The latent heat of vaporization of water is 2260 joules per kilogram.
The heat of vaporization of water is 2260 joules per kilogram.
The heat of vaporization is about 2258 joules/gram46 x 2258 = 103,868 J ~ 103.9 kJ(Using the rounded value 2260 joules per gram, about 104 kJ)it is already at boiling point so in theory zero energy is required.I think the question is asking what energy is required to free the water from its liquid state to the vapour state.At 101325Pa, (1 atmosphere in old money), the latent heat of evaporation of water is 2256.7 J/g.So 2256.7 * 46 = 103808 Joules
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
The process of vaporization, where water changes from liquid to gas, requires 2260 Joules of energy per gram to occur.
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.
The heat of vaporization of water is 2260 joules per kilogram.
When 1 gram of boiling water at 100°C condenses to water at the same temperature, it releases energy in the form of latent heat of vaporization. The latent heat of vaporization for water is approximately 2260 joules per gram. Therefore, when 1 gram of steam condenses, about 2260 joules of energy is transferred to the surroundings.
Vaporization is the change of liquid water to water vapor. Vaporization requires addition of the latent heat of vaporization to liquid water. The latent heat of vaporization supplies the liquid water molecules with enough energy to become vapor molecules. The latent heat of vaporization at 1.0 atmosphere pressure is about 1000 Btu per lbm ( 2260 kJ per kg ).
To evaporate all the water on Earth, approximately 1.2 x 10^21 joules of energy is required. This estimate is based on Earth's total water volume of about 1.386 billion cubic kilometers and the latent heat of vaporization of water, which is around 2.26 million joules per kilogram. This immense amount of energy highlights the significant role water plays in Earth's climate and energy balance.
When water vapor condenses, it releases the latent heat of vaporization, which is around 2260 joules per gram. This heat energy warms the surrounding air as it is released during the condensation process.
Perspiration coats the outside of the skin with moisture (water). The water evaporates. Evaporation requires heat to be absorbed by the water (the latent heat of vaporization). Heat is removed from the body surface to provide the water with the latent heat of vaporization.
When one gram of water vapor condenses into liquid water, it releases about 2260 joules (540 calories) of heat energy. This process is called the latent heat of vaporization.
During condensation, approximately 2260 joules of energy are removed from 1 gram of water. This value represents the latent heat of vaporization, which is the energy required to convert water from a liquid to a gas. When water vapor condenses back into liquid water, this amount of energy is released into the surroundings.
The specific heat of water is 4186 joules per kilogram degree Celsius.
To calculate the heat released when 1000 g of steam at 100°C condenses to water at the same temperature, we use the latent heat of vaporization for water, which is approximately 2260 J/g. Therefore, the heat released can be calculated as follows: Heat released = mass × latent heat of vaporization = 1000 g × 2260 J/g = 2,260,000 J. Thus, 2,260,000 joules of heat is released during the condensation process.
The latent heat of vaporisation of water requires more energy. This is because on melting, the intermolecular bonds in water are only weakened whereas on boiling, the bonds are completely broken, which requires a larger amount of energy.
The heat of vaporization is about 2258 joules/gram46 x 2258 = 103,868 J ~ 103.9 kJ(Using the rounded value 2260 joules per gram, about 104 kJ)it is already at boiling point so in theory zero energy is required.I think the question is asking what energy is required to free the water from its liquid state to the vapour state.At 101325Pa, (1 atmosphere in old money), the latent heat of evaporation of water is 2256.7 J/g.So 2256.7 * 46 = 103808 Joules