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Drink_Punk_Soda
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Light to heat energy formula?
#3745059  02/07/05 01:33 PM (11 years, 9 months ago) 


Hey all.
I've got a fairly simple question for a pretty complex application.
Let's say you're using a magnifying glass to focus light to a small point, in order to cause paper to combust. (Think burning ants on the sidewalk, or taking a solar hit on a bong etc.) Is there a formula I could use to figure out the approximate relationship between the light source and the heat result? For example, how many units of light energy would it take to heat the focused point of light to a certain temperature?
I'd really appreciate the help if anyone knows. Feel free to respond here or PM me. Thanks!

Kumbayah my lord, Kumbayah...

TinMan
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Using the wavelength of the light, E = hv, where v = c/lambda. Lambda in this case is the wavelength of the light, and c is the speed of light, 2.9979e8 m/sec. You can then use the value of v to compute the total energy in joules by multiplying the v value by 6.6261e34 j/sec in joules per second. This gives you the total energy of the wavelength in joules, which can be converted to BTUs or calories for a measurement of heat. If you want an actual temperature, you might have a problem.
So, E = v * (c/lambda)
Trendal could probably correct me on this if he were here.
Edited by TinMan (02/07/05 02:21 PM)

Seuss
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Assuming that you are using sunlight, and you know where you are (location wise), there are tables that tell you how many watts of power per square meter the sunlight delivers. Take that value, along with the area of the lens and you know how many watts of power you are getting. Next, take how long you are using the lens along with the power to get energy.
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Drink_Punk_Soda
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Re: Light to heat energy formula? [Re: TinMan]
#3745759  02/07/05 04:13 PM (11 years, 9 months ago) 


Alright, I think I understand what you're saying. Although I'm a bit confused by the final formula wouldn't it be E = h * (c/lambda) ? Unless v is equal to h, in which case the formula would be E = v squared.
Anyway, assuming the formula is 6.6261e34 * (speed of light/ wavelength), doesn't the intensity of the light factor in anywhere?
If not, then by the calculation (as I've interpreted it, at least), an average wavelength light (~500nm) would generate 3.9728e28 j/sec, or 1.6622e27 (thermochemical) calories/second. That seems incredibly low.. then again, I may have set my sights a bit high. On the other hand, that's about the same wavelength as the average for sunlight, so.. ?
Thanks again for the help.
Bogus

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Drink_Punk_Soda
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Re: Light to heat energy formula? [Re: Seuss]
#3745805  02/07/05 04:29 PM (11 years, 9 months ago) 


Sorry, Seuss, I didn't refresh the page before responding.
Let's simplify things a bit then, using your method. Assume I'm using a 60W bulb, with a 5" diameter lens. The area of a 5" diameter lens is ~19.625 sq. inches. So the energy would be 60W*19.625 sq.in., or 1,177.5 W/sq. in., right?
So how would I convert that to degrees F?

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cb9fl
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Is 100% of the light from the bulb reaching the magnifying glass?
Most charts showing watts/sq foot for sunlight, at least the one's I've seen, show how many watts can be created per sq foot using solar cells. That isn't what you want so disregard those.
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Baby_Hitler
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Quote:
Drink_Punk_Soda said: Sorry, Seuss, I didn't refresh the page before responding.
Let's simplify things a bit then, using your method. Assume I'm using a 60W bulb, with a 5" diameter lens. The area of a 5" diameter lens is ~19.625 sq. inches. So the energy would be 60W*19.625 sq.in., or 1,177.5 W/sq. in., right?
So how would I convert that to degrees F?
By measuring it with a thermometer.
Seriously though, what you're getting with the magnifying glass is heat, not temperature.
Theoretically, you could heat something to millions of degrees if you could stop the heat from escaping. Heat escapes by conduction, convection, or radiation. Some materials conduct heat at a different rate.
Some materials require more heat to achieve the same temperature as another material would. This is called the "specific heat" of a material.
Also, some materials radiate more heat at the same temperature than other materials, and lose more heat that way.
The best you can hope for is to estimate total watts and watts per area to get total energy, and energy density.


TinMan
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Yes sorry. Replace the v with the h.
 


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