Need help on this Light bulb Filament Temperature with emissivity and δ given

In summary, the question asks for the temperature of a 100 W light bulb's tungsten filament based on the information that it radiates 2.85 W of light, has a surface area of 0.400 mm^2, and an emissivity of 0.952. The formula P=eδAT^4 is used, but the resistance and ohm are unknown. The melting point of tungsten is 3683 K and the value of δ is 5.6696 X 10^-8 W/m^2.K^4. Further assistance in finding the correct formula and temperature is requested.
  • #1
sickid
1
0
Hi, I am stuck on this question, any help would be appreciated. Thanks!

  1. The tungsten filament of a certain 100 W light bulb radiates 2.85 W of light. (The other 97.15 is carried away by convection and conduction) The filament has surface area of 0.400 mm^2 and an emissivity of 0.952. Find the filament's temperature in kelvin. (melting point of tungsten is 3683 K) Take δ 5.6696 X 10^-8 W/m^2.K^4

I tried to use the formula: P=eδAT^4 but I still can't get the right answer. I tried to do a search on internet, come across a lot of problem related to the above mentioned but the resistence and ohm is known. If you could point me to the right way, providing the right formula would be very much appreciated.
 
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  • #2
What temperature did you get?
 
  • #3


Sure, let me help you with this question. First, let's start by understanding the given information. We know that the light bulb is emitting 2.85 W of light, which means that the remaining 97.15 W is being carried away by convection and conduction. We also know the surface area of the filament (0.400 mm^2) and its emissivity (0.952).

To solve for the filament's temperature, we can use the Stefan-Boltzmann law, which states that the power radiated by an object is proportional to the fourth power of its absolute temperature (in kelvin). In other words, the formula you mentioned, P=eδAT^4, is correct.

To use this formula, we need to rearrange it to solve for the temperature (T). It should look like this: T = (P/eδA)^1/4. Now, we can plug in the values we know: T = (2.85/0.952 x 5.6696 x 10^-8 x 0.000000400)^1/4 = 2,949 K.

However, this is not the final answer yet. We need to convert this temperature from kelvin to Celsius by subtracting 273.15. This gives us a final temperature of 2675.85°C. Since the melting point of tungsten is 3683 K (3409.85°C), this temperature seems reasonable.

I hope this helps you understand the problem and how to solve it. It's important to remember to always double-check your units and make sure they are consistent. Keep practicing and you'll get the hang of it!
 

What is the purpose of determining the light bulb filament temperature?

The light bulb filament temperature is important because it affects the efficiency and lifespan of the light bulb. By understanding the temperature, we can improve the design and performance of light bulbs.

How is emissivity related to the light bulb filament temperature?

Emissivity is the measure of how well an object emits thermal radiation. In the case of a light bulb, the filament temperature is directly related to the emissivity as a higher temperature results in a higher emissivity.

What is the significance of δ in determining the light bulb filament temperature?

δ, or delta, is the change in temperature that occurs when the filament is heated. It is an important factor in calculating the temperature of the filament because it takes into account the energy lost due to heat transfer.

What factors can affect the accuracy of determining the light bulb filament temperature?

There are several factors that can affect the accuracy of determining the light bulb filament temperature, such as variations in the material composition of the filament, the presence of impurities, and the accuracy of the instruments used for measurement.

How can the knowledge of light bulb filament temperature be applied in other fields?

The understanding of light bulb filament temperature has applications in various fields, such as materials science, electrical engineering, and lighting design. It can also be used in the development of more efficient and sustainable lighting technologies.

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