I'm specifically thinking about lime/sedimentation at the bottom of water heater, and calcification of heating elements (and not possible thermal insulation deterioration).

It is very often claimed (by both government energy efficiency experts and water-heater sellers/service people - but without sound theoretical explanation, alas) that calcified heating elements in electrical water heaters would lead to much higher energy bills for water heating (numbers often wildly ranging from 30% to over 400% percent increase in electrical power bills), and that it is imperative that water heaters get serviced (usually suggested on yearly basis) to drain out sedimentation and clean/replace the heating elements primarily for that reason (among others).

but I'm looking for theoretical/physical explanation for that (or a denial).

Now, I can see why calcification on and around heating element is bad, as it would be acting as thermal insulation between itself and water, which would:

  1. prolong the time needed for heating element to heat the full tank of water (due to slower transfer of heat)

  2. reduce life of heating element (due to it overheating, as thermostat shutoff would be delayed due to previous point)

Which are both valid reasons for regular service. However, I do not see why it would case any increase in energy usage?

I've tried looking up, and as best as I can see, the law of energy preservation should hold up. I cannot see the energy being converted to anything other than heat due to calcification of heating elements? And it that heat was used by heater, it must get transferred to the water around it eventually, right?

I see two things happening if heating element gets calcified:

  • as it is isolated from cold water around it and initial thermostat assumptions, it will overheat, which would increase its electrical resistance somewhat, and reduce it's power in Watts (for example from 2000W to 1500W), thus prolonging time needed to heat the water (and reducing its lifetime)
  • the calcification/lime would act as a "wall" which would slow down the heat transfer (which would also prolong time by which whole tank of water would be heated); however that heat energy would not be lost, as the calcification itself would continue to give off that heat energy back to the water even after the heating element was shut off from electrical power by thermostat

Are those assumptions correct, and if so, am I missing something else? Where would such energy used by heater be wasted (not transferred to heated water)? Or is "irregular maintenance of water heater leads to higher energy bills" just very popular urban legend?

So, the question is "is it true that calcified water heater elements would use up more electrical energy, and if so, why"?

(note: I initially considered posting this in diy.SE which has a lot to say about water heaters; but as I seek theoretical explanation and not practical usage, physics.SE seems much better place. Also note: english is not my primary language, so some terms might be wrong; and my physics knowledge is at [or was, two or three decades ago] at high school level)

  • 1
    $\begingroup$ This is probably on-topic here but Skeptics might actually be a better place for it. $\endgroup$ Commented Feb 26, 2014 at 1:37
  • $\begingroup$ on Skeptics: skeptics.stackexchange.com/questions/23802/… $\endgroup$ Commented Jan 5, 2017 at 15:00
  • $\begingroup$ I was with you from the first line. When I ask "normal" people this kind of questions, they look at me as I were stupid. Normal-Peoples answer would be: sediment causes that heat is not transferred properly, therefore it needs longer to heat up, hence more energy loss. People never get my point of view and why I question things like that, hence thinking I am stupid. I am with you bro. $\endgroup$
    – jjstcool
    Commented May 21, 2017 at 23:32

4 Answers 4


The electrical energy you pay for doesn't just go into the appliances; some amount of it ends up as heat in the electrical path from the meter to the loads. Also, heat from a load such as a heating element can travel along the electrical conductors away from the application (the water tank) - remember: most good electrical conductors are also good thermal conductors. So, your scaled up water heater element will still produce the same heating power as it always did, but less of the heat ends up in the water, and more ends up outside the water heater. Some of that wasted heat escapes along the wiring, and, because of the increased duty cycle to compensate for the ineffiency, the wiring spends more time carrying current and wasting energy.

Bottom line: energy is conserved in that it doesn't simply disappear... it just gets misplaced to become waste heat somewhere and provides no benefit.

  • $\begingroup$ Ok, that is understandable that some heat will be generated in wiring in the wall before the water heater; however I'd think that such losses would be insignificant (below 1%), and most definitely not in range of 30-400% increase. I mean, it would surely be noticeable if wiring in the walls were heated up to 4 times more than the water heater? And the wires going in the heater don't feel hot while it's working, while the water is very hot? $\endgroup$ Commented Feb 26, 2014 at 14:46
  • $\begingroup$ Also, would that mean that moving the water heater much closer to to power meter (few centimetres of wires instead of dozens of meters) would not only much reduce losses due to scaling (by 2 orders of magnitude - the same as length of wires reduction?), but also noticeably reduce energy bills when water heater is operating normally? $\endgroup$ Commented Feb 26, 2014 at 14:51
  • $\begingroup$ Absolutely correct reasoning! I love your kind of thought experimenting. Bottom line: One may have a strange feeling that the energy is wasted, in reality it is only to a negligible portion. $\endgroup$
    – jjstcool
    Commented May 21, 2017 at 23:38

The scale buildup will cause the heater to heat water more slowly because of the greater thermal resistance between the element and the water. This means that the heater element itself will get (much) hotter, and its resistance will increase. This reduces the power output, and the heater has a hard time giving you enough hot water. This will not increase your electric bill by very much - but it does mean that long showers will be cold showers.

By contrast, in a gas heater you have hot gas flowing through the heat exchanger; if there is additional thermal resistance, the gas that leaves the exchanger will still be hot, and you have real inefficiencies. Not maintaining your gas water heater will cost you more; but not maintaining the electrical water heater won't.

The only exception I can see is a situation where you have demand pricing: the cost of a kWh changes with time of day. Now, a scaled-up water heater may take its time heating the water, and it may have to continue heating water all day; a more efficient one can use just the off-peak rates, and cost less (same kWh, less money).

As for losses "elsewhere" in the system. If your electric heater is wired into a 40A, 220 V circuit, it will be using (in the US) 8 gage (AWG) wire. Assuming you are 20 feet from the switch panel, you have 40 feet of wire with a resistance of 0.025 Ohm. With 40 A flowing, this gives you a 1 V drop and 40 W dissipation. Compared to the 9 kW for the heater, that's about 0.5%. It is not zero, but it is not that big either. Certainly not as large as the number you are quoting.


If I remember correctly, a 1/8 inch of calcium build up on a heating coil is the same as several feet of concrete wrapped around it so you can see how much more electricity it would take to heat the water to temperature!


Its like having wheels on a car made of concrete ( What You Ask)
Answer having an element covered in calcium the element is not designed to work efficiently in that condition therefore if the hot water heater is used a lot it is going to cost you more electricity

The equation must relate to usage , also the element can short out in that state because sockets and cables will overheat


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