Oh, dude, to vaporize 2kg of water at 100°C, you'd need about 2260 kilojoules of energy. It's like making a really intense cup of tea, but instead of sipping it, you're just turning it into steam. So, yeah, it's a pretty hot process, literally.
To calculate the amount of energy require to vaporise water at 100C You need to first find the Latent heat of Vaporisation which for water is 2260kJ/kg So to find the amount of energy required you merely multiply the mass of the water in kg by 2260. Density of water is more or less 1kg per litre at room temperature however at 100C it could be less
You need o know the initial temperature of water:Heat = specific heat x difference of temperature x mass of water
Steam at 100C
Yes, the boiling of water at 100C and 1 atm is a spontaneous process.
Q = [ mI ] [ h sub sf + CshW ( 100.0 C - 0.0 C ) + h sub fg ]Q = [ 25 g ] [ 333.7 J /g + ( 4.184 J / g - C ) ( 100.0 C ) + 2257 J / g ]Q = [ 25 g ] [ 3009.1 J / g ] = 75230 J
To calculate the amount of energy require to vaporise water at 100C You need to first find the Latent heat of Vaporisation which for water is 2260kJ/kg So to find the amount of energy required you merely multiply the mass of the water in kg by 2260. Density of water is more or less 1kg per litre at room temperature however at 100C it could be less
The heat required to vaporize 5.00 g of water is given by: 2260 J/g * 5.00 g = 11300 J. Converting this to kJ gives 11.3 kJ.
Steam at 100C
You need o know the initial temperature of water:Heat = specific heat x difference of temperature x mass of water
No one can say, because that's not enough information. The more heat energy you provide, the faster the water will boil.If you provide exactly enough heat to keep the water at 100C, it won't boil at all!Every 540 calories of heat energy turns 1 gram of water to steam.600ml of water = 600 grams of water. So if you can provide 324,000 calories in 20 seconds - but I bet you can't! - you will turn all the water to steam.
Water at 100°C and 1 atm pressure is in its liquid state, commonly referred to as boiling water. At this temperature and pressure, water reaches its boiling point and begins to vaporize into steam.
I think it's the other way - 100°C liquid water causes more damage than 100°C gaseous water. Liquid water has molecules that are much more densely packed than gaseous water. Since temperature is the measure of average kinetic energy, the molecules in liquid form move just as fast as the molecules in gaseous form. So if you stick your finger in liquid water, much more molecules will hit you. Unless that extra enthalpy of vaporization causes a difference in speeds of the molecules, then that extra energy required to vaporize water will be transferred to you, and you will feel more pain.
Steam at 100C
Yes, the boiling of water at 100C and 1 atm is a spontaneous process.
100c
212F or 100C
Q = [ mI ] [ h sub sf + CshW ( 100.0 C - 0.0 C ) + h sub fg ]Q = [ 25 g ] [ 333.7 J /g + ( 4.184 J / g - C ) ( 100.0 C ) + 2257 J / g ]Q = [ 25 g ] [ 3009.1 J / g ] = 75230 J