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What is the latent heat of vaporization of water in joules per kilogram?
The latent heat of vaporization of water is 2260 joules per kilogram.
How many joules of heat are needed to completely vaporize 24.40 grams of water at its boiling point?
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.
What is a valid unit for specific heat?
Joule/kilogram-kelvin The SI unit is joules / kelvin. This is valid for an object of any size, but if you want the typical specific heat for a certain type of material, you have to standardize it, resulting in either joules / (kelvin x kilogram) or joules / (kelvin x mole).
Which substance is expected to have the highest heat of vaporization?
Water is expected to have the highest heat of vaporization among common substances.
Example of heat of vaporation?
The heat of vaporization is the energy required to change a substance from liquid to gas at its boiling point without changing its temperature. For example, the heat of vaporization for water is 2260 J/g at its boiling point of 100°C. This means that 2260 Joules of energy are needed to vaporize 1 gram of liquid water at 100°C.
What is the latent heat of vaporization of water in joules per kilogram?
The latent heat of vaporization of water is 2260 joules per kilogram.
What is the specific heat of water in joules per kilogram degree Celsius?
The specific heat of water is 4186 joules per kilogram degree Celsius.
How many joules of heat are needed to completely vaporize 24.40 grams of water at its boiling point?
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.
What method can be used to estimate the specific heat of water in joules per kilogram per degree Celsius?
One method to estimate the specific heat of water in joules per kilogram per degree Celsius is by conducting a calorimetry experiment, where the heat gained or lost by a known mass of water is measured and used to calculate the specific heat capacity.
How much energy is transfered when 1gm of boiling water at 100c condenses to water at 100c?
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.
What is the amount of heat needed to raise that temperature of 1 kilogram of water 1 degree Celsius?
The amount of heat needed to raise the temperature of 1 kilogram of water by 1 degree Celsius is 4186 Joules, which is the specific heat capacity of water.
When water vapor condenses how much heat is released into the atmosphere?
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.
What is is the amount of heat needed to raise the temperature of one kilogram of water one degree centigrade.?
The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius. To raise the temperature of one kilogram (1000 grams) of water by one degree Celsius, it would require approximately 4180 Joules of heat energy.
What is a valid unit for specific heat?
Joule/kilogram-kelvin The SI unit is joules / kelvin. This is valid for an object of any size, but if you want the typical specific heat for a certain type of material, you have to standardize it, resulting in either joules / (kelvin x kilogram) or joules / (kelvin x mole).
How many calories of heat is released when one gram of water vapor condenses into liquid water?
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.
Which phase change reguires water to gain 2260 joules per gram?
vaporization
During a strenuous workout an athlete generates 2000 kJ of heat energy What mass of water would have to evaporate from the athletes skin to dissipate this much heat?
To calculate the mass of water that would need to evaporate to dissipate 2000 kJ of heat, we need to use the heat of vaporization of water, which is 2260 J/g. By converting the energy to joules and dividing by the heat of vaporization, we find that approximately 884 grams (0.88 kg) of water would need to evaporate from the athlete's skin.
