Need to calculate how much a surface will heat up near a hot light

[anc2113]

This is killing me because I used to know how to solve this problem, and I can't remember how to make it work. Its been too long since I took a physics course.

I will be placing some plastic sheeting near a 1000W lamp. I want to calculate how hot the sheeting is going to get, to make sure its not going to melt or catch fire. I know that this involves the boltzman law, j = σT^4, but I can't figure out how to make it work.

Parameters are the following:
--at 20 cm distance, 400W/m^2
--at 30cm distance, 250 W/m^2
--the sheeting passes 90% of light, which is basically the same as having an albedo of 90%, right?

its not as simple as plugging the 400 or the 250 W/m^2 into the left hand side of the j = σT^4 and solving for T -- i get answers that don't make sense when I do this.

help much appreciated!

EDIT: This is not a homework problem, but seems a little bit pedestrian for this forum. If it belongs in a homework subforum, I'd appreciate if a mod could move it.

 

[willem2]

The surface area of a 20 cm sphere is 0.50 m^2 (4 pi r^2) so you get 2000W/m^2 at this distance, so I don't know where you got your numbers.

thermal radiation of the surroundings can also be important, at room temperature you'll have another 400 W/m^2 coming in.

 

[sophiecentaur]

The temperature reached by the plastic will depend on how well it can lose heat as well as proximity to a heat source. It will reach an equilibrium temperature when energy received from the lamp per second (power in) equals the energy lost by radiation and convection. If you put the whole assembly inside a well insulated box, you could probably melt the plastic even.

I should say that this problem is much more practical than theoretical and I wouldn't go away and leave it unattended, whatever I had calculated, until I'd let it run for a good few hours whilst I was watching it. 1KW is quite a heat source and, when you consider that domestic lamp shades tend to be rated for 60W, particularly when they are an 'enclosed' design, I think my concern is justified.
If there were plenty of air circulation then it would probably not be a hazard but you should experiment carefully. You could get some suitable (£££, possibly) sheeting for the job.

 

[AlephZero]

The ansswer depends very much on how effectively the material can lose heat through air convection etc.

Just as a warning, we have used halogen car headlight bulbs (which have an internal reflector to produce a directional beam of radiation) as a simple and easily controlled way to heat up small objects in a test rig at work. With only 200W of lamps, it's easy to heat small objects to a temperature of 1000C in a few minutes.

 

[sardar]

I understand how frustrating it can be to forget how to solve a problem that you used to know. However, I am here to help you find a solution to this problem.

Firstly, I would like to clarify that the Boltzmann law, j = σT^4, is used to calculate the total amount of radiation emitted by a surface at a given temperature. It is not directly used to calculate the temperature of a surface when exposed to a heat source.

To calculate the temperature of the plastic sheeting near the hot light, we need to consider the energy balance between the light source and the sheeting. This can be done using the Stefan-Boltzmann law, which states that the total amount of heat radiated by a surface is proportional to the fourth power of its absolute temperature.

In this case, we can use the following equation:

Q = εσAT^4

Where Q is the heat radiated by the light source (in watts), ε is the emissivity of the plastic sheeting, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the sheeting (in m^2), and T is the temperature of the sheeting (in Kelvin).

To solve for T, we need to know the heat radiated by the light source, which can be calculated by multiplying the power of the light source (1000W) by the distance from the sheeting (in meters). So, at 20 cm distance, the heat radiated would be 400W (1000W x 0.4m) and at 30 cm distance, it would be 250W (1000W x 0.3m).

Now, we need to take into account the fact that the sheeting passes 90% of the light, which means that 10% of the light is being absorbed. This can be accounted for by multiplying the heat radiated by the light source by 0.1. So, the heat absorbed by the sheeting at 20 cm distance would be 40W (400W x 0.1), and at 30 cm distance, it would be 25W (250W x 0.1).

Finally, we can solve for the temperature using the equation:

T = (Q/εσA)^1/4

So, at 20 cm