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07-27-2014, 05:09 PM #1
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xirrious
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Heat relationship.

Does someone here know how to calculate temperature based on wattage for a given surface area? Say 1mm^2 with 1w. How can the temperature be determined mathematically? We can assume a perfect absorption by the surface material for the sake of ease. Obviously temperature has much to do with how well the light induces molecular excitement. Clear materials wouldn't work here.

Thanks for any one who has some advice!

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07-27-2014, 05:24 PM #2
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Photonz
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Re: Heat relationship.

Well, 1W is 1 Joule per second. And we have the equation q=mC(delta)T. Where q stands for heat in Joules, m stands for mass of the object, C stands for heat capacity, and T stands for temperature. You'll plug in 1 for q (that's if it's only for 1 second). If you're shining it for more than 1 second, then you'll plug however many seconds in for q. I'll need to know what the material is as well as the mass of the object, not surface area. Surface area shouldn't matter assuming it absorbs 100% of the heat.
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07-27-2014, 05:43 PM #3
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xirrious
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Re: Heat relationship.

Surface area of the light meant. As in obviously more heat is incurred if the beam is focused to a smaller spot. Thanks for the equation though I am glad someone knew so quickly. Your awesome man thanks.
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07-28-2014, 12:08 AM #4
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styropyro
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Re: Heat relationship.

The surface area won't matter for this problem. As long as all of the light is hitting the object (while assuming 100% energy transfer to heat) then you just plug in values for the variables in the equation like Photonz said.
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07-28-2014, 01:08 AM #5
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xirrious
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Re: Heat relationship.

So assuming static and known mass the equation does not need a variable for the surface area that the light touches? This cannot be correct. It may be assuming a 100 percent heat transfer but in reality even the most ideal material will heat up quicker the more localized and focused the energy is. Or am I missing something?

Edit- I mean it will heat up locally relative to the beams focal point.
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Last edited by xirrious; 07-28-2014 at 01:09 AM.

07-28-2014, 07:14 AM #6
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amkdeath
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Re: Heat relationship.

Quote:
 Originally Posted by xirrious Surface area of the light meant. As in obviously more heat is incurred if the beam is focused to a smaller spot. Thanks for the equation though I am glad someone knew so quickly. Your awesome man thanks.
Actually, the same amount of energy is "incurred" regardless of the incident beam size. The only difference is the area over which that energy is distributed. Thus, if you distribute the same amount of heat over a pin-sized focus point, that individual point on whatever material you're shining the beam on will have to dissipate more energy than if it was focused over a larger area containing that point.

regards,

amk
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07-29-2014, 03:20 AM #7
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Re: Heat relationship.

Are you trying to calculate how hot an object will be after X amount of time being irradiated by Y power? Because it can't really be done. It depends on the specific heat of the object, the reflectivity of the surface, the emissivity of the object, ambient temperature, ambient airflow, ambient environment collective emissivity, ambient humidity, thermal conductivity of the object, and probably 5 or 10 other things I'm forgetting. Even if you had all of those, you'd need to work through some wicked calculus.

If you ignore heat lost to the environment, it's not so bad. That's basic calorimeter calculations. But you can't do that when the heat source is as meager as a laser pointer.
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07-29-2014, 03:28 AM #8
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xirrious
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Re: Heat relationship.

Yes that's the goal. However we can assume some variables as static. Ambient temperature and specific heat can be assumed static and "ideal" respectively. But ThanKS I didn't think far enough ahead to realize all those factors obviously come into play. I'd need to remove most variables for a calculation and as such it'd be a crude estimate.
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08-16-2014, 07:51 AM #9
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xirrious
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Re: Heat relationship.

Quote:
 Originally Posted by amkdeath Actually, the same amount of energy is "incurred" regardless of the incident beam size. The only difference is the area over which that energy is distributed. Thus, if you distribute the same amount of heat over a pin-sized focus point, that individual point on whatever material you're shining the beam on will have to dissipate more energy than if it was focused over a larger area containing that point. regards, amk
I am not referring to energy transfer I am inquiring as to how energy is related to heat using mW units. Assuming all necessary variables to reduce to a reasonable complexity static
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08-16-2014, 06:01 PM #10
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Re: Heat relationship.

Quote:
 Originally Posted by xirrious I am not referring to energy transfer I am inquiring as to how energy is related to heat using mW units.
That's like asking how many gallons a pound of radio waves sounds like.

Please look up any terms below you don't have a full understanding of, and ask a different question.

A watt is a unit of power.
A joule is a unit of energy.
A degree is a unit of temperature.
HeatING requires energy transfer.
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08-17-2014, 02:31 PM #11
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xirrious
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Re: Heat relationship.

All those terms are rather related lol.. Sorry for some sort of misunderstanding. My original question you already answered Cyp. The second question is the same as the first but I clarified that most variables could be assumed eithe ideal or static for the sake of ease, and simply making it a feasible calculation.
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08-18-2014, 08:10 PM #12
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BShanahan14rulz
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Re: Heat relationship.

HyperPhysics Concepts < good starting point for learning very basic principles

You will likely be combining several equations, many of which will involve numbers that you will have no real way of determining. For instance, what is the emissivity of an unknown aluminum alloy MCPCB with white paint, and silver lettering? What is the thermal conductivity of an aixiz module with 5.600mm brass heat source? What about if the heat source is 5.580mm, resulting in less contact area at that junction?

It would be easier to set up your intended setup, and use a thermometer to measure the rise in temperature.

08-18-2014, 08:44 PM #13
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xirrious
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Re: Heat relationship.

Thank you for a polite informative answer. I understand it's ridiculously hard to calculate however I am interested for the sake of math not practicality. Thanks again !
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"I appreciate the beauty and aroma of a rose as much as the next man, but an understanding of the physical and chemical processes that underly that beauty can only enrich the experience."

-Various hosts, diodes, and lenses awaiting assembly

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