Emission spectra of different materials

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  • #1
JohnnyGui
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I have some more detailed questions on this further on but I found it better to start with a very basic question first:

If higher temperature is correlated with shorter emitting wavelengths, how come there are incandescent light bulbs that emit yellowish light but are hotter than incandescent light bulbs that emit white light? (no color filters on the outside)
 

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  • #2
Drakkith
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If higher temperature is correlated with shorter emitting wavelengths, how come there are incandescent light bulbs that emit yellowish light but are hotter than incandescent light bulbs that emit white light? (no color filters on the outside)

Are you sure these light bulbs actually exist? A purely incandescent light bulb should have a color that depends almost solely on the temperature of the filament.
 
  • #3
sophiecentaur
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I have some more detailed questions on this further on but I found it better to start with a very basic question first:

If higher temperature is correlated with shorter emitting wavelengths, how come there are incandescent light bulbs that emit yellowish light but are hotter than incandescent light bulbs that emit white light? (no color filters on the outside)
Are you referring to the Heat Output here? Power and temperature are not necessarily related. A 5W 'pea bulb' can have a surface temperature of 4000°C and a 3kW electric kettle element will have a temperature just a little in excess of 100°C. In each case, they reach an equilibrium temperature at which the supplied (electrical) energy input becomes equal to the dissipated (thermal) energy.
Different light bulbs can differ in a similar but not as extreme way. Why? A bulb which needs to last a long time will be made with a fatter filament so that it runs cooler (redder) but the resistance is designed to produce the same (say 100W) power as a very bright (bluer) bulb.
 
  • #4
JohnnyGui
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A 5W 'pea bulb' can have a surface temperature of 4000°C and a 3kW electric kettle element will have a temperature just a little in excess of 100°C. In each case, they reach an equilibrium temperature at which the supplied (electrical) energy input becomes equal to the dissipated (thermal) energy.

But how can these elements of different temperatures (4000C and 100C) emit the same wavelength then while Wien's law says otherwise? Or did you mean that they don't emit the same wavelength?

A bulb which needs to last a long time will be made with a fatter filament so that it runs cooler (redder) but the resistance is designed to produce the same (say 100W) power as a very bright (bluer) bulb.

So you mean that if a reddish glowing light bulb and a bright (bluer) blue light bulb both use 100W, it's because the bright light bulb has a way more thinner filament and thus producing a higher temperature?
 
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  • #5
sophiecentaur
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But how can these elements of different temperatures (4000C and 100C) emit the same wavelength then while Wien's law says otherwise? Or did you mean that they don't emit the same wavelength?

So you mean that if a reddish glowing light bulb and a bright (bluer) blue light bulb both use 100W, it's because the bright light bulb has a way more thinner filament and thus producing a higher temperature?
The fact that they're elements doesn't mean they only emit a spectral line. Condensed matter emits a continuous spectrum. Wein's law only says which frequency is the spectral maximum. It's Black Body Radiation.
It's the surface area that governs the equilibrium temperature. You can get the same resistance for a whole range of surface areas.
 
  • #6
Drakkith
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But how can these elements of different temperatures (4000C and 100C) emit the same wavelength then while Wien's law says otherwise? Or did you mean that they don't emit the same wavelength?

They don't. The light bulb at 4,000C will be whitish while the 100C heating element will be emitting essentially no visible light at all. Also, remember that hot objects emit light in a broad spectrum. That's why the Sun is white. It emits light across the entire visual spectrum (along with lots of infrared light as well).
 
  • #7
sophiecentaur
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They don't. The light bulb at 4,000C will be whitish while the 100C heating element will be emitting essentially no visible light at all. Also, remember that hot objects emit light in a broad spectrum. That's why the Sun is white. It emits light across the entire visual spectrum (along with lots of infrared light as well).
They both will emit the same wavelength - just at different levels.
 
  • #8
Drakkith
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They both will emit the same wavelength - just at different levels.

Of course. That's what I was getting at with the last part of my previous post.
 
  • #9
JohnnyGui
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Yes, I was indeed talking about the peak wavelengths when I mentioned Wein's law. So I got confused when I misunderstood @sophiecentaur thinking he meant that elements at different temperatures can have the same peak wavelengths.

It's the surface area that governs the equilibrium temperature. You can get the same resistance for a whole range of surface areas.

So the surface area is inversely proportional to the resistance and thus temperature? So in theory, to calculate the temperature of a filament from the given power for the light bulb, one would have to calculate the resistance by using the surface area of the filament, to get the percentage of energy that turns into thermal energy, and then use the heating capacity of the specific material of that filament (in Joule/Kelvin) to eventually get the temperature of that filament. Correct?
 
  • #10
sophiecentaur
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I know you know. I was just spelling it out louder. :smile:
Yes, I was indeed talking about the peak wavelengths when I mentioned Wein's law.
So the surface area is inversely proportional to the resistance and thus temperature?
And earlier:
But how can these elements of different temperatures (4000C and 100C) emit the same wavelength then while Wien's law says otherwise? Or did you mean that they don't emit the same wavelength?
Why would you say that?
Google resistance of a wire and you will find that you can get a given resistance for a whole range of wire sizes. The resistance doesn't need to be related in any way to the surface area. This is very basic school physics. Read it up and do some thinking before coming back with random thoughts. It's all far too muddled at the moment.
then use the heating capacity of the specific material of that filament (in Joule/Kelvin) to eventually get the temperature of that filament. Correct?
Not correct. The equilibrium temperature has nothing to do with the heat capacity of the wire or the time it takes to reach that temperature. The equilibrium temperature depends on the surface area and the power supplied.
 
  • #11
JohnnyGui
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Google resistance of a wire and you will find that you can get a given resistance for a whole range of wire sizes. The resistance doesn't need to be related in any way to the surface area. This is very basic school physics. Read it up and do some thinking before coming back with random thoughts. It's all far too muddled at the moment.

"Very" basic school physics says that resistance in a cylindrically shaped material, e.g. a wire, is equal to:
$$R = \rho \cdot \frac{L}{A_{cross}}$$
The length ##L## and cross-sectional area ##A_{cross}## of a wire can be written in terms of the surface area ##A_{srf}##:
$$L = \frac{A_{srf}}{2πr}$$
$$ A_{cross} = \frac{A_{srf}^2}{4\pi \cdot L^2}$$
Plugging this in the equation for resistance and simplifying gives:
$$R = \rho \cdot \frac{2L^2}{A_{srf}\cdot r}$$
Resistance is therefore inversely proportional to the surface area of a wire.
 
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  • #12
Drakkith
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Note that a filament is coiled and then coiled once again (supercoiled). Some are even coiled a third time. Along with other things, this has the effect of reducing the effective surface area of the filament (the surfaces facing the inside of the coil will absorb almost as much light as they emit) while keeping its length and resistance the same.
 
  • #13
sophiecentaur
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"Very" basic school physics says that resistance in a cylindrically shaped material, e.g. a wire, is equal to:
$$R = \rho \cdot \frac{L}{A_{cross}}$$
The length ##L## and cross-sectional area ##A_{cross}## of a wire can be written in terms of the surface area ##A_{srf}##:
$$L = \frac{A_{srf}}{2πr}$$
$$ A_{cross} = \frac{A_{srf}^2}{4\pi \cdot L^2}$$
Plugging this in the equation for resistance and simplifying gives:
$$R = \rho \cdot \frac{2L^2}{A_{srf}\cdot r}$$
Resistance is therefore inversely proportional to the surface area of a wire.
I can't fault the Maths of that but the 'basic' Physics makes no sense in the context of this thread. You have ignored the presence of L in your conclusion. You are implying that R is fixed by A and it certainly is not. You still seem to be defending your original ideas, rather than using what you have read here to get a better understanding. The Wein's Law bit is not relevant, bringing in the thermal capacity is irrelevant and you are missing the point that Resistance depends on more than a single dimension. Rearrange those thoughts in your mind and you should get somewhere with this. If you want proof that you are going in the wrong direction - just buy a 40W halogen lanp and a 40W conventional tungsten filament lamp. They run at different colours and, if you examine the filaments, you will find they are different shapes for the same (nominal) resistance - different diameters and different lengths.
 
  • #14
JohnnyGui
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You have ignored the presence of L in your conclusion. You are implying that R is fixed by A and it certainly is not. You still seem to be defending your original ideas, rather than using what you have read here to get a better understanding.

What? I never implied that R is fixed by A. I was merely stating in my previous post that A is related to R, that's it. I'm very aware that there are other properties that define R, such that a 40W halogen and a 40W tungsten filament would need different diameters and lengths to have the same (nominal) resistance.
 
  • #15
sophiecentaur
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What? I never implied that R is fixed by A. I was merely stating in my previous post that A is related to R, that's it. I'm very aware that there are other properties that define R, such that a 40W halogen and a 40W tungsten filament would need different diameters and lengths to have the same (nominal) resistance.
I do not understand what you are asking, in that case. If you know the theory, then where is there any surprise or paradox?
 
  • #16
JohnnyGui
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I do not understand what you are asking, in that case. If you know the theory, then where is there any surprise or paradox?

I had a feeling there was a misunderstanding. I was asking about the factors that determine heat generation and heat loss (##A## being one of them) to understand what determines the end temperature of an electrical powered material, so I can conclude the peak wavelength of its emission using Wien's law, in case they can be estimated as black bodies.

You have answered that thermal equilibrium is determined by surface area and the given power. From what I understand, a certain given power can generate different amounts of heat depending on the material. I stated that heat capacity (the amount of energy needed to elevate a material 1K in temperature) is one of the factors that influence that, and this link supports that. So I'm not sure what you meant by saying that thermal equilibrium has nothing to do with heat capacity.
 
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  • #17
Drakkith
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You have answered that thermal equilibrium is determined by surface area and the given power. From what I understand, a certain given power can generate different amounts of heat depending on the material. I stated that heat capacity (the amount of energy needed to elevate a material 1K in temperature) is one of the factors that influence that, and this link supports that. So I'm not sure what you meant by saying that thermal equilibrium has nothing to do with heat capacity.

That link talks about the heat capacity because it gets into the oscillation of the temperature of the filament due to being powered by an AC power source. We can probably assume the temperature oscillates around some mean value and just use that as the temperature.
 
  • #18
JohnnyGui
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That link talks about the heat capacity because it gets into the oscillation of the temperature of the filament due to being powered by an AC power source. We can probably assume the temperature oscillates around some mean value and just use that as the temperature.

I'm stumped. If you add energy to any material, and you'd want to calculate the mean end temperature, ignoring heat loss for the moment, how would you calculate that?

Likewise if you give power to any material. I understand you can use the Stefan-Boltzmann law to calculate the temperature but who said that the given power to that material is automatically the same power that the material emits in radiation?

So if one asks; an object with surface A and a certain weight is given energy of 100W, calculate its radiative power. I'd first have to calculate the temperature of the object after giving 100W using its specific heat capacity, put that to the 4th power and multiply it by the surface A (and emissivity) to get its radiative power (neglecting heat loss for the moment)..
 
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  • #19
sophiecentaur
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In an equilibrium condition, what else can happen to added energy than being lost to the surroundings as heat?
If you are using AC power, there is a continually changing power input so the temperature will constantly be fluctuating - chasing equilibrium. In that case, the rate of change of temperature will depend on thermal capacity. Small filaments flicker more. But that is all another level of complication. Sort out the DC situation first.
 
  • #20
Drakkith
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We can assume the heat capacity of a filament is very small. Small enough so that it heats up to equilibrium very quickly and we don't need to worry about it for this question.

Under that assumption, the power dissipation by the filament is done mostly through radiation. So 100 W in, 100 W out as radiation. Another assumption, but getting into the different modes of heat loss for a light bulb is unnecessary.
 
  • #21
JohnnyGui
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We can assume the heat capacity of a filament is very small. Small enough so that it heats up to equilibrium very quickly and we don't need to worry about it for this question.

Under that assumption, the power dissipation by the filament is done mostly through radiation. So 100 W in, 100 W out as radiation. Another assumption, but getting into the different modes of heat loss for a light bulb is unnecessary.

Ah, so it does depend on the heat capacity, but because it's so small, it just can be neglected in case of a filament.

For the sake of exercise, I've tried to prove that for a small heat capacity, the input power would be more or less the same as the radiation power, using the Stefan-Boltzmann law.
$$(\frac{P_{in}}{C \cdot kg})^4 \cdot A \cdot \epsilon \cdot σ = P_{out}$$
Where ##kg## is the weight of the filament (which is obviously very small) and ##C## the heat capacity per kg, such that the fraction within brackets would give the temperature. I can't see how a small ##C## would make ##P_{in} ≈ P_{out}##.

I have a hunch I should put a heat loss factor in there as a function of ##A## to adjust the temperature, but I'm not really sure how.
 
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  • #22
nasu
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The smallness of heat capacity has nothing to do with the equilibrium temperature. It has some influence on how long it takes to reach the equilibrium.
You don't need to prove the equality of powers, this is the condition for equilibrium or constant temperature, not something resulting from an equation.
If the power are not equal you either increase the internal energy of the filament and so the temperature increases or you decrease the internal energy and the temperature decreases. Equilibrium means neither increase nor decrease of temperature so the two powers must be equal.

When you start heating, the power in is larger than the power out and so the energy accumulates in the filament and the temperature increases. Bu the power out depends on temperature difference between filament and medium (as well as area of filament) so as the temperature increases the power out increases and it goes like this until the power out catches with the power in and the equilibrium is reached.
If the area of the filament is larger, the power out equals power in at a lower temperature. (for the same power in).

For a filament heated electrically and loosing energy by radiation you have the ##P_{in} ## as the nominal power of the bulb, as given by the manufacturer. This takes into account the variation of resistance with temperature and provides the actual power used when the filament is hot. There is no point to express in in terms of resistance as the resistivity depends on temperature (unless you want to use the bulb at non-nominal conditions).
For the power dissipated by radiation you could use ## P_{out}=\sigma T^4 A##
The equilibrium temperature will be reached when the two powers are equal.
You can see that the equilibrium temperature, for a given input power, depends on the area of the filament.
 
  • #23
sophiecentaur
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I'm stumped. If you add energy to any material, and you'd want to calculate the mean end temperature, ignoring heat loss for the moment, how would you calculate that?

Likewise if you give power to any material. I understand you can use the Stefan-Boltzmann law to calculate the temperature but who said that the given power to that material is automatically the same power that the material emits in radiation?

So if one asks; an object with surface A and a certain weight is given energy of 100W, calculate its radiative power. I'd first have to calculate the temperature of the object after giving 100W using its specific heat capacity, put that to the 4th power and multiply it by the surface A (and emissivity) to get its radiative power (neglecting heat loss for the moment)..
This post needs tidying up if you are challenging this standard book work.
Para 1. Of course you can also lose heat by contact with a moving fluid (convection) or by straight conduction. The radiated power is still determined by the SB relationship. However, there will be absorption of EM from surrounding matter. The simple case of an isolated radiator only applies in some situations.
Para 2: Watts are Power, not Energy and please use Mass when you mean Mass.
Whatever can you mean by "neglecting heat loss" when the thread is all about heat transfer. If there is no heat loss there is no limit to the temperature that's achieved. The temperature rise rate will be constant and proportional to input power. But this is irrelevant.
 
  • #24
JohnnyGui
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I think I've finally got it now. For some reason, I wasn't considering the radiation power through the SB formula as a factor that takes out energy and thus can lower the temperature of a material. Thus, the final temperature can't be just determined by the ##P_{in}## but by the equilibrium depending on how high the radiation power is (##P_{out}## if this is the only way of energy loss).

Since both ##P_{in}## and radiation power are dependent on T, heat capacity merely has an influence on the time until a temperature equilibrium has reached. The surface ##A## of a material is independent from temperature and thus a larger surface can influence temperature in such a way that a lower equilibrium temperature will be reached as well as reaching that equilibrium faster..

With a bit of help from @nasu I was able to deduce a formula that determines the change in temperature per unit time. Furthermore, I was trying deduce a formula that plots the temperature of a material that merely loses heat energy through radiation power (i.e. a filament). Let's assume that ##P_{in}## doesn't change with temperature (I know it's incorrect but to keep it simple).I came up with:
$$T(t) + \frac{(P_{in} - T(t)^4 \cdot A \cdot σ)dt}{C \cdot m} = T(t + dt)$$
My problem is, that eventually ##T(t)## will be a function of the outcome of this formula. I don't really know how express this to plot it. Is that even possible, if a variable in a formula is dependent on the outcome?

Whatever can you mean by "neglecting heat loss" when the thread is all about heat transfer. If there is no heat loss there is no limit to the temperature that's achieved. The temperature rise rate will be constant and proportional to input power. But this is irrelevant.

You're completely right. I totally missed this.
 
  • #25
Drakkith
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My problem is, that eventually T(t) will be a function of the outcome of this formula. I don't really know how express this to plot it. Is that even possible, if a variable in a formula is dependent on the outcome?

I think it would be a differential equation.
 
  • #26
JohnnyGui
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I think it would be a differential equation.

Sorry for the late reply. Looking at my deduced formula in my post #24, specifically at the ##dt## in it, I have a hunch I should integrate that formula to give the total rise/decline in temperature in a specific chosen time duration, and add that to the initial temperature ##T(t)##. Such that:
$$T(t) + \int_t^{Δt} \frac{(P_{in} - T(t)^4 \cdot A \cdot σ)t}{C \cdot m} dt = T(t + Δt)$$
But I'm still stuck here with the fact that the ##P_{out}## which is the radiation power, is a function of T as well. Not sure how to continue further with this...
 
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  • #27
JohnnyGui
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Guys, I need help. Now that I understand that temperature gets in balance because of losing energy in the form of radiation power, I now have a difficulty understanding how a material then increases in temperature in the first place.

Let's say a material with a surface ##A## of 1 receives energy ##P_{in}## and loses energy solely in the form of radiation ##P_{out}##. As @nasu said, ##P_{in}## will first be larger than ##P_{out}## at the start. This means that the material will increase in temperature. However, even if the temperature increase is very small (##dT##) by a factor of ##x##, this means that the ##P_{out}## will increase by a factor of ##x^4## since radiation power is proportional to the 4th power of temperature. So, if ##P_{in}## is proportional with T increase, this translates into ##P_{out}## being proportional to ##T## increase to the 4th power, leading to the ##P_{out}## being initially larger than ##P_{in}##. How then can there be temperature increase in the first place?
 
  • #28
Drakkith
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So, if ##P_{in}## is proportional with T increase, this translates into ##P_{out}## being proportional to ##T## increase to the 4th power, leading to the ##P_{out}## being initially larger than ##P_{in}##.

How is ##P_{in}## related to ##T##?
 
  • #29
JohnnyGui
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How is ##P_{in}## related to ##T##?

Thanks for your reply. I'd say:
$$ΔT \cdot C \cdot m + T(t)^4 \cdot A \cdot σ = P_{in}$$
This indicates that ##P_{in}## must be higher than the radiation power ##P_{out}## for a temperature increase of ΔT. But I just don't understand how ##P_{in}## gets larger than the radiation power in the first place. What is it that make the ##P_{out}## lag behind at the start and give ##P_{in}## the chance to rise?
 
  • #30
Drakkith
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I have no idea if your equation is correct, but I know that when you first turn on a light bulb the temperature is very low and ##P_{out}## is very small. The power you put into the circuit is used to heat the filament until the radiation output equals the input power.
 
  • #31
Charles Link
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Sorry for the late reply. Looking at my deduced formula in my post #24, specifically at the ##dt## in it, I have a hunch I should integrate that formula to give the total rise/decline in temperature in a specific chosen time duration, and add that to the initial temperature ##T(t)##. Such that:
$$T(t) + \int_t^{Δt} \frac{(P_{in} - T(t)^4 \cdot A \cdot σ)t}{C \cdot m} dt = T(t + Δt)$$
But I'm still stuck here with the fact that the ##P_{out}## which is the radiation power, is a function of T as well. Not sure how to continue further with this...
Your equation is basically a differential equation ## \frac{dT}{dt}=\frac{P_{in}-\epsilon (\sigma (T^4-T_{ambient}^4)) ( A)}{Cm} ##. I think you included an extra "t" in the integral that shouldn't be there, but it does require a ## T_{ambient} ## term.
 
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  • #32
JohnnyGui
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I have no idea if your equation is correct, but I know that when you first turn on a light bulb the temperature is very low and ##P_{out}## is very small. The power you put into the circuit is used to heat the filament until the radiation output equals the input power.

I deduced that formula from my post #21 that Nasu verified for me in a PM. I indeed understand that the ##P_{in}## should be larger than the ##P_{out}## at the start but I don't understand why. My instinct says that, as soon as there's power ##P_{in}## and temperature increases proportionally to ##P_{in}##, then ##P_{out}## should rise accordingly to ##T^4 \cdot A## and thus win over ##P_{in}## already from the start.
But what makes ##P_{out}## not immediately react to that temperature increase and thus resulting in temperature increase?
 
  • #34
Drakkith
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I deduced that formula from my post #21 that Nasu verified for me in a PM. I indeed understand that the ##P_{in}## should be larger than the ##P_{out}## at the start but I don't understand why. My instinct says that, as soon as there's power ##P_{in}## and temperature increases proportionally to ##P_{in}##, then ##P_{out}## should rise accordingly to ##T^4 \cdot A## and thus win over ##P_{in}## already from the start.

I don't understand how you've come to that conclusion. If ##P_{in}## is 100 watts, and ##T## is 300 K, then what's your equation say happens at ##t = 0## and then at a very small ##Δt##? You can assume whatever values for the other variables you'd like.
 
  • #35
Charles Link
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I don't understand how you've come to that conclusion. If ##P_{in}## is 100 watts, and ##T## is 300 K, then what's your equation say happens at ##t = 0## and then at a very small ##Δt##? You can assume whatever values for the other variables you'd like.
I edited the Post #31 one more time. Please look and see if that is what you need. And initial conditions are that ## T=T_{ambient} ##.
 

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