Let's assume 1kWh of electricity is used to run a fan heater. Heat pumps are more efficient, but have a range of different efficiencies, so perhaps aren't a good base comparison.

Assuming a good quality log burning fireplace, how much firewood (by weight) does it take to produce the equivalent of that 1kWh of heating? This stove claims combustion efficiency of 89.1%, but 70% is perhaps a reasonable standard to work with.

Obviously water content of the wood would make a big difference too, so let's assume either 10% or 20%, whichever is easier. I understand 20% is considered to be dry, when dealing with firewood.

  • When comparing a space heater to a heat pump, is coefficient of performance a better metric than simple efficiency? 100% of the electricity used by a space heater results in heating, while the same is not true of a heat pump. But, the heat pump produces more heat with the same amount of electricity, because it takes the "free" exterior heat for some its input energy.
    – LShaver
    Commented Dec 21, 2016 at 19:02
  • getting a usefull result could be impossible. you would need to know the temperature of the air that is replacing the air going up the chimmeny.As this air ultimatly has to come from outside, outside temperature needs to be factored in. however this air circulation is lowering the moisture in the house which also has a bearing Commented Dec 8, 2017 at 17:25
  • The "This stove" link is broken.
    – Tim
    Commented Apr 27, 2019 at 5:59
  • @Tim I've fixed the link.... thanks for that, and thanks for your answer! Commented Apr 28, 2019 at 10:52
  • @HighlyIrregular No worries. Just curious: You haven't accepted an answer in the 4 years that this question has been up. Are all of the answer missing something in particular that you may be after?
    – Tim
    Commented Apr 28, 2019 at 21:15

5 Answers 5


Per Google, 1 kWh = 3412 BTU. (And a fraction, which I'll neglect.) Here's a table (one of many found with Google) of BTU/cord for various kinds of wood: https://chimneysweeponline.com/howood.htm It varies a lot by species, but let's say 20 million BTU per cord, and 3500 lbs/cord as average for hardwoods. So 1 lb of average hardwood gives you 5714 BTU. Multiply by the efficiency of your stove (70 - 89.1%) gives 4000 - 5091 BTU. So it takes roughly 0.67 - 0.85 lbs (304 to 386 grams) of firewood to equate to 1 kWh of electric heat.

  • You've ignored the fact that burning wood for heat requires managing exhaust. You won't get the heating figures you calculate in real life because some amount of heat is lost up the chimney (whereas there is no exhaust at the point of heating for resistive electrical heating). Commented Dec 9, 2017 at 0:25
  • 1
    Chimney losses are included in stove efficiency. In fact, they're a major part of that efficiency.
    – jamesqf
    Commented Dec 9, 2017 at 3:12
  • This may be what some folks claim but it's not really true. There's no way to rate a stove for chimney losses because every home and chimney is unique. The efficiency rating has something to do with the lab in which the stove is certified - optimal conditions that don't reflect anyone's actual home. Commented Dec 9, 2017 at 13:37
  • @Jean-Paul Calderone: Which, I assume, is part of the reason why wood stove efficiencies are given as a range.
    – jamesqf
    Commented Dec 10, 2017 at 4:50
  • My understanding is that they are given as a range because that is the range required by EPA for certification. If you get your stove certified, then you are in that range by definition. Then you can skip the separate testing phase (with its extra expense) for determining your product's actual efficiency. Commented Dec 13, 2017 at 13:51

I would present a different analysis to jamesqf, because his does not take into consideration the latent heat of evaporation, or at least does not mention it.

If we're assuming a regular hardwood used typically for firewood (here in the UK at least), which has a net calorific value of 4.06 kwh/kg at a moisture level of 20%, if this is burnt in a stove with an efficiency level of 70% then it is safe to calculate that it would take a 0.352kg piece of firewood to produce 1kWh of electricity.

If the moisture content of the firewood was 10% then it would take a 0.307kg piece of firewood to produce 1kWh of electricity.

In calculating these figures I have used a net calorific value for dry firewood of 5,251. However, and this is a big HOWEVER, this calculation assumes that the 1kWh produced can be perfectly turned into electricity by a thermoelectric generator. This is a naive assumption, hence I would like to present a probable calculation about how much firewood we would need to produce 1kWh of electricity, because that's really what the question is asking. The problem however is that currently I know too little about this process, let me do some research and I may be able to provide a better answer in the not too distant future.

  • Can you provide a source or two for your answers?
    – LShaver
    Commented Apr 27, 2019 at 15:35
  • @LShaver I don't work in this industry any longer however I seem to remember that the calorific values quoted were an average from many credible sources. Commented Apr 28, 2019 at 15:27

Efficiency numbers are, unfortunately, misused/abused by companies that manufacture and sell wood heaters.

"Combustion Efficiency" is not actually a direct measure of how much heat is produced by burning wood. It's a direct measure of "how completely (or cleanly) the wood has burned". A stove that has an 89.1% combustion efficiency burns 89.1% of all the available fuel (under test conditions). 10.9% of the fuel either remains behind in the firebox (usually in the form to charcoal) or escapes up the chimney (usually in the form of carbon monoxide). In any case, it's not a measure of heat but, because the number tends to be higher than the other types of efficiency ratings, marketing departments love to use it.

If you want to know how much heat is produced, then you need to look at either "Net Efficiency" or "Gross Efficiency" ratings.

"Net Efficiency" ratings describe how much heat you can theoretically extract from firewood using the appliance in question. Contrary to what most people understand by 'net' it does not factor in how much of the heat is wasted vapourising the moisture in the wood.

"Gross Efficiency" ratings describe how much heat you actually extract from firewood using the appliance. The energy taken to vapourise moisture has been taken into consideration.

Wood heaters in Europe are tested according to EN standards, and those tests use firewood that has an 18.5% moisture content. If your heater is described as having a Net Efficiency, and burns regular (dry) firewood, you can convert that to a Gross Efficiency by multiplying by 0.91 (e.g. Net 81% * 0.91 = Gross 73.7%).

The 0.91 conversion number comes from 'Table E4: Efficiency conversion factors' in the 'The [UK] Government’s Standard Assessment Procedure for Energy Rating of Dwellings'. The latest version is SAP2012. Direct link to PDF. Table E4 is on page 64.

Gross Efficiency is what you are after to do the math. Gross Efficiency is useful. Once you have a Gross Efficiency figure, the rest is easy.

Each kilogram of (EN Standard) dry wood contains 5.14 kWh of "potential heat" so 1 kWh of energy is contained in (1/5.14=) 0.195 kg of wood.

If your wood heater's Gross Efficiency was (say) 70% then the actual amount of wood that you would need to produce 1 kWh of heat is (0.195/0.70=) 0.279 kg.

tl;dr: 1 kWh of heat can be extracted from 0.195/GrossEfficiency kg of dry firewood.

PS: In many Western counties, wood heaters have to undergo independent testing before they can be sold to ensure that they comply with local regulations. The test results may be commercial-in-confidence, but reputable wood heater manufacturers/vendors — with nothing to hide — should be willing to provide a copy of either the entire report or an extract. This document is almost always more useful than the glossy brochure the Sales department provides as it contains all of the efficiency ratings that were measured/calculated.

PPS: I am not trying to suggest that Combustion Efficiency figures are useless. They are useful if what you are primarily concerned about are emissions. And whilst a "complete burn" will release more heat than an "incomplete burn", if your heater needs to draw in twice as much air to do so then you are going to lose a larger fraction of heat through the flue — and your house may actually end up colder. It's called the "Efficiency Paradox". Some modern heaters — with multi-stage burns, convoluted gas paths, and low flue temperatures — resolve the paradox and feature very good combustion and gross efficiencies.

PPPS: If, for whatever reason, you don't want to use the EN's 5.14kWh value, you can go to the Phyllis2 database, check the box next to "untreated wood" (or expand the category and select one or more entries as a subset), then click the button labelled "Average of Selected". Results will be shown on the right. Scroll down to the Calorific Values section. Look at the row marked "Net calorific value (LHV)" and the column marked "Mean" (it's the only bold one). That gives you MJ/kg. Divide by 3.6 to get kWh/kg. At the time of writing, the average for all untreated wood was 18.88MJ/kg, so (18.88/3.6=) 5.24kWh/kg. As more and more samples are added to the database, expect the values to refine over time.

  • +1. Can you provide a source or two for your numbers?
    – LShaver
    Commented Apr 27, 2019 at 15:35
  • 1
    The 0.91 net-to-gross factor is now referenced. The 5.14kWh/kg figure (although widely used in places like soliftec.com/fuelproperties.htm) is behind standards paywalls, so I wasn't able to get a link to that. Added a postscript showing how folks can access such a figure for themselves using a live database.
    – Tim
    Commented Apr 27, 2019 at 22:39

The link below gives you net calorific values and energy densities for different woods with different moisture content.


  • 1
    This is only one component of the answer and the OP is still left with not knowing how to calculate. And the link is incorrect and only for Finnish wood. And link-only answers are unwanted on SE sites because of possible link rot.
    – user2451
    Commented Dec 13, 2017 at 14:04

Let’s look at this anther way.

  • Your house will have some insulation
  • There will be heat input from the people and electrical equipment in the house.
  • There will be heat lost from the fireplace including when the stove is not in use.

Therefore having a fireplace will increase the number of hours a year you need to heat compared to electrical heating.

So the questions as asked does not lead to a useful result, in some homes the wood burned will have a negative effect on the heat needed, assuming that the fireplace would not otherwise be there.

  • 4
    Might be useful for those considering installing a fireplace, but this definitely doesn't answer the question! Commented Jul 13, 2015 at 1:20
  • Note also that any practical wood heating would not use a fireplace, but an efficient wood stove, or fireplace insert (for those of us who bought houses with decorative fireplaces already installed). From personal experience, I can definitely state that a wood-burning fireplace insert does not cause significant heat loss, as the only times my furnace comes on is when I'm away for a day, or on cold nights when I don't want to get up to chuck in another log or two at 4 am.
    – jamesqf
    Commented Jul 14, 2015 at 21:20
  • 1
    @jamesqf, JUST having cold air coming down the chimney due to convection currents, will make any wood burning stove/inset cold. This has been shown to be an issue in VERY well insulated homes in the UK. Commented Jul 15, 2015 at 16:35
  • @Ian Ringrose: I think I'll put my practical experience first. Now it may be true that there is a bit of overall heat loss through the chimney that would make the house a bit colder when the stove is not burning, but if you heat with wood, there are very few such times. Chimney losses would be part of overall stove efficiency numbers.
    – jamesqf
    Commented Jul 15, 2015 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.