540 calories are needed to turn one gram of water at 100 degrees celsius to steam.
The answer is 55,117 kJ.
The boiling point of ethanol is 78 C but it can evaporate slowly at just room temperature. You can set it on fire and it will vaporize even more quickly.
Heat of vaporization
It takes 80 calories per gram to increase the temperature of water by one degree. 4000 * 80 * 100 = 32000000 calories.
The energy needed to completely vaporize a mole of a liquid
1oo calories for 1 g
The amount of energy needed to vaporize 175 g of water depends on the temperature of the water. However, we shall assume it is 100 degrees C. We multiply 175 by 539 and get 94,325 calories. (Notice the small c). We could express it as 94 Calories if we were talking about the stuff on your dining room table.
This is the latent heat of vaporisation of water, which at standard pressure, is 539 calories (per gram).
The answer is 55,117 kJ.
The boiling point of ethanol is 78 C but it can evaporate slowly at just room temperature. You can set it on fire and it will vaporize even more quickly.
Gasoline has a less boiling point (72 degree Celsius) While water has a bigger boiling point (100 degree Celsius)
The boiling point is the temperature where a substance BEGINS to vaporize. So all of the water doesn't necessarily need to boil off instantly. To vaporize, molecules of water need to have energy. Only at the boiling point do they have enough energy to boil away, and when they do, they carry this energy with them. This means that a constant supply of heat for a certain period of time is needed for all water in a sample to boil off.
1 cal/gdegC x 225g x (100 - 50.5)degC heat to the boiling point 540 cal/g x 225g heat to vaporize the water 0.5 cal/gdegC x 225 g x (133-100)degC heat to chang temp of steam = 11137.5 cal + 121500 cal + 3712.5 cal = 14850 cal
there are no calories in water you idiot
The needed energy is 10 calories.
The number of calories required will depend on the mass of water which is to be heated.
Heat of vaporization