Water Heating Calculator for Time, Energy, and Power
Only input whole numbers, do not use a comma or point.
The calculators support Celcius/Centigrade, Fahrenheit, Watts (w), Kilowatts (Kw), Btuh, Joule, British termal unit (Btu), liter, gallon, kg, lb, cubic inch, cubic foot etc.
The calculators assume 100% efficiency and no loss of energy during the heating process.
Improved calculators that support comma, dot, and efficiency, are available here.
Water Heating Time Calculator
This calculator tells you how long it takes to heat water from start to end temperature with a given heating power.
Water Heating Energy Calculator
This calculator tells how much energy will be consumed to heat the water from the start to end temperature.
Water Heating Power Calculator
This calulator tells you how much minimum heating power is required to heat the water within a specified amount of time.
Improved calculators that support comma, dot, and efficiency, are available here.
Posted on
Last updated on
Comments
is there a formula to calculate the temperature required for 5 gallons of water that will increase the temperature of a stainless steel 15.5 gallon tank from 46F to 161F. The volume in gallons (5), ambient temperature (46F) and target temperature (161) are variables that will change.
Very useful! Thanks:-)
great calculators 🙂
Thanks! This was super helpful in calculating the power needed to produce my hot water via solar photovoltaics. For less than the price of a new heater tank, I now have free hot water. Sweet!
Very nice and handy, Thanks !
How do I calculate btu size for heating a swimming pool?
I’m looking at a 50,000 btu boiler with a heat exchanger to heat a 33,000 gallon pool to 90 degrees f.
I’ve run then calculation through the formula above, and it’s coming up with a 600,000 btu boiler to raise the water temp (68f) in 5 hours to 90f, which seems to be a really high number?
Hi “New pool, cold water !”
I have done the calculations in the calculator above for heating 33’000 US gallons from 68F to 90F in 5h (300min) and got a result of approximately 1’200’000 btuh.
I have done the calculation again manually in an Excel spreadsheet and got the same result.
What formula did you use to calculate the water heating power?
I am looking for how much energy is need to maintain water temperature. This is for a hot tub, I do not need to increase or decrease the temperature. I just want to maintain the temperature. 250 us gallons at 105f with moderate insulation… would 100 watts DC heater maintain this temperature?
Hi Willaim,
that is impossible to calculate with the information given. It very much depends on the surrounding temperature, the heat conductivity of your hot tub material, and the shape of the tub.
But I have an idea. If you can test how fast the water cools down in your tub, then you can calculate how much heating power is required to prevent that from happening. Let’s make a calculation example for a tub with 250 us gallons that cools down from 105f to 103f in 2 hours. Using the “Water Heating Power Calculator” above (250 us gallons, start temperature 103f, end temperature 105f, 120min) tells us that a heating power of 611 watts is required. If you use a 611 watts heater in this example, then it will be about enough to reverse this cool-down from 105f to 103f by heating the water back up from 103f to 105f during the same period (realistically, of course, the water does not cool down, but stays at the same temperature).
That’s just an example, I have no idea how fast a tub cools down. Also, you should have some extra power available, just to be sure. And for testing/sampling the cool-down, measure a larger drop in temperature, such as 10f, else the result will be unreliable (reading 1f or 2f changes off a small thermometer may be very inaccurate). And as always, double check with some other source and my help is without any guarantee or similar.
Ace. Thanks 🙂
Thanks. Very useful 🙂
Possibly very useful.
Would this formula be correct to use for a heating loop?
I have a heating loop with a average deltaT of 1°C over a period of 22hours and a rated flow of 27,5 l/s.
Using this calculator i get 1381 kWh
is this the formula used?
volume in litres x 4 x temperature rise in degrees centigrade / 3412
Thanks!!! I am tasked to maintain a 110L aquarium at 22-24C outdoors in Texas. Your calculators are a big help in figuring the size of the heat sump.
Great help to calculate water flow and energy consumption!
Problems with limit calculations: It complains if the end temp is 212°F or even 211°F, saying it must be less than boiling. It complains when volume is 0.25 US gallon, saying volume must be greater than zero.
Hi John,
it takes the same amount of energy to heat water from 48 degrees to 52 degrees as it takes to heat water from 58 degrees to 62. But when the state of water changes from solid to fluid (e.g. -2°C to + 2°C) or from fluid to gas (e.g. 98°C to 102°C) this does not hold true any more. It would be more complicated to build calculators that can handle that and I have not done so. This is why the calculators complain in these situations.
The calculators cannot handle points or commans, only whole numbers. So they may complain when you input commas or points.
To determine BTU lost can I use the “Water Heating Energy Calculator” only backwards? Start temp as final and end temp as start.
Hi Larry, the energy that is used to heat water from 40 degrees to 60 is the same amount that is lost when the water cools down from 60 to 40 (only changed sign).
I’m trying to calculate the amount of BTU output my homemade outdoor wok burner produces with its jet burner. I heated 1pint (1 lb) of water starting at 75F degrees to 158F degrees in 60 seconds. It went from 75F to 212F(boiling) in one minute thirty three seconds. Yes, I know it is smoking fast.
I’m struggling to calculate how many BTUs that is per hour. I’ve come up with two separate answers: 41,000 BTUs/Hour and 298,800 BTUs/Hour. I’m not confident either are correct.
All I know is my wok is very very hot and almost unusable. Can you help me find the answer given the information above?
Reuben
Houston, TX (essentially sea level if that matters)
rmccain2030@gmail.com
Thanks for the Calcs! How might this apply to a stratified water tank? i.e. – 9600 gallons in an 8′ x 8’x 20′ tall square tank. Perhaps with the top 1/3 of the tank heated to 180 deg f, the middle third heated to ~120deg f, and the lower 1/3 left as-is….
Hi,
Howmuch joule are needed to boil water.
So from going from liquid to gas.
I already know howmuch +1°C needs, but now still need to know howmuch energie is needed for vaporize, So I can calculate howmuch energie it takes when , I put it in hot liquid metal.
Thank you! I’ve been looking for a simple way to ballpark the size of a solar thermal hot water system – and this makes it dead simple!
This is the best set of calculators on the internet 🙂
Thanks again!
hello, thanks for this,
is this calculation valid only when submerging the heating device in the water, or also when the water is circulating through a heating pipe using a pump?
Wouldn’t it be faster in the latter case, because the water flow would exchange heat more efficiently than just by convection?
I’m trying to figure out what would it take to heat my pool using a heating pipe after the filtration pump.
Thanks
Hi Giulio
I do not think it makes much of a difference in practice.
As for efficiency, an electrical water heater can convert electricity with almost 100% efficiency to heat that can be transferred to the water. The insulation of where the heating device sits and whether any heat can escape into something other than the water (like a wall, concrete etc.) is more important for efficiency.
As for effectiveness, a large enough pipe with a pump can probably in theory transfer more heat per second (meaning power, e.g. watts) to water. When using a simple coil submerged in water, then it will start to heat the water just around it. Since water gets lighter when heated it will start to ascend around the coil and cold water will flow to the coil from below. This way a circulation develops in the water container given the shape of the water container permits it. I think of it this way: the power limit of heating coil simply submerged in water is reached when the water starts to boil around the coil while the water in other places is still much colder. Is that realistic?
Please remember, that the calculators above assume 100% conversion efficiency. For an electrical heater that can be a good assumption, but not for a gas heater.
Also, the calculator is only correct in the theoretical case that while the water is being heated, no heat escapes from the water to the environment (air or walls etc.). For that reason, these values are the minimum and some margin should be added to them. How much depends on the shape of your pool, the quality of the insulation of the pool, the difference of the desired water temperature to the environment and how long you are willing to wait until it is heated.
Nice calculator buttt…… you’ve got the abbreviations for the metric units incorrect. A Watt is named after James Watt hence the abbreviation is a capital W.
A thousand of something is a “kilo” for example a kilogram. In turn the abbreviation, for one thousand, is k.
Bringing these two together, a kilowatt is a “kW”, not as you have, “Kw”. Over an hour it would be “kWh”.
Hello Antonio,
Appreciate the calculators but i was wondering if you could tell me the exact equations used for the “Water Heating Energy Calculator” or give some sort of excel sheet to double check it. I am guessing its q=m*cp*dt but when I do the calcs by hand, i get something off so i would like to know my mistake
How much heat energy in joules is necessary to raise the temperature Of 7000 kg (7 M3) of water from 20 °C to 80 °C?
1 litre of water is 1kg
1 meter cube = 1000 Ltrs
M = 7 meter cube = 7000 Ltrs or 7000 Kg
The heat capacity Cp of water is 4.186kJ/kg-C
ΔT = 80-20 = 60 C
So, the energy required to raise the temperature of 7000 kg of water from 20C to 80C is:
Energy E = m•Cp•ΔT = 7000 x 4.186 x 60 = 1758120 kJ
or = 488 .36 KW
Power = Energy / time
And this is 488.36 Kilo Watts of power (since 1J/s = 1W)
1758120 kilojoule/hour = 488.36666667 kilowatt hour
488 kwh for 1 hour
For 4 hours 488/4 = 122kw
For 8 hours 488/8 = 61kw
This calculator is great. I’m curious, though. I just recently learned about hybrid heat pump water heaters for the home that have much higher heating efficiency than standard electric. They are saying the modern heat pumps can give you 3.8 watts worth of hot water (or something close to that) for each watt that the heat pump uses. It would be really cool to see a calculator that takes this new technology into account.
This is great. I am assuming calculations take into account the specific heat capacity of water? I am wondering though, doesn’t heat capacity increase with temperature, potentially throwing off the calculation?
@Anna
The calculators use the specific heat capacity of water of 4186 J/kg/°C (Joules per kilogram per degree Celsius). The heat capacity is largely constant in the temperature range that the calculators work (34-210°F or 1-99°C). It is the way it is taught at school. For practical purposes, it should be precise enough. For rocket science one might want to calcualte it more accurately.
Great calculator. Straight to the point. Thank you
I entered 688 litres start temperature 29 C end temp 37 C with 1600 watts of power. The answer was 4 hours.
If i change the power to 1.6kw . The answer is 6 hours 24 minutes.
An I missing something here?
The calculators cannot deal with dots or comma. It makes 1 out of 1.6
I have a system with 2500 lit of water.
The water is recirculated from the system to a storage tank, using a pump 30.000 lit/h.
The tank is heated by steam using heat exchanger at 74 degrees of Celsius.
The water returns from system to tank at 70 degrees of Celsius.
My question is how I can calculate the heat power required to maintain the temperature forward and return at the same level (eg 74 degrees of Celsius).
Thank you
I need to heat a swimming pool that holds 500 m3 of water
how long will take to raise the temperature from 20 C to 30 C ?
To achieve this I will use 4 exterior stainless steel tanks of 72 m3 each, heated with 1500 watts glow plugs.
any comments or ideas? … please?
excellent piece of kit. I have saved the page to my home screen. 🙂
As another calculator for us math dummies could be total kw x cost of electrical supply
eg 250kw x .26c = $65
At any rate a really good calculator q
Very good and saved me a lot of time and effort, some useful questions and answers to. 🙂
Fluid behaviour and characteristics change with the ambient. we used to apply charts upon empirical results (at the laboratory) for accurate calculations. Formulas and equations are more for theoretical calculation but never in reality in engineering terms.
thanks for the calculations I seem to have lost the place I used to use
do you have the calculation for heat loss
ie) how long it will take for 1000L to drop from 20 deg to 17 deg if ambiant temperature is 4 deg and or only had 300w heater etc
Thanks for this calculator. I use it all the time to figure things out for various DIY heating ideas. It is very usefull and nicely straightforward. My only wish is that there was a way to say the calculations in metric instead of always defaulting to imperial.
Other than that a very handy tool and thank you for creating and hosting it.
I have a 4000 gallon exterior ornamental fish pond that I would like to keep above freezing. The pond stays completely circulated, so there would be heat losses for the circulation and for the ambient air temperature. The calculator says to raise 4000 gallons by 5 degrees with a 1500 watt heater, it takes 32 hours. Any way to take into account the many variables that a pond is subject too and decide whether or not a 1500 watt heater will do anything overall for the pond, or just keep the spot where it is from freezing? I am in Virginia. Thanks
For those into mental maths and the metric system, energy required in Wh is equal to liters times degree change, plus a sixth. Eg. 20L, up 30deg, requires 700Wh.
an extremely well insulated camper R24 which is probably impossible…. how many gallons of water and btus does it take to keep the camper above 50 degrees for 8 hours? Everyone has these super tiny wood stoves that only last maybe 4 hours….How about a twenty gallon boiler and the heated water could be the secondary heater? Or tiny pellet stove run of batteries to keep feeding the pellet stove during the night…
how many gallons of 174 deg f water does it take to heat
90 gallons of 38 deg f water to 82 deg f
Is it possible to see the equations used by the calculators? Just interested in the math behind the numbers
Please see the improved water heating time calculator linked in the article above.
I need to heat 50 kg of water from 60c to 90 c every minute I should take heater of what kw it is close loop circulation
I used 33°F as the start temp, and I get an alert box saying starting temp must be above freezing. Last I checked it was 32°F?
If i need to calculate for coold water i receveid errors.
The calculators use a simple constant for how much energy is needed to heat water by 1 degree (4168 joules/kg/°C). In reality, the amount of energy is slightly different at different starting temperatures, but the constant is accurate enough for practical purposes. The problem is that this constant does not work well around freezing. The result would be inaccurate.
Very helpful calculator – It has been my reliable ‘go to’ water heating calculator for some time. Thank you very much!
GREAT calculator! After freaking out that my new hot tub was not hearing properly (8hrs of heating). The calculator helped me relax given it should that 19hr to hear 576 gal/50deg to100deg with 4kW heater. I wonder how much heating I could get running pumps full blast. As in the tiny amount of her transfer due to circulation..LOL
Very easy way to calculate energy vs time.
I need 208 gpm @ 45 kw
how to calculate the tank capacity and temperature rise.
Thanks, this is very nice and helpful.
May I get the equations that are used to perform the calculations or the references please?
Thanks for these extremely useful tools . I refer to them all the time for different calculations for various heating ideas. They let me know straight away wether they are viabe or not.
Much appreciation for your efforts in helping others.
Thanks!
I left school in 1986, and even though physics was one of my top subjects – after 35 years of not doing anything like this – I am a bit rusty 🙂
This is exactly what I was looking for, cheers!
Hi just wondering if any one can answer my question?
I am starting a new project and here are my requirements.
I need to maintain 1200 gallons of water in a tank at 170F
I will be using a closed loop boiler at 400MBH input 96%
With a exterior plate heat exchanger single pass with 2” in/out connection rated at 150PSI 230F.
Boiler exit temp to heat exchanger 190F
Tank water temp.at initial start ambient 70F entering heat exchanger.circulating pump for tank loop 3/4Hp at 35GPM
Boiler has built in circulator maximum length of pipe 2” to tank one way is 30’
Will this work to maintain my tank temperature?
The tank holds water with some cleaning agents for steel plates processing and degresing
John
Math checks out. Thanks for providing a simple tool. Including the formulas and values used in the calculation would be a bonus.
In the first calculator I used 2400 watts and then 2.4 KW. I should have gotten the same result, but I did not.
75 cubic meters in the pool. Pool equipped with semi-solar cover, keeping heat in. Have heat pump 13,0 kW. wish to heat the water by just 4 degrees C, ie 19 t0 23C. Using geothermic drilling, how far down does driloling need to be?
Thanks
Thanks for this. Another handy calculator would be one that takes water volume, start temperature, power and time and outputs end temperature. That is, “how much can I warm my water given that I can run my water heater for a certain number of hours a day”.
this is handy, but I keep getting conflicting answers from other sources. I am trying to heat up a large water tank using diversion from my solar panels. I can feed direct DC into the heating elements using contactors, or I can power the same heating elements on ac using just the extra power after topping off batteries (turn elements off or on based on SOC of battery bank.
Other sources have said that feeding the heating elements directly with dc heats faster as the power is continuous where as powering it via an inverter means that it only gets 50% of the power due to the sign wave dropping voltage as it crosses over 0° at 60 htz. (full power at either end of the sine wave at 180°)
Is the watts figures for AC., or DC?
edit: I have 8Kw of solar panels and only need 3.5kw in the summer months. the other 4.5kw are needed during the rainy season to run a splitpack to dry out the cabin and during the winter due to lower solar production.
Thanks, Very useful and great calculater
A very nice easy to use calculator.
Awesome to use.
Thanks
Antonio. This is a marvellous tool, thank you so much.
I am testing a variety domestic thermal solar options to heat out house.This is so we no longer need dead trees to be cut down for fire wood to heat the house and they can remain standing for birds to nest in.
Many of us back yard inventors are not so good with numbers and this calculator makes it possible for me to compare, evacuated tube, parabolic trough, flat plate, and polly pipe, collectors very easily.
By making this available you have made a great contribution to preserving the planet.
Best Wishes Bruce G. Perth Australia
I have a gaservecation boiler (wood) with a 4000ltr acumerlator tank i also have a 1500 ltr lpg tank which at the movement serves 1 fire in the lounge decor really now the wood is getting a bit of a choir and i want to install a lpg boiler to heat the tank (4000ltr) asap and like batch heat i have a 30kw boiler the price i pay per litre gas is 43p how much would it cost to heat my tank from 30c to 85c
How about adding < 1 gal or liter… ounces would be preferred. If I want to use 16oz, I could not enter 0.125 or 0.25 for a larger volume… I was just trying to do a cup of tea, which depending on where you are is anywhere from 6 to 12oz on average.
I wanted to have the modification be simple and as broad reaching as possible.
The calculator does not support numbers with comma or dot. Please use the improved calculators linked in the article above.
There appears to be rounding errors in the calculator. Using the ‘Water Heating Time Calculator’, look at what happens when you go from parameters of 50/100/101.8/563 to 50/100/102/563. It jumps from 14 minutes to 27 minutes, showing the lack of precision…
The calculator does not support numbers with comma or dot. Please use the improved calculators linked in the article above.
Great resource!
Quick question. I used your formula and my answer differed slightly from the one your online calculator gave me.
I think you are rounding your answers UP to the nearest minute, rather than to the closest minute.
Is that the case??
Rounding up would make sense to do for a number of reasons. I guess most importantly that the actually time to heat the water would be longer than your formula suggests due to heat lost to the surrounding environment.
Please try the improved calculators linked in the article above.
Very nice formula. I used it to calculate how long it’d take to boil the fish in my aquarium if my heater gets stuck on.
I have a 18hp/12 kw engine on my boat. It provides 2.1 kw of electrical energy and is used to heat a 50gal (uk) calorifier from 10c to 65c. By my calculations the electrical generation and hot water is consuming about 9.5kw of engine power, am I correct in my thinking.
Thank you.
A great help. I wanted to find the best size water element to heat my 250 litre hot water service to 58 degrees from a 6.6kW PV array. I settled on 1800 Watt.
Very many thanks for this wonderful ‘tool’ that has given me something ‘concrete’ on which to argue my case.
My new swimspa is struggling in trying to maintain 37 degrees c. My plan was to use the spa at 37 degrees for the older members of the family, and as the day went on to allow it to cool for younger members to swim in a cooler temperature. Pleasingly, and even in the depths of winter there is very little loss during the day provided the spa is kept covered. Overnight it consumes 21kW of lecky and sometimes will increase the temp from say 35 degrees (loss during the day when used) to 36 or 36.5 – sometimes not even that.
Using any of the three calculations above for heating 7200 litres with the inline factory fitted 3kWh heater gets me no where near what it should.
Desperate for some additional thoughts please, the supplier and manufacturer haven’t any thoughts and are so far passing the buck!
Other owners of similar Swimspa say do not have a problem – I suspect they have more pennies than I do!
TIA
Nigel
Hello,
Why must the end temperature be less than 212f (boiling)?
Thank you,
Bill
Because it the amount of energy needed to increase the water temperature by 1 degree is fairly constant for liquid water. But that does not hold true for changes from frozen to liquid or from liquid to steam. The calculator cannot handle these more difficult to calculate cases.
Fabulous calculator. I’ve been using it for several years now comparing different Time, Energy, Power scenarios. For example: comparing the energy saved by using a kettle to heat water for washing dishes compared to running the hot tap and wasting 4 liters of water before any hot water comes out. I only need one liter!
Thank you!
I want to keep a gallon of water (33 degrees F) from freezing at 15 degrees F ( or x degrees below 0 deg C)
Thanks, great that you would make this and I can find it from the UK – a real time saver
Thanks for adding metric
i have a combi boiler in the UK which has poor performance in winter as the input water is around 7 C. Just by using my old attic tank as a source reservoir and using attic space ambient ( varies from 40C in summer to 15 C winter). Plugging those numbers in realise around 50% cut in energy use. I needed confirmation my project was worth the hassle. Expected gas prices in the future £2000 / year. Can even put more time and money into it – Many thanks
I am looking at heating water to 160f from 130f but has a continuous draw of 30 deg.. I am looking to maintain at least 140f. I can heat any amount of water just need to no where to start. Can you help?
Hello! Before seeing your excellent calculators online, I wrote my own algorithm for calculating how many hours is required to heat 1032.8 U.S. gallons of water from 43 deg F to 84 deg F with 11 kW of power. Your calculator gives 9 hr 25 min and my calculator gives 9 hr 24 min — so the model and the maths seem to be the same. My problem is that when I actually heat up the water in my very large hot-tub from 43 deg F to 84 deg F using g an 11 kW heater, it takes far longer than our calculated times. Our calculations show that the heating rate is 4.3 deg F/hr, whereas in actuality the “experiment” shows that it is 2.1 deg F/hr. So there appears to be only half the heating power going into the water. The tub is a cylindrical cedar tub in the basement within a surrounding temperature of about 40 deg. Does this make any sense to you–that half the heat generated is lost? Best regards, Robert
HI WILLIAM:
I just read your reply to William on April 24th, 2018. That’s exactly what I need to do—and I am doing it as we speak! Since my plot of water temperature vs. time (during heating) is a straight line, I am assuming that this is telling me that most of the heat loss from the tub is due to convection and not to radiation (would you agree?) So I am going to see how much the temperature of the water drops in cooling by about 5 deg F and will then calculate how many Watts would be needed to reverse this cooling effect–as you suggested. Stay tuned…but I doubt if it’ll be enough to explain the apparent loss of heat I am experiencing according to the calculations. It’s a pleasure discussing this with you! Robert
Hello again, William. So I did your suggested experiment letting my tub cool. In two hours the water temperature changed by less than 1 deg F–so the loss of heat through cooling cannot explain the roughhly factor of two discrepancy between theory and experiment. Any more ideas…..?
Robert
Hi Robert
It may be that much more energy is needed to heat the pool water up by 1 degree than for preventing it from dropping 1 degree (in the same amount of time). Possible explanation: when first heating it up, all the surroundings of the pool also need to be heated up and that consumes energy.
That’s possibly the explanation. So it seems that half the kW being pumped into the water are raising the temperature of the plumbing, cedar walls of the tub, and cold air surrounding the tub (about 40 deg F yesterday). Yay! you solved my issue! When I put in an average heat loss correction factor of 100% my calculated temp rise per hour falls to 2.17 deg F/hr–which is exactly what I measure!
I’m trying to calculate the heater size, in watts, to keep an 1100 gallon water storage tank from freezing over the winter. The tank will likelihood be holding 350 to 850 gallons.
Daytime temps are usually in the mid-30°s, rarely below 25°F for more than 72 hours. Nighttime temps average in the high-teens but can be 0°F on occasion.
TIA
Antonio
Well played. Neat calculator. And a number of helpful responses.