How come a light bulb does not explode?

[yasar1967]

I have two questions actually:
1. According to Stefen's Law formula power created by an conventional tungsten bulb can be calculated and temperature of the filament is about 3500 K.
I know it has low pressure gas inside a bulb but doesn't that mean sooner you switched on the bulb the pressure inside will reach roughly about 11 times higher! (PV=nRT... I can assume dilute gas is used and ideal gas assumption is correct)
2. Filament will carry the temperature of the gas to 3500K eventually when the system reaches equilibrium. From then that hot gas will heat the room by conduction(through bulb's thin wall) and then through convection. Then glass should melt and even if it doesn't , one bulb should produce enough heat to warm the whole room! But that doesn't happen why? where does that 3500K temperature goes to? I'm not looking at the radiation side of the problem you see. What I'm interested in here is a filament which is at 3500K and that piece of metal heats the gas inside to 3500K?? or does it?

 

[MATLABdude]

For the first question, consider this: is the lightbulb under more stress when there's a high vacuum (ie low pressure) or lower vacuum (ie higher pressure) inside the light bulb? Assume that the vacuum inside the lightbulb isn't so great (I couldn't find any numbers easily, so we'll use qualitative from my experience) and the pressure is 1% of the outside. Clearly, even when the gas heats up, it'll still be a vacuum.

As for the second question, equilibrium. The lightbulb might be putting out, say, 90 W of thermal energy. The lightbulb's temperature rises when you first turn on the bulb, but it plateaus (and fairly quickly). As the temperature increases, the rate of heat loss also increases. The heat is carried away into the air in the room, and then carried away into the rest of your house, and then carried away to the outside. At some point, the rate of heat production matches the rate of heat loss (from the surface of the bulb) and the bulb reaches a state of equilibrium.

Now, if you were to wrap the bulb in a very good insulator (say, a thick asbestos or fibreglass blanket) the rate of heat loss is reduced and the temperature increases. Enough to reach the melting point of the glass? Maybe, but I won't be trying that experiment any time soon (nor should you, since the glass might buckle and implode, it'll get even hotter than it normally does, and you might end up with an electrical short for your troubles!)

 

[Andy Resnick]

The key is that incandescent bulbs are well-evacuated. High pressure arc lamps, OTOH, can be pressurized to 200 atms, and those can indeed explode upon failure.

 

[Phrak]

I have two questions actually:
1. According to Stefen's Law formula power created by an conventional tungsten bulb can be calculated and temperature of the filament is about 3500 K.

I thought it was much higher--like 5000K. Do you have a reference for this so I can check it out?

 

[D H]

Tungsten melts at about 3700 K. The filament in a bulb is heated to less than this. The 3500 K cited in the original post is a bit high. This site, http://hypertextbook.com/facts/1999/AlexanderEng.shtml, claims 2800 - 3300 K; wikipedia claims a range of 2000 K to 3300 K for the filament temperature.

 

[fluidistic]

See here https://www.physicsforums.com/showthread.php?t=302552. I asked exactly the same question almost exactly 1 year ago.

 

[russ_watters]

For the second question, a mediocre window has an insulation R value of about 0.2 K-m^2/W, which means a single typical 2 sq meter window with a temperature difference of 5K between inside and outside will dissipate 50 watts.

So in winter, a medium-sized room with two outside walls and a handful of windows will dissipate a continuous 1000-2000 watts.