If a machine is never 100% efficient transforming energy into work because part of the energy is converted into heat, does it mean an electric heater is 100% efficient? [@[email protected]](h
If a machine is never 100% efficient transforming energy into work because part of the energy is converted into heat, does it mean an electric heater is 100% efficient? [@[email protected]](h
If a machine is never 100% efficient transforming energy into work because part of the energy is converted into heat, does it mean an electric heater is 100% efficient? @[email protected]
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You know how you turn on an electric heater and the filament begins to glow? That is energy being converted to light, so not 100% efficient.
Doesn’t the light turn into heat anyway as soon as it’s absorbed?
I mean if you want to go that route, we could just say that every speaker, light source, motor, etc is 100% efficient at generating heat because all of its energy output will eventually become heat.
That is also completely true, but meaningless because heat generation is not the purpose of these devices. However, if you use them in a building heated by a thermostat-controlled electric heater, you’re effectivhly running them for free.
I‘m was using two old servers with folding@home running as space heaters in the winter. I got them for dirt cheap and thought if I convert electricity into heat, I might as well do something good with it. Also nice opportunity to run a minecraft server for the kids during that time.
Energy can neither be created nor destroyed so in the grand scheme of things, everything is 100% energy efficient one way or another.
I suggest we submit proposals to define "100%" and "efficient" before we design the experiment
The visible part of the spectrum is likely going to be absorbed somewhere far away from the place you're trying to heat up. Also, I'm not educated enough to tell if there will be further losses of energy
If it’s in a room the visible radiation will still just heat up the room. If you’re using it outdoors and point it away then yeah you’ll have some waste.
Not sure if visible radiation that leaves through a transparent window will still heat up only inside the room, that what I meant. Probably should have phrased it better
that's only true if you shine it out a very large the window
normally windows cover a quite small fraction of a rooms surface area
but sure, if a few fractions of a percent leave through a window, i guess its technically not a 100% effective space heater, if we define the work as heating only the relevant room.
and that light hits stuff and gets converted to heat.
That heat also powers certain chemical reactions happening on the surface of the hot wire. It’s not a lot of energy, but it’s still something. Light and sound tend to be converted back to heat at some point, but chemical transformations can be more stable, which would result in a tiny loss of efficiency.
Black-body radiation is an interesting argument against 100% efficiency, but couldn't you just extrapolate and argue that the emission will be converted back to heat once it stops reflecting and becomes absorbed?
That's like arguing that trickle down economics is efficient because the money eventually gets into the hands of the poor.
That's like arguing that 99% of the light off a heating element is a laser beam directed straight into deep space.
It depends on the framing of the question a bit. If we are defining 100% efficiency as 100% of electrical energy being converted into kinetic energy (heat) by the device, then that is a no. Some percentage is emitted as EM radiation instead of heat. If they were so then a light bulb or a bomb is a 100% effective heater as well.
It depends on what you consider the room: Both a light bulb and a bomb would deliver all their energy around a fully enclosed room. Incandescent bulbs are indeed effective heaters, LEDs just deliver much less energy. And a bomb, by design, is hard to contain in a room.
Not all electric heaters use that kind of filament
This is true. I assume the ceramic heaters also emit light, just not in the visible spectrum.