# Energy required to heat tank - help request

• posted on October 18, 2003, 5:03 pm
can someone tell me the calculation I need to work out how many kWhs are needed to raise say 1000 litres of water by say 20 deg C, please?
With thanks,
Nick

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• posted on October 18, 2003, 5:31 pm

needed to raise

This is an American site, but it give a calculator called Watt-U-Use which you can download and try.
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• posted on October 18, 2003, 5:36 pm

Or this site gives you a table of all types of appliances that you have in the house. You add the quantity of each ones you have to the list and click the calculate button at the bottom to read your total energy use if all are on at the same time.
http://www.majorservices.co.uk/consult/calculators/gen_calc.htm
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• posted on October 18, 2003, 6:31 pm

Thanks for the replies - will take a look
Nick

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• posted on October 18, 2003, 5:43 pm
froggers wrote:

Need help with your homework ;-) (no offence intended - but this sort of question is rather too open ended to give any useful answer to because there are too many unknowns)
The quick mathematical answer would be:
Specific heat capacity of water 4200 J/kg/deg c Density of water 1000kg/m^3
So:
1Kg of Water needs 4200 Joules of energy to raise its temp by 1 deg C 1L of water has a weight of 1Kg
Hence:
Total energy required = 1000 x 4200 x 20 = 84,000,000 Joules
1 Watt = 1 J / sec therefore 1 kWhr in Joules = 1000 x secs per hour 1 kWhr is equal to 1000 x 3600 = 3,600,000 J
so 84,000,000 / 3,600,000 = 23.33 kWhs of energy
Notes:
In the real world this answer is going to be wrong for various reasons:
You did not specify details of the tank (what is it made of - how much mass does it have etc) - that will require extra energy to heat as well.
You did not specify the heating method - which may affect the result (i.e. an immersion heater will have some heat capacity of its own).
You did not state how well lagged the tank was and hence its rate of heat loss to the surroundings - that will require extra heat input to compensate.
You did not specify the initial temperature of the water, hence one can not estimate the rate of heat loss at any point in the process. The rate of loss (assuming the insulation values remain constant) will be proportional to the temperature difference between the water and the ambient temperature - hence it varies and is unknown. Also if the initial temp was below 0 or above 80 then answer is very wrong due to the state changes required (solid to liquid or liquid to gas)
You did not state if the tank was open or closed and hence if a change in pressure needs to be taken into account, if there will be convection losses from the top surface of the water etc.
And so on....
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Cheers,

John.

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• posted on October 18, 2003, 5:44 pm

needed to raise

I'll have a go: Specific heat capacity of water at around room-temperature = 4200 J/(kg.k) Mass of 1 litre of water = 1kg
Energy required = specific heat capacity x mass x temperature rise = 4200 x 1000 x 20 = 84000000 J = 23.3 kWh (because 1 kWh = 3600000 J)
That sounds about right. This assumes that the tank is fitted with perfect insulation, which I am sure you can buy from Screwfix. If you don't have perfect insulation, it will take a bit longer.

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• posted on October 18, 2003, 5:57 pm

The specific heat capacity of water is 1 cal/g/C (this must be the definition of 1 calorie?).
1 kWh = 3,600,000 J = 860,000 Cal.
I presume you know how to turn litres of water into kg?
(time was I could have done all this in my head, but now I had to look it up, pity about the mixed units, I'm sure you'll manage)
tim

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• posted on October 19, 2003, 8:59 am

Nick,
Third time I've posted this, so ISP server must be US.....
Here it is:
No. of kW Quantity of water in litres X Temp rise degrees C X Specific heat of water ________________________________________________________
Number of seconds in an hour
1000 X 20 X 4.2 _____________ 3600
= 35 kW/hr
So, it takes 35 kW one hour to raise 1000 litres 20 degrees C
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• posted on October 19, 2003, 3:10 pm
IMM wrote:

You seem to have come late to this party... and brought the wrong answer with you as a guest! ;-)
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Cheers,

John.

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<%-name%>
• posted on October 19, 2003, 3:20 pm

Correct equation; punched in the figures too quick. Should be 23.33 kW
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• posted on October 21, 2003, 7:31 am
Thanks to all who have taken the time to reply - I had checked the results / workings myself so no problem with that wrong answer, and now I can do the other similar calculations.
Thanks again
Nick

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• posted on October 21, 2003, 10:05 am

/ workings

What might they be? :)
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