Finding the power of my camping stove

 I poured one litre of water into the kettle and timed how long it took to boil with a full gas cannister.

The mass of water was 1kg. The water was at ambient temperature of about 10 degrees Celsius and reached 100 degrees Celsius for boiling. Energy absorbed is m.c.delta(theta) = 1 x 4200 x 90 = 378000J.

It took 7 minutes 25 seconds or 445 seconds to boil. So the effective power of the stove in this circumstance is 378000/445 = 850W. I say "effective" power because I'm only interested in the useful energy transformation; the energy wasted heating the surroundings is of no interest here.