I Emission spectra of different materials

[Charles Link]

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.

 

[JohnnyGui]

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?

 

[Charles Link]

Please read my edited post #31. I think that is what you need.

 

[Drakkith]

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.

 

[Charles Link]

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 $T=T_{ambient}$.

 

[Drakkith]

I edited the Post #31 one more time. Please look and see if that is what you need. And initial conditions are that $T=T_{ambient}$.

Okay, I'm showing that if we let $P_{in} = 100 W$ then $\frac{dT}{dt} = 0$ when T = 3250.36, assuming $ε=0.2, σ=5.67*10^{-8}, A=8*10^{-5}, C=0.1339776, m=0.042g$

You can actually ignore $T_{ambient}$ and the result is only off by 0.06 Kelvin since $T^4$ becomes much larger than $300^4$.

A few sources and other random bits of info on this topic I found on google:
http://hypertextbook.com/facts/2004/DeannaStewart.shtml
http://faculty.trinityvalleyschool.org/hoseltom/labs/Lab-28-(Light Bulbs).pdf
https://www.physicsforums.com/threads/surface-area-of-a-bulb-filament.721597/

 

[Charles Link]

Okay, I'm showing that if we let $P_{in} = 100 W$ then $\frac{dT}{dt} = 0$ when T = 3250.36, assuming $ε=0.2, σ=5.67*10^{-8}, A=8*10^{-5}, C=0.1339776, m=0.042g$

You can actually ignore $T_{ambient}$ and the result is only off by 0.06 Kelvin since $T^4$ becomes much larger than $300^4$.

A few sources and other random bits of info on this topic I found on google:
http://hypertextbook.com/facts/2004/DeannaStewart.shtml
http://faculty.trinityvalleyschool.org/hoseltom/labs/Lab-28-(Light Bulbs).pdf
https://www.physicsforums.com/threads/surface-area-of-a-bulb-filament.721597/

I think you might even find that the emissivity of a tungsten filament increases as the temperature increases. (I don't have data to support that, but I do know the resistance goes up considerably with temperature.) Raising the emissivity would lower the operating temperature. I think T=2800 K-3000 K might be more accurate, but T=3200 K is a reasonably good number for the first iteration.

 

[Drakkith]

I think you might even find that the emissivity of a tungsten filament increases as the temperature increases. (I don't have data to support that, but I do know the resistance goes up considerably with temperature.)

From what I found, the emissivity of a tungsten filament behaves quite erratically in both its spectral and temperature dependences. I looked it up and found this source: http://pyrometry.com/farassociates_tungstenfilaments.pdf
In some spectral regions the emissivity decreases with temperature, while in others it increases.

I ignored the emissivity change associated with the increase in temperature since it appeared fairly small and then used an emissivity of 0.2 since most of the radiation is emitted in the IR region of the spectrum. The emissivity in this region is about 0.2.

 

[Charles Link]

From what I found, the emissivity of a tungsten filament behaves quite erratically in both its spectral and temperature dependences. I looked it up and found this source: http://pyrometry.com/farassociates_tungstenfilaments.pdf
In some spectral regions the emissivity decreases with temperature, while in others it increases.

I ignored the emissivity change associated with the increase in temperature since it appeared fairly small and then used an emissivity of 0.2 since most of the radiation is emitted in the IR region of the spectrum. The emissivity in this region is about 0.2.

The spectral peak will be at $\lambda_{max}=1.0$ micron (approximately) since by Wien's law $\lambda_{max} T=2898$ micron degree K. Very nearly 1/4 of the energy lies to the left of this peak and 3/4 lies to the right (longer wavelengths). From looking at the curves, this would suggest $\epsilon=.3$ might be a better number than $\epsilon=.2$, but even $T=3250$ K is a reasonably good calculation.

 

[Drakkith]

The spectral peak will be at $\lambda_{max}=1.0$ micron (approximately) since by Wien's law $\lambda_{max} T=2898$ micron degree K. Very nearly 1/4 of the energy lies to the left of this peak and 3/4 lies to the right (longer wavelengths). From looking at the curves, this would suggest $\epsilon=.3$ might be a better number than $\epsilon=.2$, but even $T=3250$ K is a reasonably good calculation.

Indeed. Interestingly, the link in my previous post finds that the temperature of a tungsten filament in one of their test bulbs is approximately 3250 K. A pretty amusing coincidence!

 

[Charles Link]

Indeed. Interestingly, the link in my previous post finds that the temperature of a tungsten filament in one of their test bulbs is approximately 3250 K. A pretty amusing coincidence!

There also is no perfect answer because if you lower the voltage a 100 watt bulb could instead put out e.g. 80 watts and it would of course be running at a lower temperature. The manufacturer can make adjustments (e.g. in the filament length and diameter to adjust the resistance), and thereby could lower the operating temperature. Operating at a slightly lower temperature could e.g. increase bulb lifetime, etc.

 

[JohnnyGui]

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.

Ah, I forgot that it's about the difference w.r.t. the initial temperature $T_{ambient}$. I now understand that $P_{out}$ needs time to compensate $P_{in}$. Thanks!

I've got a few questions if you don't mind:

1. So is it correct to say that to calculate the end temperature after a certain $Δt$ time starting from the $T_{ambient}$ I'd have to use:
$$T_{ambient} + \int_t^{Δt} \frac{P_{in} - A \cdot σ(T-T_{ambient})^4}{C \cdot m} dt = T(Δt)$$

2. I noticed that plotting the differrential equation $\frac{dT}{dt}=\frac{P_{in}-\epsilon (\sigma (T^4-T_{ambient}^4)) ( A)}{Cm}$ gives a good representation on which parameters influences the time duration until a temperature equilibrium is reached ($dT/dt = 0$). Changing the heat capacity $C$ shows that the temperature equilibrium doesn't change but it does change the declining slope of the graph (next derivation of $dT/dt$ changes) as well as the starting position from which the graph declines.
I have understood so far that a lower heat capacity would make a material reach the temperature equilibrium faster, but I see that a lower heat capacity also makes the graph start declining from a higher temperature change. Shouldn't that higher starting position work against the shorter time to reach the equilibrium so that eventually the time duration until equilibrium stays the same?

3. When I increase the surface $A$ in the differential equation from question 2, it shows that the temperature equilibrium ($dT/dt=0$) is lower, which is what I expected. However, it also shows that the graph starts declining from a higher temperature increase. How can a larger surface cause an initial higher temperature increase per $dt$?

 

[Charles Link]

To answer (3), if you start with a $T$ above ambient, (with $P_{in}=0$), a larger area will cause a faster $|dT/dt |$ because it is radiating more. But notice dT/dt is negative. It is cooling faster. What you did is kept the mass the same and changed the shape to increase the surface area. A sphere has minimum surface area for a given mass or volume. A flatter ribbon will have more area for given mass, etc. $\\$ Editing... To answer (2), basically a lower heat capacity wil allow it to get back to ambient quicker from a higher starting point. You can see this with a sheet of aluminum foil that has a low heat capacity. If it just came out of the oven, you can grab it, and it holds so little heat, unlike a pan at the same temperature. $\\$ Editing some more... For (1) the upper limit on your integral should be $t+\Delta t$, or the lower limit should be zero.

 

[JohnnyGui]

To answer (3), if you start with a $T$ above ambient, (with $P_{in}=0$), a larger area will cause a faster $|dT/dt |$ because it is radiating more. But notice dT/dt is negative. It is cooling faster. What you did is kept the mass the same and changed the shape to increase the surface area. A sphere has minimum surface area for a given mass or volume. A flatter ribbon will have more area for given mass, etc. $\\$ Editing... To answer (2), basically a lower heat capacity wil allow it to get back to ambient quicker from a higher starting point. You can see this with a sheet of aluminum foil that has a low heat capacity. If it just came out of the oven, you can grab it, and it holds so little heat, unlike a pan at the same temperature. $\\$ Editing some more... For (1) the upper limit on your integral should be $t+\Delta t$, or the lower limit should be zero.

Thanks a lot for your answers!

Aplogies regarding (3), I just noticed that I was putting $T$ values lower than the initial $T_{ambient}$. Looking at $T$ values higher than the initial $T_{ambient}$ shows that it indeed doesn't change the maximum initial temperature increase.

To answer (2), basically a lower heat capacity wil allow it to get back to ambient quicker from a higher starting point. You can see this with a sheet of aluminum foil that has a low heat capacity. If it just came out of the oven, you can grab it, and it holds so little heat, unlike a pan at the same temperature

Do you mean that even though a lower heat capacity gives a higher starting point by a certain factor, the factor by which the slope declines more (cooling) is stronger?

Editing some more... For (1) the upper limit on your integral should be t+Δtt+Δt t+\Delta t , or the lower limit should be zero

Of course, I didn't pay attention.

 

[Charles Link]

You are actually missing one term in your thermal transfer problem here=you are assuming a material with very high heat conduction so that when heat radiates away, the substance is at the same temperature everywhere throughout the entire volume, i.e. it maintains thermal equilibrium throughout without any temperature gradient between the outer surface and the interior. For metals such as your filament, this assumption is reasonably good. $\\$ And yes, a lower heat capacity means it will cool much quicker because radiating away "x"= $\Delta Q$ amount of heat will cause its temperature to change more: $\Delta Q=c \Delta T$. A lower heat capacity $c$ will mean larger $|\Delta T |$.

 

[JohnnyGui]

You are actually missing one term in your thermal transfer problem here=you are assuming a material with very high heat conduction so that when heat radiates away, the substance is at the same temperature everywhere throughout the entire volume, i.e. it maintains thermal equilibrium throughout without any temperature gradient between the outer surface and the interior. For metals such as your filament, this assumption is reasonably good.

So if there is a temperature gradient between the outer surface and the interior, having a lower heat capacity doesn't necessarily shorten the time duration until thermal equilibrium is reached because a lower heat capacity also gives a higher starting point from which the material needs to cool down?

 

[Charles Link]

So if there is a temperature gradiënt between the outer surface and the interior, having a lower heat capacity doesn't necessarily shorten the time duration until thermal equilibrium is reached because a lower heat capacity also gives a higher starting point from which the material needs to cool down?

Your equation is sufficient. If you try to do the general case of baking pans of different materials coming out of the oven, the surface could cool quicker from both radiative cooling and cooling from the air than the interior for a material with low thermal conduction. Your light bulb filament equation also assumes only radiatiive cooling occurs, but that is ok too.

 

[JohnnyGui]

And yes, a lower heat capacity means it will cool much quicker because radiating away "x"= ΔQΔQ\Delta Q amount of heat will cause its temperature to change more: ΔQ=cΔTΔQ=cΔT \Delta Q=c \Delta T . A lower heat capacity cc c will mean larger |ΔT||ΔT| |\Delta T | .

Hmm, I just tested this out and it shows that if a heat capacity is lower by a factor x, the starting point from which a material needs to cool down is higher by a factor x and the factor by which the slope declines faster is also by a factor of x (I did this by looking at the ratio of the next derivation of $dT/dt$ with 2 different heat capacities). I might be reasoning this the wrong way but doesn't this mean that the net time duration until a thermal equilibrium is reached does not change when heat capacity differs since the same factors cancel each other out? Or did you mean that this is indeed the case for a filament?

 

[Charles Link]

The heat capacity does not affect the operating temperature of the filament. The starting point will be the same, but a lower heat capacity means it will cool down faster. It will also heat up faster when starting at $T=T_{ambient}$ and applying a $P_{in}$. $\\$ And your "factor x" calculation is correct. You can rewrite your differential equation (and edited with a small $c$ ) as $(c \, m)( \frac{dT}{dt})=P_{in}-\epsilon \, \sigma (T^4-T_{ambient}^4)A$ and for the same right side of the equation, lowering the heat capacity $C=c \, m$ by some factor will cause $\frac{dT}{dt }$ to increase by the same factor. $\\$ (Editing: I needed to make a correction here: The change in heat of the material $\Delta Q=C \, \Delta T=c \, m \, \Delta T$, where $C=c \, m$ is the total heat capacity, and $c$ is the heat capacity per gram. You can change the total heat capacity $C$ by simply using more material " $m$ " (but keep surface area $A$ constant). Alternatively, you can change the type of material to change $c$ which will also cause $C$ to change ). $\\$ Your differential equation is really a very good one for getting good estimates on the temperature behavior of the filament.

 

[JohnnyGui]

The starting point will be the same

I think I chose the wrong wording here. What I meant with starting point is the amount of $dT/dt$ increase at the moment when the material is at $T_{ambient}$ and starts receiving $P_{in}$. So a lower heat capacity will make the material initially rise faster and then more slowly to the thermal equilibrium (which is the operating temperatue as you call it).

And your "factor x" calculation is correct. You can rewrite your differential equation (and edited with a small cc c ) as (cm)(dTdt)=Pin−ϵσ(T4−T4ambient)A(cm)(dTdt)=Pin−ϵσ(T4−Tambient4)A (c \, m)( \frac{dT}{dt})=P_{in}-\epsilon \, \sigma (T^4-T_{ambient}^4)A and for the same right side of the equation, lowering the heat capacity C=cmC=cm C=c \, m by some factor will cause dTdtdTdt \frac{dT}{dt } to increase by the same factor.

Yes, indeed. So a larger $dT/dt$ by a factor of $x$ caused by a lower heat capacity declines faster down to $0$ (thermal equilibrium) by that same factor of $x$. Hence me concluding that the time duration until $dT/dt = 0$ is the same for lower heat capacities: there's an initial larger $dT/dt$ that needs more time to get to 0 but at the same time the decline (cooling down) is steeper by that same factor. Apologies if I missed your verification on this but is this correct for a filament?

Your differential equation is really a very good one for getting good estimates on the temperature behavior of the filament

I'm glad I got it (mostly) correct :)

 

[Charles Link]

Looks like you have a good handle on it. I'd enjoy seeing a graph of the complete solution to the differential equation as a function of time for a couple of cases.

 

[JohnnyGui]

Looks like you have a good handle on it. I'd enjoy seeing a graph of the complete solution to the differential equation as a function of time for a couple of cases.

Something that I'm indeed trying to figure out. I merely concluded this since the next derivative of $dT/dt$ as well as $dT/dt$ itself are both proportional to the heat capacity.
I'm probably doing this the hard way but I would first want to solve the differential:
$$\frac{P_{in}-\epsilon \cdot \sigma \cdot A (T-T_{ambient})^4}{c \cdot m} = 0$$
To get $T$. Which is in this case equal to:
$$T = (\frac{P_{in}}{\epsilon \cdot \sigma \cdot A})^{\frac{1}{4}} + T_{ambient}$$
After solving for T I'd then have to know the time until that $T$ is reached. The only formula so far that has a time parameter in it is the integral that I gave previously, so that:
$$\int_0^{Δt} \frac{P_{in} - \epsilon \cdot A \cdot σ(T-T_{ambient})^4}{c \cdot m} dt = (\frac{P_{in}}{\epsilon \cdot \sigma \cdot A})^{\frac{1}{4}}$$
However, I don't have any idea how to get $Δt$ out of this equation. Nor do I know what to fill in for $T$ in the intergral formula.

An alternative is maybe taking the next derivation of $dT/dt$ as a function of time, but that would only show the steepness of the decline without taking into account how high $dT/dt$ initially is at the start. It's very probable I'm doing this all the wrong way.

 

[Charles Link]

I am no expert in differential equations, but I think it is likely that the exact and final $T$ will not be reached in a finite time. In any case the function $T(t)$ could be found by numerical methods (again of which I am no expert), and I think it might be the case that $T(t)$ approaches its final value asymptotically but never quite gets there. $\\$ Editing... One thing you can do with this differential equation is write it in the form $\frac{dT}{T^4-B}=dt$ and try to integrate both sides. I will need to study it further=I don't know that it has a closed form solution... Upon taking the difference of squares in the denominator and using partial fractions, perhaps it does have a closed form solution. Perhaps someone else can also comment on it. @fresh_42 Might this one have a closed form? $\\$ Editing... $\int \frac{dx}{x^2+1}=tan^{-1}x$ and $\int \frac{dx}{x+a}=ln|x+a|$ so I think a closed form might exist. If I get some extra time I may try to write out the complete solution... $\\$ Editing some more... It's likely to yield a complex expression for $T(t)$ on one side of the equation that is equal to $At$ on the other. The asymptotic solution will then be found by setting $t=+\infty$ and the result will be that the term inside the natural log must be zero so that it also is equal to infinity. In this sense, I don't think this "closed form" solution will be of much help...

 

[JohnnyGui]

I am no expert in differential equations, but I think it is likely that the exact and final TT T will not be reached in a finite time. In any case the function T(t)T(t) T(t) could be found by numerical methods (again of which I am no expert), and I think it might be the case that T(t)T(t) T(t) approaches its final value asymptotically but never quite gets there.

I think this explanation has made me notice my error in reasoning that time duration until thermal equilibrium doesn't change with lower heat capacities; it actually does.
The fact that a lower heat capacity gives a higher initial $dT/dt$ as well as a stronger decline in $T$-increase over time (until thermal equilibrium has been reached) doesn't mean that a lower initial $dT/dt$ with a weak decline in $T$-increase takes the same amount of time to reach thermal equilibrium. Even if $dT/dt$ and the steepness are both changed by the same factor.

When plotting the differential equation with 2 differen heat capacities (starting from T=300), this gives:

As shown, a higher heat capacity $C$ gives a high initial $dT/dt$ along with a rather abrupt decline in $T$-increase, such that the temperature increase abruptly changes to a more or less horizontal line (the asymptote you mention) i.e. the thermal equilibrium.
A lower heat capacity $C$ has a lower initial $dT/dt$ and also a less abruptly declining $T$-increase (the decline is less steep). Plotting temperature over time according to these descriptions gives:

Which obviously shows that it takes more time for the lower heat capacity to reach the same thermal equilibrium. Come to think of it, it's actually quite obvious. I guess I was focussing too much on the differential curves and taking way too farfetched conclusions.

I appreciate the effort you're taking with trying to prove this mathematically. Kind of feeling guilty now that I've drawn you into my confusion and trying to clear this up the hard way.

 

[JohnnyGui]

@Charles Link

If you don't mind, I have some final questions on emissivity to wrap this all up and continue on with my life :P.

1.
If I understand correctly, partial emissivity (<1) means that the temperature and surface do not fully represent the radiation power according to the Stefan-Boltzmann formula.
What I’d really like to know is what the physical causes are that let materials not fully emit radiation energy according to that formula. Is it caused by the fact that some materials lose thermal energy in other forms than radiation? If so, how does that take place?

2.
I understand that a black body emits the maximum radiation energy according to the Stefan-Boltzmann law. In addition, a black body also absorbs all wavelengths. Is it possible for any other material to absorb all wavelengths at full capacity and yet still have a partial emissivity less than 1? (e.g. putting the absorbed energy also in other forms than radiation energy)

 

[Charles Link]

To answer question (2): In general the emissivity at a given wavelength for an opaque material is equal to the absorbance $A=1-R$ where $R$ is the reflectivity. This can get complicated by surfaces having reflectivities that depend on the angle of incidence, etc., so it is more applicable to diffuse reflectors that absorb a certain fraction of the incident light. In any case, if $A=1$ for all wavelengths and all angles, then the emissivity $\epsilon=1$. $\\$ To try to answer question (1): A good example of a substance that is at temperature $T$ and has low emissivity is a gas. The reason that it has low emissivity is basically that the photon modes don't have sufficient coupling to the particles of the gas. The same is true for transparent materials=they also have very low emissivity. For a semi-transparent material, in a somewhat simplified manner, you can write the conservation of energy as $A +R+T=1$ (where $T$ is the transmissivity) and $A$ will again be the emissivity.

 

[JohnnyGui]

To answer question (2): In general the emissivity at a given wavelength for an opaque material is equal to the absorbance A=1−RA=1−R A=1-R where RR R is the reflectivity. This can get complicated by surfaces having reflectivities that depend on the angle of incidence, etc., so it is more applicable to diffuse reflectors that absorb a certain fraction of the incident light. In any case, if A=1A=1 A=1 for all wavelengths and all angles, then the emissivity ϵ=1ϵ=1 \epsilon=1 .

Thanks. What surprises me is that the radiation power in the SB-formula is determined by the amount absorbance of wavelengths, since it has an emissivity factor in it. What if the rise in temperature of a material is caused by something else than the amount of absorbed wavelengths? Why would that then need an emissivity factor?

 

[Charles Link]

Thanks. What surprises me is that the temperature in the SB-formula is determinede by the amount absorbance of wavelengths, since it has an emissivity factor in it. What if the rise in temperature of a material is caused by something else than the amount of absorbed wavelengths? Why would that then need an emissivity factor?

The emissivity affects how much energy is radiated. With the filament, you supply electrical power to raise the temperature. If you supply zero electrical power, the filament will be in thermal equilibrium with the walls of the room=which, regardless of their emissivity, the room, because it is enclosed will look like a blackbody insofar as how much thermal energy (in the form of electromagnetic waves) is in the air (i.e. being radiated and/or reflected off of the walls). The tungsten filament will be in equilibrium even with emissivity of .2. At equilibrium with zero input power, it will radiate at the same rate that it absorbs energy and be at the same temperature as the surface temperature of the walls of the room. Even if the room is in a total vacuum (no air), it will still have the electromagnetic wave energy density characteristic of what occurs inside any enclosure at temperature $T$. $\\$ And if you drill a small hole of area $A$ in the wall, the energy coming out of that aperture will be exactly like that of a blackbody of area $A$ at the same temperature as the surface temperature inside the walls of the room (We're assuming the light bulb is turned off.) . Meanwhile, if a light beam from the outside goes into the aperture, it bounces around inside the room and gets absorbed and doesn't find its way back out=thereby the aperture can be considered to have absorbtivity=1. (I don't want to use the letter $A$ again, because I just used it for area.) That's why the aperture is said to have an emissivity of 1. (If the walls are so hot that they are red hot, the aperture would appear to be reddish in color and look exactly like the walls.) (We're assuming the walls inside the box are hot, but are insulated on the outside so that they are cool. The only thing that will look red from the outside is the aperture of area $A$. Anyway, that's a quick introduction to some of the details behind the science of "blackbodies".

 

[JohnnyGui]

The emissivity affects how much energy is radiated. With the filament, you supply electrical power to raise the temperature. If you supply zero electrical power, the filament will be in thermal equilibrium with the walls of the room=which, regardless of their emissivity, the room, because it is enclosed will look like a blackbody insofar as how much thermal energy (in the form of electromagnetic waves) is in the air (i.e. being radiated and/or reflected off of the walls). The tungsten filament will be in equilibrium even with emissivity of .2. At equilibrium with zero input power, it will radiate at the same rate that it absorbs energy and be at the same temperature as the surface temperature of the walls of the room. Even if the room is in a total vacuum (no air), it will still have the electromagnetic wave energy density characteristic of what occurs inside any enclosure at temperature $T$. $\\$ And if you drill a small hole of area $A$ in the wall, the energy coming out of that aperture will be exactly like that of a blackbody of area $A$ at the same temperature as the surface temperature inside the walls of the room (We're assuming the light bulb is turned off.) . Meanwhile, if a light beam from the outside goes into the aperture, it bounces around inside the room and gets absorbed and doesn't find its way back out=thereby the aperture can be considered to have absorbtivity=1. (I don't want to use the letter $A$ again, because I just used it for area.) That's why the aperture is said to have an emissivity of 1. (If the walls are so hot that they are red hot, the aperture would appear to be reddish in color and look exactly like the walls.) (We're assuming the walls inside the box are hot, but are insulated on the outside so that they are cool. The only thing that will look red from the outside is the aperture of area $A$. Anyway, that's a quick introduction to some of the details behind the science of "blackbodies".

Thanks for the detailed explanation.

So let's say I have a mirror, which has an emissivity of basically 0, and I heat it up to the same temperature as a black body. Does this mean that the mirror doesn't emit radiation at all even if it has the same temperature as the black body?

 

[Charles Link]

The answer is yes. Incidentally, at my workplace, using a thermal imager to test this concept , we once put a gold-coated plate in an electrical frying pan. (gold has reflectivity nearly 1 and thereby emissivity near zero throughout the infrared). The gold plate appeared very cold=(room temperature, caused by reflected energy from the walls of the room), while another object, (with emissivity close to 1), appeared to be nearly at the temperature of the frying pan. As cold as it looked according to the thermal imager, you couldn't touch the gold plate=it was hot !