First post here, so much to learn - About Heat & Power

Main Question or Discussion Point

First post here, so much to learn -- About Heat & Power

I would first like to thank those who participate on this forum, I have referred to it several times in the past and have found it highly informative.

I am not a teacher, just an old man intrigued by learning, I am mostly retired and have a few hobbies to keep me off the streets. :D I have a large group of friends who also enjoy discussing things we know little about but relish the exchange of ideas, it keeps our geriatric minds working.

There is one topic we have beat to death, wattage and heat. I have searched this forum on this topic and did not really find a direct answer to the question. I have hopes that we can settle it with the combined knowledge here and that will allow us to move on to more helpful topics in our group.

In this example we have two lights, a 100 watt metal halide and a 100 watt LED. Assuming both lights are pulling the same 100 watts of energy from the wall in a one hour period (measured with a killawatt meter) is it reasonable that both fixtures are also creating 100 watts of heat as the eventual outcome of the energy consumed and the work done? I understand that the LED is more efficient at creating light than the halide and will produce more light but will both units end up producing the same amount of heat?

Answers and Replies

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Nugatory
Mentor
In this example we have two lights, a 100 watt metal halide and a 100 watt LED. Assuming both lights are pulling the same 100 watts of energy from the wall in a one hour period (measured with a killawatt meter) is it reasonable that both fixtures are also creating 100 watts of heat as the eventual outcome of the energy consumed and the work done? I understand that the LED is more efficient at creating light than the halide and will produce more light but will both units end up producing the same amount of heat?
Light is a form of energy, as is heat, so total energy coming out of the light bulb is the sum of the amount of heat produced and the amount of light produced - and that sum will of course be 100 watt-hours for both bulbs. However, the LED will deliver more of that energy in the form of light than the halogen unit.

On the other hand, no matter how it's produced, the light will eventually hit something and be absorbed, and when light is absorbed it generally turns into... heat. So no matter what happens in between, you'll end up with 100 watt-hours of heat in the end.

phinds
Science Advisor
Gold Member
2019 Award
Welcome to the forum.

What do you think "efficiency" means in the context of lighting? That should answer your question.

I see I was beat to the punch, BUT ... I do think Nugatory's answer misses the direct point of your question, which I believe is about the IMMEDIATE generation of heat vs light.

Gentleman, I appreciate the confirmation. College was a long time ago but the learning has yet to cease. Sadly the older we get the more our brains resemble a wiffle ball.

In essence, wattage consumed will eventually be returned as heat, in the same amount as the wattage drawn. 100 watts of energy in will equal to 100 watts of heat out. Are there any instances where this is not true?

Nugatory
Mentor
In essence, wattage consumed will eventually be returned as heat, in the same amount as the wattage drawn. 100 watts of energy in will equal to 100 watts of heat out. Are there any instances where this is not true?
Running down every last little bit of the energy may be a bit difficult (consider that some tiny fraction of the light may pass through a window, through the earth's atmosphere, head off into interstellar space, and finally be absorbed somewhere in the Andromeda galaxy) but if you can account for all of it, you will end up with 100 watt-hours of energy in and 100 watt-hours of energy out - this is just the law of conservation of energy at work.

Philip Wood
Gold Member
Sorry to be a dissenting voice, but a filament lamp gives out (mainly) light and infrared. If you are classifying infrared as heat, then you must do the same with light. There is no fundamental difference between them, though our retinas are sensitive to the range of wavelengths we call 'light'. Our skin is sensitive to both, if the intensity is high enough. Filter out the infrared and let only visible light fall on our skin and it'll start to feel warm.

Thank you, and yes I do understand that in "real world" scenarios there may be a fraction lost but somewhere along the journey to Andromeda it will give off it's energy as heat.

Phillip it is my understanding that no matter what form the energy is converted to, the result will be heat, no matter what form it has changed to, motion, light, chemical, it all ends up as heat in the end. It is a hard concept for a lot of people to wrap their head around, I think you are saying the same thing.

Philip Wood
Gold Member
I stick to the thermodynamic concept of heat as energy in transit down a temperature gradient (whether it's conducted, convected or radiated). I won't use 'heat' to mean energy residing in a body; that's internal energy.

Nugatory
Mentor
Sorry to be a dissenting voice, but a filament lamp gives out (mainly) light and infrared. If you are classifying infrared as heat, then you must do the same with light. There is no fundamental difference between them, though our retinas are sensitive to the range of wavelengths we call 'light'. Our skin is sensitive to both, if the intensity is high enough. Filter out the infrared and let only visible light fall on our skin and it'll start to feel warm.
Good point, and I must confess that I was counting the infra-red as "light", as it is electromagnetic radiation; but of course the efficiency rating they print on the package the bulb came in is using a different definition, counting only the visible light as "light" and everything else including the infrared as "heat".

This doesn't change the basic principles though:
- If 100 watt-hours go in, 100 watt-hours must come out in one form or another
- The higher-efficiency bulb will release a larger fraction of that 100 watt-hours as visible light
- It pretty much all ends up as heat eventually

russ_watters
Mentor
In essence, wattage consumed will eventually be returned as heat, in the same amount as the wattage drawn. 100 watts of energy in will equal to 100 watts of heat out. Are there any instances where this is not true?
What do you mean by "any"? Any light bulb type or any use of electricity? Because there are many uses of electricity that convert electrical energy into types of energy besides light and heat. Including ones that utilize electrical energy to move thermal energy (heat).

That was the question Russ. I have never been a believer in absolutes. I had a feeling it was possible to use wattage that did not result in a heat exchange but could not think of any examples. I am here to learn and that is my only absolute. I still don't know of any examples or understand the one you cited but I am a believer in possibilities.

Nugatory
Mentor
That was the question Russ. I have never been a believer in absolutes. I had a feeling it was possible to use wattage that did not result in a heat exchange but could not think of any examples.
You could shine the light on a photoelectric cell, use the current that is generated from the cell to charge a battery or operate an electric motor that compresses a spring. These processes wil turn the energy of the light into potential energy.

Integral
Staff Emeritus
Science Advisor
Gold Member
As Russ said, any use of electricity to do mechanical work does not necessarily get turned into heat. Suppose you use a pump to fill your swimming pool. If you pump the water from 30m underground you have done a lot of work that does not show up as heat.

An even better example is a hydroelectric dam. They really need to run the dynamos at a steady load, so when external demand is low they run pumps to pump water from below the dam back to the reservoir where it can be used again to produce power.

Does that pump not consume wattage and that wattage consumed converted into heat? Both a 100 watt submersible pump and a 100 watt electrical heater consume and release heat back to the water as 100 watts. Even the work being performed by the pump, moving water, creates friction which also creates heat. Have I misunderstood something?

Integral
Staff Emeritus
Science Advisor
Gold Member
For the dam pushing water back to the top of the dam, sure there is some heat generated but not kW of heat, there is kWs of real WORK done. A motor is very efficient at turning electricity into motion and NOT heat. In most motors just a few % of the applied wattage is lost to heat, the rest does mechanical work.

That is why I gave the example of filling a swimming pool. You have increased the gravitational potential energy of the water by doing mechanical work on it. That mechanical work cannot show up as heat also.

I am having a hard time with the concept of mechanical work not showing up as heat. Motion doesn't generate heat? That pump pushing water up 30 meters is consuming x watts but is not generating the same x wattage as heat? That sounds contradictory to post #2 by Nugarory, which is in line with my own experiences with submersible pumps, which do heat the water, and non-submersible pumps where the body of water being pushed is not in contact with the motor, just the spinning impeller, and those too heat the water.

If we placed a speaker and a heater in two separate sealed boxes and supplied each with 100 watts of power, would not both of them impart 100 watts of heat to the box? The heater by creating heat directly and the speaker from both heat generation from the efficiencies of power conversion and friction from air and speaker movement? The same energy is going in but taking different forms, but the differing forms are still generating heat, whether it be from motion or direct heat generation.

I do not wish to sound argumentative, just trying to understand how motion does not generate heat and where the excess wattage that the pump used is going?

Philip Wood
Gold Member
SirReal63. First para: Most of the energy supplied to the submersible pump goes to internal energy in the surrounding water, making the water slightly hotter. A fraction of the energy would finish up as grav. P.E., as Nugatory says, if the pump is being used to raise water to a higher reservoir.

Second para: All the energy supplied to the heater and speaker will eventually go to internal energy (see post 8) in the walls and water, making them slightly hotter.

If the water and walls are not insulated, and were originally at the temperature of their surroundings, then, having been made very slightly hotter, they will very slowly give out heat to their surroundings. This usage of "heat" accords with my scruples mentioned in post 8, because I'm now talking about energy in transit.

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russ_watters
Mentor
Pumps can be as much as 60-80% efficient, so it is possible that pumping water up from a well can result in most of the electrical energy becoming gravitational potential energy and not heat.