What is the Best Wavelength for Heat Transfer in a Black Body Radiator?

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

I am not sure about the relevant equations to be used here. At first sight, I thought I had to apply Stefan's law but then, the question states the power consumed by bulb, not the radiated one. I am completely clueless about what to do here. The first line states the peak wavelength, I have seen peak wavelength being used in Wien's displacement law but I am not sure about this.

Any help is appreciated. Thanks!

 

[haruspex]

You know the power going into the bulb. How does that relate to the power coming out of the bulb?
Can you determine the temperature? What then does Stefan's Law tell you?

 

[Saitama]

You know the power going into the bulb.

Yes, its 60 watts (0.5*120)

How does that relate to the power coming out of the bulb?

And that's where I am stuck. Can you please give me some hints?

Can you determine the temperature?

Temperature of bulb? How?

 

[SteamKing]

Temperature of bulb? How?

Check out Wien's Displacement Law. It contains a neat relationship between max. wavelength and temperature.

http://en.wikipedia.org/wiki/Wien's_displacement_law

 

[Saitama]

Check out Wien's Displacement Law. It contains a neat relationship between max. wavelength and temperature.

http://en.wikipedia.org/wiki/Wien's_displacement_law

Ah yes, so the temperature of bulb is $2.898 \times 10^3 K$.

What next?

From Stefan's law,

$$P_e-P_a=\sigma A(T_e^4-T_a^4)$$

where $P_e$ is the rate at which heat energy is emitted, $P_a$ is the rate at which heat energy is absorbed. $T_e$ is the temperature of body and $T_a$ is temperature of air.

Now how do I find $P_e$ and $P_a$?

 

[SteamKing]

I think for your light bulb, Pa = 0 and you are given the information on the power consumption of the bulb in the OP. I say Pa = 0 because the energy input to the bulb is converted to light, and air is transparent to light (i.e., light radiates without being absorbed by the air).

 

[Saitama]

I think for your light bulb, Pa = 0 and you are given the information on the power consumption of the bulb in the OP. I say Pa = 0 because the energy input to the bulb is converted to light, and air is transparent to light (i.e., light radiates without being absorbed by the air).

Something like $60=\sigma A((2.8998 \times 10^3)^4-(300)^4)$, right?

 

[SteamKing]

That would be my guess. You know the Boltzmann constant σ (or can look it up.)

 

[Saitama]

That would be my guess. You know the Boltzmann constant σ (or can look it up.)

Thanks a lot SteamKing! That's the correct answer. :)

But I feel that the question didn't give enough information. Don't you think the problem should have specified the nature of bulb? I mean I had to consider it as a black body to reach the correct answer.

 

[SteamKing]

I'm not sure what you mean by 'the nature of the bulb.'

Hot objects radiate energy as a black body at a certain temperature. The only thing missing that I can see was the emissivity value of the filament. Since, for this problem, the emissivity was apparently assumed to be 1, then you are dealing with a perfect black body radiator.

 

[haruspex]

I'm not sure what you mean by 'the nature of the bulb.'

Hot objects radiate energy as a black body at a certain temperature. The only thing missing that I can see was the emissivity value of the filament. Since, for this problem, the emissivity was apparently assumed to be 1, then you are dealing with a perfect black body radiator.

I agree the emissivity should be taken as 1 here, in the absence of any other indication, but in general objects can radiate better at specific wavelengths. A filament of a different metal would have a different colour at the same temperature.