ as a function of $T$, if it's not constant). In the scenario under discussion in this thread, we don't.