Cylinder power and time. Say if you had a 300 litre cylinder, with a 3kW element, specific type of thermal insulation, and had to increase it from 30 degrees celcius to 60 degrees. Surely there must be a formula on calculating this value? As we know the volume of a Cylinder (in meter cubed) is = pi * diameter squared * height / 4...Then by including the losses (efficiency), element power rating, temperature increase required and maybe some other factors will give us the answer??? Theres also the time, how long would it take to heat the water to 60 degrees from 30 degrees? Thanks for your help, murphfan51 ysics 101 was a few years ago but....... The specific heat of water is 4.186 joules/gram-C Q = mc(DeltaT) T1 = (30c) start temperature T2 = (60c) finish temperature Q = m ( 4.186 joules/gram-C ) (T2-T1) Q = 300 litres * 1000 grams * (4.186j/g-C)(30C) Q = 300,000*4.186*30 = 37674000 Q = 37674000 joules for a 30 degree C rise in temperature A joule is a watt-second so: 37674000 watts will heat 300 litres up by 30C in a second. or 627900 watts will heat 300 litres up by 30C in 60 seconds. or 10465 watts will heat 300 litres up by 30C in 1 hour. or 3488 watts will heat 300 litres up by 30C in 3 hours. I think that's right..... murphfan51 30-03-2009, 08:31 PM Ah, thank you, that looks about right. What about efficiency? I will show