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Im looking at installing an electric towel radiator. The heating element is 200W.

Dose that mean that it would both :

  1. use 200W of electricity

  2. provide 200W of heat output to the room ?

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  • \$\begingroup\$ Of course you would want to actually measure the W used by the heater. Power on devices like this can be more marketing than it is an accurate specification, but it's probably close. \$\endgroup\$ – JPhi1618 Aug 29 '19 at 17:47
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Yes, electric heaters are basically a resistor.

A resistor converts electric energy into heat, it does that with 100% efficiency.

That might sound weird but think about it this way: if a resistor was 90% efficient, where would the 10% "lost" power go?

Nearly all not 100% efficient devices lose the wasted energy as heat. Generating heat is the sole purpose of a (resistor) heater. So even if the heater was only 90% effcient, that 10% would still be heat, making the efficiency 100%.

So indeed the heater consumes 200 W (when it is in operation) and it will then also emit 200 W of heat.

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    \$\begingroup\$ honestly? If a resistor is 90% efficient, it means it loses 10% of its supplied power as other type of energy - magnetic field in coiling, mechanical (vibrations), light, chemical etc. FWIW, it's actually the case for many heating elements - they produce visible light (and possibly other radiation that is neither infrared nor visible, e.g. UV). Because of that, there's a distinct error in saying "it does that with 100% efficiency". Nearly 100% efficiency - yeah, sure. 100% efficiency - nah ah, no way. \$\endgroup\$ – user20088 Aug 28 '19 at 21:25
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    \$\begingroup\$ The visible light (and UV) is radiative heat. It contributes to the heating. Vibrations and mechanical fields can be energised but with a constant supply of energy >1W (99% rounding point) are either going to dump that into heat or get really noticable really quickly! They would probably cause the heater to disintigrate pretty quickly, dumping its energy as heat. The sound, once heard, will be damped and become heat. State change and chemical energy can absorb significant energy (such as in a steam generator) but if your heater is puttiny >1W into that, you've got a weird heater. \$\endgroup\$ – Dan Sheppard Aug 28 '19 at 21:44
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    \$\begingroup\$ @vaxquis: visible light is one way that thermal energy in the heating element escapes. Magnetic vibration and direct EM emissions at 60Hz and harmonics are non-heat outputs, but the light energy went through heat. Err, but yes the OP said "200W of heat output to the room", while this answer only talks about converting electrical energy to heat without considering getting it transferred to the room. So very close to 100% is very likely for that, but any light that escapes out a door window (instead of being absorbed by objects in the room and heating them) is lost. \$\endgroup\$ – Peter Cordes Aug 28 '19 at 21:44
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    \$\begingroup\$ @vaxquis: The only way you "lose" efficiency is by some of the energy turning into heat when you don't want it to. No matter what you propose as the "inefficiency" of your heater, the "lost" energy eventually (rather, quickly) ends up as heat. \$\endgroup\$ – R.. GitHub STOP HELPING ICE Aug 29 '19 at 1:31
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    \$\begingroup\$ @vaxquis: Right. Thus the parenthetical "rather, quickly" - energy is going to have a hard time getting out of the room before it turns to heat, and much more is going to be lost once it's already been turned into heat. \$\endgroup\$ – R.. GitHub STOP HELPING ICE Aug 29 '19 at 14:30
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Yes (maybe a couple of watts more for electrical losses in the cabling from the energy meter) and yes (unless there are losses through the wall to the outside world). Of course it may only produce 95% of the rated output - nothing is perfect of course and, it may indeed produce 105% of the rated labelled power.

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    \$\begingroup\$ To clarify, he means the device may exceed its rating, e.g. draw 210W from the wall and produce ~210W of heat, not that it could be over 100% efficient \$\endgroup\$ – Reroute Aug 28 '19 at 12:58
  • \$\begingroup\$ @Reroute actually, "Efficiency" in a closed system can be pretty much directly equated to heat loss--since a heater is all heat loss it's actually pretty near 100% efficient (at being inefficient). \$\endgroup\$ – Bill K Aug 29 '19 at 21:00
  • \$\begingroup\$ @BillK near 100%, not exceeding it, what part of my clarification is incorrect? \$\endgroup\$ – Reroute Aug 30 '19 at 11:17
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Dose that mean that it would both:

  • use 200w of electricity

Yes. It would consume 200 W of electrical power.

  • provide 200w of heat output to the room ?

It would give of 200 W in the form of heat.


'W' for watt, 'V' for volt, 'A' for ampere.

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Possibly the most universal law in physics, even before the constant speed of light, is conservation of energy. Energy in + energy stored = energy out. So if there's 100W going in, and there's not a significant amount of energy being stored, then energy in = energy out. Always. Every time. And not just heaters -- lights, refrigerators, light bulbs*, motors, the Starship Enterprise**, etc.

If there's 100W going into the room, and it's not coming out in the form of light or radio waves or mechanical energy (or, presumably, subspace beacons in the case of the Enterprise), then it's going to heat up the room. Period. End of story. (And, sadly, it's why perpetual motion machines don't work).

Which is a really long way of saying that, yes, your 100W heater will consume as much power is it gives you in heat. And if it doesn't quite match its ratings and it consumes more power, then it'll deliver more heat, and visa-versa if it consumes less power.

* Confusing "100W equivalent" ratings on LED and CFL bulbs notwithstanding -- I'm talking about real energy.

** Although the Starship Enterprise won't fit well into a room that is appropriate for a 100W heater, and it usually comes with a lot of stored energy in the form of antimatter.

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  • \$\begingroup\$ it's not coming out in the form of light or radio waves or mechanical energy - well, in the ideal world it wouldn't. In the real world with non-zero self-capacitance/inductance of the elements and EM radiation from heating happening spontaneously in a way we can't prevent - the energy does indeed leak. \$\endgroup\$ – user20088 Aug 28 '19 at 21:29
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    \$\begingroup\$ +1 for the Star Trek reference. You can also use a phaser to heat things up. \$\endgroup\$ – manassehkatz-Moving 2 Codidact Aug 29 '19 at 4:19
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    \$\begingroup\$ @vaxquis if the room is closed then all the self-capacitance/inductance and probably all the EM radiation will remain in the room and eventually end up as heat. Electric heating is that unique scenario where nearly all the loss is actually counted as efficiency. \$\endgroup\$ – Bill K Aug 29 '19 at 21:03
  • \$\begingroup\$ @BillK nearly, yes. 100%, no. That was my point all along - you can't prevent the 'leakage' of useful energy, even in a case of simple heater. The fact that the leakage is negligible for any practical application doesn't change the fact that the leakage exists. If the heating coil of the heater turns red, and my neighbour can see through the window that it's glowing red, then the energy leaked. If he can detect the EM field of the coil with a meter - it leaked. If the energy supplier can detect that I turned the heater on via detection of back EMF - the energy has leaked. \$\endgroup\$ – user20088 Aug 30 '19 at 12:48
  • \$\begingroup\$ My point was just that If you have a relatively closed room (Closed curtains) nearly all leakages will degrade into heat. I'm not sure I can imagine what could escape a metal heating unit that blocks most EM and then escape the closed room it sits in. \$\endgroup\$ – Bill K Aug 30 '19 at 16:30
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Both.

Heat is the easiest thing to make. In fact it's the ultimate destination of virtually all energy conversion, because of how entropy works.

Every watt ends up turning into heat, or rarely, light. For instance a 100W incandescent makes about 98 watts of heat, and 2 watts of luminous light, which turns into heat after it hits wall surfaces.

A 15 watt (100W equivalent) LED makes 13 watts of heat and 2 watts of light.

One of my running jokes is that some people like to build heaters out of resistors, I like to build them out of Bitcoin miners. The electricity bill and useful heat will be the same either way, only one of them also gives you bitcoin.

So whether your towel heater computes Bitcoin or not, the answer is, 200W in, 200W out. It's not going anywhere else.

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    \$\begingroup\$ For heating, heat pumps reach efficiencies surpassing 300%. So no, your bitcoin heater is trash compared to a proper heating system. (But if you are going to put a resistance heater then yeah, your system at least secures payments meanwhile) \$\endgroup\$ – Oxy Aug 29 '19 at 14:55
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To add to the existing answers, your radiator is likely to have a thermostat to control the radiator temperature (not the room temperature), so the average power may be somewhat less than 200W.

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