I want to create a solar powered kettle but first, I need to know how much power will be required.

This means, I think, I'll need to know how much water I want to bring up to a certain temperature.

I'm trying to understand how to work this out, and I'm frankly stumped.

Is it possible to work this out, if I knew I wanted to bring 0.5litres to 60 degress C, how much power would I need to achieve this within 30 minutes?

My end goal here is to find a suitable solar panel powerful enough.

  • $\begingroup$ A good place to start is solar water heaters. A simple internet search using your favorite search engine can lead to a plethora of infomation $\endgroup$ – Mahendra Gunawardena Jun 7 '15 at 11:53
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    $\begingroup$ Why use an electric solar panel to convert solar energy into electricity into heat? You can use a (thermal) solar hot water system, made of mirrors and black piping, to convert solar energy directly to heat, without using electricity. $\endgroup$ – Li-aung Yip Jun 8 '15 at 13:39
  • $\begingroup$ @Li-aungYip While Watts/m^2 of insolation do indeed provide better energy acquistion thermally than electrically, the use of PV systems for water heating is gaining increasing popularity. Numerous studies have been carried out and in many situation a good economic case can be made for doing this. A good PV panel has a 20+ year service life (I have one 30+ years old still going albeit not well), energy losses from panel to heater can be low, no tank high on roof , no water pipes, pumps, corrosion, leaks, ... .PV energy can be used for heating when not needed for other use. Temperature can ... $\endgroup$ – Russell McMahon Jun 8 '15 at 16:19
  • $\begingroup$ @Li-aungYip ... be high regardless of available insolation (reduce flow rate to suit). Heating is at point of need with no pipe lagging or losses. No "dead" cold water in pipes at start of day. And more. I'm investigating aspects of solar PV at present and will be 'playing' with small thermal evacuated tube solar heating "sometime" - but PV is far nicer to work with overall. $\endgroup$ – Russell McMahon Jun 8 '15 at 16:22
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    $\begingroup$ @MyDaftQuestions Yes PV = Photovoltaic = std electrical output solar. (There are other solar to electrical output systems). Li-Aung-Yip discussed PV so I responded (as I'm doing just this now) BUT eg Olin assumed thermal solar heating. It's not clear which you mean. Both have their place. What is your application? $\endgroup$ – Russell McMahon Jun 9 '15 at 11:14

The amount of water and how much you want to heat it only tells you the energy required, not the power. Power is energy per time, so it matters how fast you want to get there.

You say you want to heat 500 ml of water by 60°C in 30 minutes. Do the math.

The specific heat of water is 4.18 J/g°C. The total energy required is therefore:

(500 g)(60°C)(4.18 J/g°C) = 125 kJ

This energy spread out over 30 minutes (1800 seconds) is:

(125 kJ)/(1800 s) = 70 W

However, that is only the power that has to go into the water. Your 500 ml of water sitting in some container at near boiling temerature is most likely loosing more than that to the environment. Put another way, your heater has to not only provide enough power to raise the temperature of the water, but also to overcome losses of this heat to the enviroment. At only 70 W into the water, the heat loss will be significant.

With a lot of care spent on insulation, maybe 150 W is enough. It's really hard to guess since we have no idea what your mechanical system looks like. A few 100 W would be better.

Also, consider that 30 minutes is a long time to wait for a cup of boiling water. If the system can just do that in full sun, then it won't work in anything less than ideal circumstances. 500 ml of water is also not a lot of stuff to heat. I think you are grossly underspecifying your heater. You may get 500 ml to boil in 30 minutes, but I expect you won't be happy with the performance of the heater overall.

Figure you get about 1 kW per square meter of full sunlight, so a few 100 W doesn't really take a large collector. However, you have to pay attention to the aligment, and focus the rays on a small spot roughly the size of the bottom of the kettle.

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    $\begingroup$ It happens that I've been 'playing' extensively in this area of late so the fact that your answer was wrong was instantly obvious while I had to read through the text to see why. Easily done. Your answer needs adjusting by a factor of 4.18^2, for reasons which will be obvious when you look. Doh! I personally have an ongoing allotment of such inversions to use up :-). $\endgroup$ – Russell McMahon Jun 8 '15 at 16:13
  • $\begingroup$ @Russell: Doh indeed! 4 W sounded very little to me, but I convinced myself that the numbers didn't lie. Sorry for the confusion. $\endgroup$ – Olin Lathrop Jun 8 '15 at 16:43

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