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.
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
1Kg of Water needs 4200 Joules of energy to raise its temp by 1 deg C
1L of water has a weight of 1Kg
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
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
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....
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.
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
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
= 35 kW/hr
So, it takes 35 kW one hour to raise 1000 litres 20 degrees C
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