2755W @ 230V enough to boil 30L?

Hey guys.

I have 31L stainless steel pot and I have insulated it with a mattress
for camping/yoga. I recently bought 3000W/240V heating element but
I have 230V which is 2755W/230V. So I'm wondering if 2755W is
enough to boil around 30L within reasonable amount of time?

It will definetly hold a boil, it will vary how long it takes to get there though. I have had 53L boiling only using 55% power on a 5500w element with no insulation. I think it should be reasonable to use your Element for 30L

Easy enough to figure how long it will take. 1 BTU is equivalent to 1055 joules and will raise one pound of water 1 °F. As a watt is 1 joule/sec, 2755W is 2755/1055 = 2.61 BTU/sec or 156.7 BTU/min. You have 30L of water weighing 66 lbs, Thus it will take 66/156.7 = 0.42 minutes to raise the temperature 1 °F. From 72° to 212°F is 140° and it would thus take 59 minutes to raise 30L of water from room temperature to boiling. This asumes no losses through the vessel or the surface of the water. There is actually quite a bit of loss through the surface so times will be noticeably longer if you don't cover the pot (and preferrably, insulate it e.g. by putting a throw pillow on it).

It's also pretty easy to calculate the amount of water vaporized in a certain time interval once boiling starts. The heat of vaporization for water is 2260 kilojoules per kilogram. 2755 watts is 2.755 kJ/sec so it will take 2260/2.755 = 820 seconds (14 minutes) to boil off a kg (~ 1L) of water.

Wow, amazing calculation.

Ok, it takes less time to heat to 68C and not so very long time from 68 to 100C

You want to boil 30 liters in a 31 liter pot? That's cutting it pretty fine, don't you think? Boil overs are not a pretty sight.

No, probably not boil 30L but at least I need it to get 68C within a reasonable time. If I can get 30L to boil I probably can do it with 27L or little less.

Just noticed where you are posting from. Had I noticed that before I would have kept it metric.

The specific heat of water is about 4.186 kJ/°C-kg

2775 watts is 2.775 kJ/sec and thus will raise 1 kg (1L) of water 1° C in 4.186/2.775 sec.

I always like to point out that James Prescott Joule, after whom this unit of energy is named, did his fundamental research in a laboratory attached to his father's brewery.

Thank you for your time, these calculations should come in very handy when I start using the element.