5

I think this calculation uses the specific heat capacity for water, which is from memory a bit over 4. Can someone please do the math?

Also, if we assume electricity costs $0.30 per kilowatt hour, what is the cost of heating a full 180 litre hot water cylinder?

I'm trying to gain an understanding of the energy usage and costs in our home so that I can compare with other options for a sensible balance of cost and sustainability.

Thanks!

Improve this question
2
  • I'm not quite sure how this question relates to sustainability; would you mind adding a bit of context to the question? Right now it reads more like a physics homework question. Is your aim here to provide people with a general ballpark cost of how much it costs to consume all the stored hot water in a household hot water tank? – Nic♦ Mar 2 '20 at 15:52
  • 1
    @Nic I've edited the question. Generally when we run the hot tap, we don't know how much water we've used, but we do know when the cylinder has been emptied when people spend too much time in the shower. It's useful to know the energy use and cost of doing that, and it makes a good comparison with other usage too. – Highly Irregular Mar 4 '20 at 22:14
19

Let's look at how much electricity and money can be saved by using solar thermal to heat a cylinder of hot water, rather than by using an electrical resistance heater.

Assuming a 100% efficient resistance heater (that's near enough to real values), and ignoring tank losses during the heating period.

180 litres of water is near enough 180kg.

The specific heat capacity of water is approx 4.2 J / g.K

The increase in temperature is 45 Kelvin (60-15)

So the energy needed is 180,000 x 45 x 4.2 = 34MJ

1kWh = 1000W x 3600s = 3.6MJ

Hence, the energy needed is 34/3.6 = 9.5 kWh. At $0.30/kWh that's $2.85

If you want to account for the heating efficiency, just divide by it. So if your heater is 100% efficient, you divide by 1 - which is why we ignore it in the above calculation. If your heater is 95% efficient, you divide by 0.95, to get 10 kWh. This calculation works whatever the type of heater it is.

Improve this answer
2
  • 3
    Just a comment, increase of 1 Kelvin is perfectly equal to 1 degree Celsius, so it's possible to use degrees Celsius in this calculation with the same result. – sharptooth Oct 2 '13 at 7:24
  • It's not just the efficiency of the heater but also the insulation of the tank that needs to be considered. This is harder to calculate as it depends on temperature differential between the tank and it's surroundings as well as the time taken. The process is more efficient using a more powerful heater as it takes less time. Which is why the EU trying to limit the power of kettles is utterly stupid... – George of all trades Feb 18 '17 at 9:07
7

Here's the simplest way to do conversions like this:

180 liters * 45 degrees Celsius = 8100 kilo-calories.

Google knows how to do conversions: 9.41400 kilowatt hours

And it knows how to do arithmetic: 2.8242

Improve this answer
0
0

Start with the volume of 180 litres. Multiply by 4. Then multiply by the target rise on temperature and divide the product by 3412 to give the KWhs needed. Then divide that answer by the KWh output of your heat source to get the time required to achieve your target or alternatively multiply the KWhs required by your local unit cost of electricity to calculate the appx. cost of your water heating. Example:- Volume of water....180 litres Target rise in Temp .........45 degrees Celsius Local cost of electricity ...$0.30 180 X 4 X 45 = 32,400 32400 / 3412 = 9.495KWhours @ $0.30 per KWh costs $2.85 If you are using a say 3KW heat source then time taken will be 9.495 divided by 3= 3hours and 10 mins appx. Gerry

Improve this answer
1
  • 8
    It would help if you explained your magic numbers (4, 3412), use correct units, and edit your wall of text. – user2451 Aug 6 '16 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.