The specific heat capacity of water is 4.18 J g

^{−1}K

^{−1}. The tank has a capacity of 115 litres. A litre of water is equal to a kilogram, so to raise the temperature of the tank by 1° C takes:

^{−1}

A Joule is equal to 2.7778×10

^{−7}kWh, which gives:

^{−7}= 0.1335 kWh C

^{−1}

The recommended temperature for a hot water tank is 65° C, and a reasonable mains water temperature is 15° C, so to heat a full tank of water takes:

My current electricity tariff is 11.52 pence per unit (1 unit = 1 kW of electricity consumed), so heating a full tank costs me about 70 pence. This all assumes 100% efficiency everywhere, so the actual cost is a bit more.

Anyway, back to the original question. Logically, for my plan to save significant amounts of money I'd have to lose all the stored heat from the hot water tank while the immersion heater is switched off (i.e. the losses from the tank would have to be at least 6 kWh per day). Recently I did leave the heater off for 24 hours (give or take), so armed with these numbers I should be able to work it out.

And the answer is: not significantly. To reheat the tank after 24 hours took about 40 minutes. That's a lot less than 2 hours. The actual energy consumption for that was:

That's pretty good, given that the rated heat loss from the tank is 2.3 kWh per day (though it's currently quite warm here which will help).

I think I'd still save some electricity - the rate of heat loss will be higher the hotter the tank is, so it'll take more energy to continuously maintain the tank temperature - but probably not a massive amount. Maybe next I'll get hold of a thermocouple or something and log the temperature loss over 24 hours.

Of course, this does rely on me not going horribly wrong somewhere in the maths :)

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