What is the heat required to evaporate water?
540 calories per gram
energy known as the latent heat of vaporization is required to break the hydrogen bonds. At 100 °C, 540 calories per gram of water are needed to convert one gram of liquid water to one gram of water vapour under normal pressure.
How much energy does it take to vaporize water?
Water has a heat of vaporization value of 40.65 kJ/mol. A considerable amount of heat energy (586 calories) is required to accomplish this change in water.
How do I calculate the energy required to evaporate water?
The specific latent heat is different for solid to liquid transition and liquid to gas transition. For example, if we want to turn 20 g of ice into water we need Q = 20 g * 334 kJ/kg = 6680 J of energy. To turn the same amount of water into vapor we need Q = 45294 J .
How many BTUs does it take to condense water?
When one pound of water boils or evaporates, it absorbs 970 BTU’s at a constant temperature of 212° F. (at sea level) and to condense one pound of steam to water 970 BTU’s must be extracted from it.
What is the heat of vaporization of water at boiling point in Btu lb?
970.3 BTU per Lb.
The latent heat of vaporization of water is at 212°F, 970.3 BTU per Lb. (Column 7).
What does high heat of vaporization mean?
One unique property of water is its high heat of vaporization. Heat of vaporization refers to the energy required to convert one gram of liquid into a gas at boiling point. This required energy will break down the intermolecular attractive forces in water.
How much energy does it take to evaporate a gallon of water?
It takes 8,092 BTUs to evaporate one gallon of water. Natural gas has a heating value of 1,000 BTUs per cubic foot (1 Therm = 100,000 BTUs). Approximate cost of natural gas is $0.50 per Therm.
How much energy is needed to evaporate 100g water?
However, 540 calories of energy are required to convert that 1 g of water at 100˚ C to 1 g of water vapor at 100˚ C. This is called the latent heat of vaporization. On the other hand, you would have to remove 80 calories from 1 g of pure water at the freezing point, 0˚ C, to convert it to 1 g of ice at 0˚ C.
How do you calculate the energy required to evaporate 1 kg of water?
Originally Answered: what is the energy needed to evaporate 1 kg of water? 1 liter of water has a mass of 1 kg. If the water starts off at 100 degrees C, then the energy required to evaporate it would be 540,000 calories or 2.3 x 10^6 Joules.
How many BTU does it take to boil a gallon of water?
8.33 BTU
One BTU is the amount of heat energy required to raise one pound of water by 1ºF. Water weighs 8.33 pounds per gallon so we can calculate that one gallon of water requires 8.33 BTU to raise the temperature 1ºF.
How do I calculate BTU of water?
BTU = Flow Rate In GPM (of water) x (Temperature Leaving Process – Temperature Entering Process) x 500.4*Formula changes with fluids others than straight water.