# I Emission spectra of different materials

1. May 1, 2017

### JohnnyGui

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)

2. May 2, 2017

### Staff: Mentor

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. May 2, 2017

### sophiecentaur

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. May 2, 2017

### JohnnyGui

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?

Last edited: May 2, 2017
5. May 2, 2017

### sophiecentaur

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. May 2, 2017

### Staff: Mentor

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. May 2, 2017

### sophiecentaur

They both will emit the same wavelength - just at different levels.

8. May 2, 2017

### Staff: Mentor

Of course. That's what I was getting at with the last part of my previous post.

9. May 2, 2017

### JohnnyGui

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.

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. May 2, 2017

### sophiecentaur

I know you know. I was just spelling it out louder.
And earlier:
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.
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. May 2, 2017

### JohnnyGui

"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.

Last edited: May 2, 2017
12. May 2, 2017

### Staff: Mentor

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. May 3, 2017

### sophiecentaur

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. May 3, 2017

### JohnnyGui

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. May 3, 2017

### sophiecentaur

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. May 3, 2017

### JohnnyGui

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.

Last edited: May 3, 2017
17. May 3, 2017

### Staff: Mentor

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. May 3, 2017

### JohnnyGui

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)..

Last edited: May 3, 2017
19. May 3, 2017

### sophiecentaur

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. May 3, 2017

### Staff: Mentor

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.