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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.
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How much energy is required to vaporize 2 kg if aluminum?
To calculate the energy required to vaporize 2 kg of aluminum, we use the heat of vaporization of aluminum, which is approximately 10,900 J/kg. Therefore, the energy required is 2 kg × 10,900 J/kg = 21,800 J, or 21.8 kJ. This is the amount of energy needed to convert 2 kg of aluminum from a liquid to a vapor at its boiling point.
How much energy is required to vaporize 2kg of aluminum?
To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10.5 MJ/kg. For 2 kg of aluminum, the energy required would be about 21 MJ (megajoules). This calculation assumes that the aluminum is already at its melting point and that no heat losses occur during the process.
How much energy is required to vaporize 1.5 kg of aluminum (refer to table of latent heat values.)?
To calculate the energy required to vaporize 1.5 kg of aluminum, we need to use the latent heat of vaporization for aluminum, which is approximately 10,900 J/kg. The energy required can be calculated using the formula: Energy = mass × latent heat of vaporization. Thus, for 1.5 kg of aluminum: Energy = 1.5 kg × 10,900 J/kg = 16,350 J. Therefore, 16,350 joules of energy is required to vaporize 1.5 kg of aluminum.
How much heat energy is needed to vaporize g of water at its boiling point of K?
1oo calories for 1 g
How much heat is required to vaporize 2 kg of aluminum?
To vaporize aluminum, you need to know its heat of vaporization, which is approximately 10,700 kJ/kg. Therefore, to vaporize 2 kg of aluminum, you would require about 21,400 kJ (2 kg × 10,700 kJ/kg). This calculation assumes that the aluminum is already at its melting point and that the vaporization occurs under standard atmospheric conditions.
How much energy is required to vaporize 1.5 of aluminum?
1650kj
How much energy is required to vaporize 2 kg if aluminum?
To calculate the energy required to vaporize 2 kg of aluminum, we use the heat of vaporization of aluminum, which is approximately 10,900 J/kg. Therefore, the energy required is 2 kg × 10,900 J/kg = 21,800 J, or 21.8 kJ. This is the amount of energy needed to convert 2 kg of aluminum from a liquid to a vapor at its boiling point.
How much energy of is required to vaporize 1.5 kg of aluminum?
1650kj
How much energy is required to vaporize 2kg of aluminum?
To vaporize aluminum, you need to consider its heat of vaporization, which is approximately 10.5 MJ/kg. For 2 kg of aluminum, the energy required would be about 21 MJ (megajoules). This calculation assumes that the aluminum is already at its melting point and that no heat losses occur during the process.
How much energy is required to vaporize 1.5 kg of aluminum (refer to table of latent heat values.)?
To calculate the energy required to vaporize 1.5 kg of aluminum, we need to use the latent heat of vaporization for aluminum, which is approximately 10,900 J/kg. The energy required can be calculated using the formula: Energy = mass × latent heat of vaporization. Thus, for 1.5 kg of aluminum: Energy = 1.5 kg × 10,900 J/kg = 16,350 J. Therefore, 16,350 joules of energy is required to vaporize 1.5 kg of aluminum.
How much energy is required to vaporize 1.5 kg of aluminum apex answer.com?
The energy required to vaporize 1.5 kg of aluminum can be calculated using the formula: energy = mass * heat of vaporization. The heat of vaporization for aluminum is around 10,000 J/g. So, the energy required would be 1.5 kg * 10,000 J/g = 15,000,000 J or 15,000 kJ.
Most methods of large scale desalinization are not cost effective because?
too much energy is needed to vaporize water
How much heat energy is needed to vaporize g of water at its boiling point of K?
1oo calories for 1 g
How much heat is required to vaporize 2 kg of aluminum?
To vaporize aluminum, you need to know its heat of vaporization, which is approximately 10,700 kJ/kg. Therefore, to vaporize 2 kg of aluminum, you would require about 21,400 kJ (2 kg × 10,700 kJ/kg). This calculation assumes that the aluminum is already at its melting point and that the vaporization occurs under standard atmospheric conditions.
How much energy will be needed to vaporize 175g of water?
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.
How much energy is needed to vaporize 2 kg of gold?
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.
How many watts is needed to melt one pound of aluminum?
How fast the energy is provided (power, in joules/second or watts) is irrelevant, as long as not too much energy gets radiated away. What you really need to know is how much energy (in joules) is needed.
