Not sure exactly what you are asking but it takes 1 btu to raise the temprerature of 1 pound of water by 1 degree F. ( 4.2 joules to raise 1 lb by 1 degree Celsius )
so it depends on how much water we are talking about.
there is also a transition to boiling where the water has reached 212 'F and has not yet begun to boil... it then takes an additional 970.3 btu's ( per lb ) for the water to achieve boiling, this is called the latent heat of vapourization
.. hope this helps
212-65=147. 147(20)=2940 btu needed is this right?
The latent heat of vaporization of water at 212 degrees Fahrenheit at atmospheric pressure involving only vaporization and no saturation is 970 BTU/lb, so to evaporate 15 lbs of water we need 970 x 15 = 14550 BTU's
160 btu's. It takes 1 btu (British Thermal Unit) to change the temperature of 1lb of water 1°. This is referred to as sensible heat. This 1 to 1 ratio does not hold true when changing the state of that water to ice or to steam. Changing the state 1lb of water at 212° to 1lb of steam at 212° you must add 970 btu's and to change 1lb of water at 32° into 1lb of ice at 32° you must add 144 btu's. This is referred to as latent heat.
If you're in Celsius, 13459.5 Btu's. If you're in Fahrenheit, 12632.5 Btu's.
First convert 1 lb of water to lb-moles which is 0.055 lb-moles (you'll need this later). This problem can be broken into 3 steps:(1) You need to detemine how much heat is needed to raise room temperature water (68oF) to 212oF. This can be used using the heat capacity of water which at room temperature is 1 Btu/lboF. So the amount of heat needed for this is:Q1 = m*Cp*ΔT= (1 lb)*(1 Btu/lboF)*(212 - 68oF)= 144 Btu(2) Next you need to account for the phase change. The water changes to steam at 212oF. You use the heat of vaporization which you can look up in any Chemistry or Chemical Engineering Handbook. The Hvap that I found is 17493.5 Btu/lb-mole.Q2 = n(lb-moles)*Hvap= (0.055 lb-moles)*(17493.5 Btu/lb-mole)= 972.64 Btu(3) Next you need to find out how much heat is needed to raise the temperature of the steam from 212 to 213oF. You can look up the heat capacity of steam at 212oF to be 0.485 Btu/lboF.Q3 = m*Cp*ΔT= (1 lb)*(0.485 Btu/lboF)*(213-212oF)= 0.485 BtuTo find the total heat needed add Q1+Q2+Q3 (144+972.64+0.485) =1117.12 Btu
1 BTU is required to raise 1lb of water 1 degree F in 1 hour. 212-75=137 degrees 600 lbs water x 137 degrees= 82,200 BTU's required to change 75 degree water to 212 degree water. To change 212 degree water to 212 degree steam it requires 970 btu's (latent heat of vaporization) per lb of water 970 btu x 600 lbs water = 582,000 btu Answer - 582,000 btu+ 82,200 btu = 664,200 btu's
212-65=147. 147(20)=2940 btu needed is this right?
It takes 970.4 BTUs to convert 1 lb of water at 212°F to 1 lb of steam at 212°F. This process is known as the latent heat of vaporization.
212 - 80 = 132 degrees temperature increase x 1 pound water = 132 BTU
The latent heat of vaporization of water at 212 degrees Fahrenheit at atmospheric pressure involving only vaporization and no saturation is 970 BTU/lb, so to evaporate 15 lbs of water we need 970 x 15 = 14550 BTU's
To raise the temperature of water from 32°F to 212°F, you would need 180 BTUs per pound of water. This calculation is based on the specific heat capacity of water, which is 1 BTU/lb°F.
To calculate the BTU needed to heat 10 gallons of water from 32°F to 212°F, you can use the formula: BTU = gallons × 8.34 (weight of water per gallon in pounds) × temperature change (in °F). The temperature change is 212°F - 32°F = 180°F. Thus, BTU = 10 gallons × 8.34 lbs/gallon × 180°F = 15,012 BTU.
It takes about 55 BTU'S It takes 180 BTU'S to raise 32 degrees to 212 degrees.
160 btu's. It takes 1 btu (British Thermal Unit) to change the temperature of 1lb of water 1°. This is referred to as sensible heat. This 1 to 1 ratio does not hold true when changing the state of that water to ice or to steam. Changing the state 1lb of water at 212° to 1lb of steam at 212° you must add 970 btu's and to change 1lb of water at 32° into 1lb of ice at 32° you must add 144 btu's. This is referred to as latent heat.
I had the same question and the only answer I could find was this: To raise the temperature of 1 LB of water from 40 ºF to 41 ºF would take 1 BTU To raise the temperature of 1 LB of water from 177 ºF to 178 ºF would also take 1 BTU However, if you tried raising the temperature of water from 212 ºF to 213 ºF you would not be able to do it. Water boils at 212 ºF and would prefer to change into a gas rather than let you get it any hotter. Something of utmost importance occurs at the boiling point of a substance. If you did a little experiment and added 1 BTU of heat at a time to 1 LB of water, you would notice that the water temperature would increase by 1 degree Fahrenheit each time. That is until you reached 212 ºF Then something changes. You would keep adding BTU's, but the water would not get any hotter! It would change state into a gas and it would take 970 BTU's to vapourize that entire pound of water. This is called the Latent Heat of Vapourization and in the case of water it is 970 BTU's per pound.
10 x (70 - 50) = 200 Btu
How hot and how fast? 5,000 btu's will heat it. Not very hot and it will take forever. If you are using it for residential hot water application. Roughly 40,000 BTU's