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One question which comes up time and again is “How many kW do I need to heat up my tank?”

Or phrased a different way, “How long is it going to take my ? litres of solution to raise ? °C using my ? kW heater?”

If we can calculate the** volume of water** and the **required temperature rise**, we can answer these questions using the following formula.

It is used to calculate the power of heating element needed to heat a specific volume of water by a given temperature rise in 1 hour.

**volume in litres x 4 x temperature rise in degrees centigrade / 3412**

(4 being a factor and 3412 being a given constant)

for example 100 litres of water, to be heated from 20**ºC** to 50**ºC**, giving a temperature rise of 30**ºC** would give –

100 x 4 x 30 / 3412 = 3.52

meaning that the water would be heated in 1 hour by 3.5kW of applied heat.

Also we can use this information to extrapolate both ways. To heat the same water volume in half the time (30 minutes) would need twice the heating power, ie, 7kW.

Converesely, if we only use half the heating power, 1.75kW, it will take twice as long to heat up to desired temperature, ie, 2 hours.

If we only have a 1kW element available, we will expect a heat up time circa 3.5 hours.

Also we can use this formula as the basis of similar calculations for heating oil. Generally speaking, oil heats up in about half the time of water, due to its viscosity & density. However, oil requires a much lower watts density element than water, as described here in the “How to choose an oil heater” article.

Another variant of this formula, given here at the excellent website Sciencing.com gives the following varaint of the formula & subsequent explanation-

*Pt = (4.2 × L × T ) ÷ 3600*

*Calculate Kilowatt-Hours*

*Calculate the kilowatt-hours (kWh) required to heat the water using the following formula: Pt = (4.2 × L × T ) ÷ 3600. Pt is the power used to heat the water, in kWh. L is the number of liters of water that is being heated and T is the difference in temperature from what you started with, listed in degrees Celsius.*

*Solve for Thermal Power*

*Substitute in the appropriate numbers into the equation. So imagine you are heating 20 liters of water from 20 degrees to 100 degrees. Your formula would then look like this: Pt = (4.2 × 20 × (100-20)) ÷ 3600, or Pt = 1.867*

*Divide by Heater Element Rating*

*Calculate the amount of time it takes to heat the water by dividing the power used to heat the water, which was determined to be 1.867 with the heater element rating, listed in kW. So if your heater element rating was 3.6 kW, your equation would look like this: heating time = 1.867 ÷ 3.6, or heating time =0.52 hours. Therefore, it would take 0.52 hours to heat 20 liters of water, with an element with a rating of 3.6 kW.*

Which made better sense in my little brain when I put a multiplication sign between P and t, allowing 30+ year old math class memories to clarify that if you move the Power (P) or the Hour (t) to the other side of the equals symbol, we gotta divide by that number also. “Change the side, change the sign” Thanks Mr Phipps, some of it actually stuck, hope you are still above ground, happy & healthy.

*P x t = (4.2 × L × T ) ÷ 3600*

…which doesn’t usually “show” as t = 1 hour, as in kW(1)h.

Hope you found this useful.

Any feedback, suggestions, improvements, etc, PLEASE COMMENT, I promise to read ’em.

mist42nz says

thanks for such a concise answer. I’m calculating how much kW I will need for a solar Hot Water system. Such systems get 4 – 5 hours of sun a day, and have to heat around 300litres. A tube is rated around 60W, but only get about half that on poor days.

Using your information it’ll be a snap to calculate.

ajay says

Well your answer is not helpful please try to clean the two things..1) what is 4 and what is 3412

John Doe says

I have no idea why you would want to first multiple by four and then divide by 3412 instead of just dividing by 853.

Ahmad Morsal says

is the calculation heat up time using immersion heater and catridge heater is same

France says

excuse me , may i ask you how to calculate every things concerning heat?

M.Farrukh says

Excellent explanation thanks

Petson Banda says

4 is which factor? And 3412 is which constant?

Nasir Mehmood says

we need heat pump heater calculations. we will be grateful.

Peter Hämmerl says

I use another formula, resulting about the same:

To heat 1 Liter of water 1 Kelvin needs 1.16Wh.

Nerven Maistry says

how do i contact jamie by email?

anacy says

why 3412?

Ian Demicoli says

Thank you very helpful