540 calories are needed to turn one gram of water at 100 degrees celsius to steam.
1650kj
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.
To calculate the heat needed to vaporize sweat, you would need to know the specific heat of vaporization of sweat. Once you have that information, you can use the formula Q = mL, where Q is the heat needed, m is the mass of the substance (converted from volume using its density), and L is the specific heat of vaporization.
The heat of vaporization of ethanol is approximately 840 kJ/kg. To find the total heat required to vaporize 1.25 kg of ethanol, you can multiply the mass by the heat of vaporization: 1.25 kg * 840 kJ/kg = 1050 kJ.
Ethanol has a higher boiling point than methane because ethanol molecules are larger and have stronger intermolecular forces such as hydrogen bonding, making it more difficult for the molecules to break free from each other and vaporize.
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).
1650kj
To melt 10 grams of ice at 0 degrees Celsius, it would require 80 calories of heat energy per gram, so a total of 800 calories (80 calories/gram * 10 grams = 800 calories) would be needed.
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
Gasoline has a less boiling point (72 degree Celsius) While water has a bigger boiling point (100 degree Celsius)
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.
Changing water pressure can affect the boiling point because it alters the equilibrium between liquid and vapor phases. Increasing pressure raises the boiling point, as more energy is needed to overcome the higher pressure. Decreasing pressure lowers the boiling point, as it requires less energy to vaporize the liquid.
The boiling point of a substance decreases as the atmospheric pressure decreases. This is because lower atmospheric pressure reduces the pressure pushing down on the liquid, making it easier for the liquid to vaporize. Conversely, higher atmospheric pressure increases the boiling point of a substance as more pressure is needed to overcome the atmospheric pressure and cause the liquid to vaporize.
To calculate the heat needed to vaporize sweat, you would need to know the specific heat of vaporization of sweat. Once you have that information, you can use the formula Q = mL, where Q is the heat needed, m is the mass of the substance (converted from volume using its density), and L is the specific heat of vaporization.
Water boils at 100 degrees Celsius, so 87 degrees is 13 degrees away from boiling.