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.
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.
1650kj
The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
Since gasoline and ethanol do not have exactly the same boiling point, it is possible to separate them by means of fractional distillation. But if you are just looking for cheap vodka, it's really not worth the trouble. Just buy vodka at the liquor store, it's much easier that way.
The energy required to vaporize a material can be calculated using its heat of vaporization. For gold, the heat of vaporization is approximately 330 kJ/mol. Since gold has a molar mass of 196.97 g/mol, 2 kg of gold is equal to 10.15 moles. Therefore, the energy needed to vaporize 2 kg of gold is approximately 3.35 MJ.
The amount of heat required to boil alcohol (ethanol) depends on the quantity being heated and its initial temperature. On average, it takes about 207.3 kJ of heat to vaporize 1 mole of ethanol.
1oo calories for 1 g
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.
1650kj
The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10,700 kJ/kg. For 2 kg of aluminum, the total energy required would be 2 kg × 10,700 kJ/kg, equating to about 21,400 kJ. Therefore, around 21.4 million joules of energy is needed to vaporize 2 kg of aluminum.
Since gasoline and ethanol do not have exactly the same boiling point, it is possible to separate them by means of fractional distillation. But if you are just looking for cheap vodka, it's really not worth the trouble. Just buy vodka at the liquor store, it's much easier that way.
The energy required to vaporize a material can be calculated using its heat of vaporization. For gold, the heat of vaporization is approximately 330 kJ/mol. Since gold has a molar mass of 196.97 g/mol, 2 kg of gold is equal to 10.15 moles. Therefore, the energy needed to vaporize 2 kg of gold is approximately 3.35 MJ.
The amount of heat absorbed by 1 kg of liquid to vaporize it depends on what that liquid is, (water?), and what the temperature of the liquid is at the start of the process. Obviously, it will take more heat to vaporize 1 kg of liquid that is at 0º than it will to vaporize 1 kg of liquid that is at, say, 30º.
To calculate the energy released when 1.56 kg of ethanol freezes, first convert the mass of ethanol to moles using its molar mass. Then, use the heat of fusion of ethanol to determine the energy released using the formula: Energy released = moles of ethanol x heat of fusion.
Heat of vaporization at 100 degrees C is 40.65 kJ/mol. 100g water * 1mol/18.015g = 5.55 mol 40.65*5.55 = 225.6452956 = 226 kJ if three sig figs.
too much energy is needed to vaporize water