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I have two portable electric heaters and their power is exactly the same: 2000W. One is using ceramic technology, and the other resistor technology. The one with ceramic technology produce a lot more heat.

Considering they have the same power, where has the energy not converted into heat (by the less efficient heater) gone?

I didn't expect there would be so much difference, as heat is usually one of the unwanted byproducts of machines intended for different purposes, like motion. And I know from the thermodynamics laws that energy is not destroyed or created. So where has the missing heat gone? And how can two heaters with the same power produce different heat outputs?

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    $\begingroup$ Perhaps one produces more hot air and the other more infrared light. The experience will be quite different. Indeed, how do you quantify "heat". $\endgroup$
    – my2cts
    Apr 10, 2021 at 23:37
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    $\begingroup$ The way I quantify it is by looking at the time it takes to bring my entire room to 20 degrees and the time is considerably different. Ceramic heathers are also described online as more efficient, but still don't get why... $\endgroup$ Apr 10, 2021 at 23:49
  • $\begingroup$ Oh, I see you did specify ceramics vs resistance heater. $\endgroup$
    – DKNguyen
    Apr 11, 2021 at 5:41
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    $\begingroup$ Have you determined the actual power rating of the specific heaters by independently measuring the current and voltage drop? $\endgroup$
    – DJohnM
    Apr 11, 2021 at 6:36
  • $\begingroup$ Where are you located? 2 kW from a portable heater is not possible in some locations... $\endgroup$
    – DJohnM
    Apr 11, 2021 at 6:39

3 Answers 3

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A ceramic heater works by convective heat transfer, i.e., the energy consumed directly goes towards increasing the kinetic energy of air molecules, which are then blown into the surrounding space. The warmth felt due to such a heater is due to molecules in that warm air colliding with your skin, causing random motion of molecules in your skin due to the collisions.

In contrast, an infrared heater with a tungsten filament or other resistive element that grows red hot works by emitting electromagnetic radiation in the infrared range. The energy consumed by an infrared heater mainly goes into producing that electromagnetic radiation. When that electromagnetic radiation interacts with the molecules of your skin or other surface, it causes random motion of the molecules in your skin (heat) by causing the molecules to transition to a higher-energy vibrational state.

Which one will feel warmer sooner depends on the circumstances. An infrared heater can be more efficient by transferring heat energy directly to your skin, which is where you want it to be, instead of wasting energy by warming air molecules that might not even ever wind up subsequently hitting your skin, for example if you're in a drafty garage. On the other hand, the electromagnetic radiation emitted by an infrared heater travels in a straight line, so you might not be directly heated by the electromagnetic radiation if you're in a part of the room where there isn't an uninterrupted straight line between the heating element and your skin.

If you're measuring heater effectiveness by measuring how long it takes to get the air in a room up to a certain temperature, a convective heater is likely to win, because heating air is what a convective heater does directly. An infrared heater heats whatever the radiation hits, which means you could for example be wasting energy by heating a wall or something directly, instead of nearly all of the energy going into heating air.

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  • $\begingroup$ I have a number of tungsten filament space heaters. None of them have an exposed filament that could radiate heat to the surroundings. All of them have fans that blow room air over the filaments to extract energy from the filament by forced convection. $\endgroup$
    – DJohnM
    Apr 11, 2021 at 6:36
  • $\begingroup$ Electromagnetic radiation can excite electrons into different orbitals, but not at IR (or visible light) frequencies, only at higher frequencies: UV, X-rays, gamma rays, which are quite to very dangerous. IR radiation can excite higher mechanical (rotational and vibrational) molecular states though. $\endgroup$
    – doetoe
    Apr 11, 2021 at 11:15
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    $\begingroup$ @doetoe and DJohnM: Thanks for your comments. I have tweaked the answer to reflect your comments. $\endgroup$
    – Red Act
    Apr 11, 2021 at 14:29
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    $\begingroup$ @EricDuminil I just set the "community wiki" flag on this answer, so you can tweak the answer to fix the problematic wording yourself, if you'd like. $\endgroup$
    – Red Act
    Apr 14, 2021 at 1:24
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Pretty much all the energy consumed by any electric appliance is converted to heat, light or sound... and most of the light and sound ends up as heat before it can escape your home. (Heat pumps don't count. They cheat.)

Both of your heaters are therefore equally efficient heat producers, as is your refrigerator, your computer, your TV, etc. The difference is whether they put the heat somewhere where you will feel it, or somewhere where it will be lost or wasted.

If your hot-wire heater doesn't do as good a job of heating up the air you feel, or the air where the thermostat is, then it's probably not too hard to figure out where it's going. The heater itself is probably toasty warm, and I bet the nearby walls are pretty warm, too. It probably also does a better job of heating up the air near the ceiling.

There is no fundamental reason why a heater with a ceramic element would be better than a heater with a metal element. It will come down to what you can do at a particular price point.

The problem of designing a space heater is basically that you need a heating element that wastes 1500W or so, you want to avoid radiating IR directly (because it heats up walls), and you want to use airflow to keep it cool (because a lot of the heat you dump into the heater itself ends up wasted). Your fan can be as big as you like, because it's a 100% efficient heater, too.

Given that problem, it helps if the heating element has a large surface area, so that you can spend 1500W in it and more easily cool it so that it never has to get red hot.

It's likely that a ceramic element with a large surface area can probably be made more cheaply than a metal element with a large surface area. For the same money, then, the ceramic heater gets better cooling and can more efficiently heat up the air, because the hot-wire heater designer needs to spend more money on elements that distribute the heat of a much hotter wire.

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  • $\begingroup$ nit: the refrigerator in your examples is one of the "cheating" heat pumps :p $\endgroup$
    – briantist
    Apr 11, 2021 at 15:00
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    $\begingroup$ Yes, but it doesn't exchange the heat with the outdoors, so in the end it's a just a heater too. $\endgroup$ Apr 11, 2021 at 15:02
  • $\begingroup$ I suppose that's true except for when you first turn one on, and once the heat pumped from the inside is "lost" (equalized with the room) it's no longer adding more heat than the energy it consumes. $\endgroup$
    – briantist
    Apr 11, 2021 at 15:06
  • $\begingroup$ All the electricity consumed by a heat pump is also converted to heat. $\endgroup$ Apr 11, 2021 at 19:08
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    $\begingroup$ because a lot of the heat you dump into the heater itself ends up wasted - how? By conducting into the floor and eventually to a basement or other room or something? Or heating the thermal mass of the heater, where it will be released over time after the human turns off the heater and leaves the room? I guess that matters if you're mostly considering short-term use, not steady-state operation. $\endgroup$ Apr 12, 2021 at 0:38
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First of all, saying "... produce heat..." is just not correct. Heat is one of the two forms of energy that can be exchanged between two systems, not produced. If you have an electric heater (with electric I mean it works only if you plug it in), and you consider as system the heater itself, it exchanges electric work and heat with the surroundings. Now, heat can be exchanged in three ways, depending on the tech of the heater one of these will be used. However I don't want to go into details of this, what is important is that the first thermodynamics principle tells: the heat power exchanged is equal to the electric power($=v(t)i(t)$) plus the derivative of the internal energy of the system itself. This holds every time instant 't' and with the conventions of positive input heat power and positive electric power as output. If the two heater have really the same electric power input (you should consider that even if the two have the same nominal electric power input they could absorb different amount of that in the real world) then the difference between the two heat power is given by the derivative of the internal energy of the heater. I think it is exactly this parameter, which also depends on the tech of the heater, which causes the two different speeds of response.
Edit: Example
if your heater is made by a pool of 100 kg of water with inside some resistive element that warms up the water, the derivative of the internal energy of the heater will be: $$\frac{dU}{dt}=cm\frac{dT}{dt}$$ $$\frac{dU}{dt}=4186*100*\frac{dT}{dt}$$ Where T is the temperature of the water in Kelvin.

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  • $\begingroup$ Could you give an example of "internal energy" of the heater? Would this be something possibly used up or consumed, such as chemical energy (a battery)? If you wanted to use the waste heat of a battery charger, I suppose it could even work in the other direction (end up with more internal energy). I'm just trying to think of a heater where some sort of "internal" energy might be gained or lost. $\endgroup$
    – Phil Perry
    Apr 11, 2021 at 22:06
  • $\begingroup$ @PhilPerry: A typical example would be a storage heater (en.wikipedia.org/wiki/Storage_heater), I guess. The energy flows are basically never balanced : it consumes a lot of electricity during the night, it releases a lot of heat during the day. It's made possible because it has a large thermal mass, and the internal energy is varying 24/7. $\endgroup$ Apr 12, 2021 at 6:56
  • $\begingroup$ I've added an example, hope it's more clear @PhilPerry $\endgroup$
    – Landau
    Apr 12, 2021 at 8:31

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