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How much energy is required to vaporize 2 kg of copper Use the table below and this equation?
To calculate the energy required to vaporize 2 kg of copper, you need to use the specific latent heat of vaporization for copper, which is approximately 300,000 J/kg. The total energy (Q) can be calculated using the equation Q = m * L, where m is the mass (2 kg) and L is the latent heat of vaporization (300,000 J/kg). Thus, Q = 2 kg * 300,000 J/kg = 600,000 J. Therefore, 600,000 joules of energy is required to vaporize 2 kg of copper.
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How much energy is required to vaporize 2 kg of copper Use the table below and this equation Q Melva?
2200 kj
How much energy is required to melt 2 kg of copper Use the table below and this equation Q mLfusion.?
414 kJ
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 many kilojoules or energy are required to vaporize 5 g of ethane?
The heat of vaporization for ethane is approximately 16.7 kJ/g. Therefore, to vaporize 5 g of ethane, the energy required would be: 16.7 kJ/g * 5 g = 83.5 kJ.
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 is required to vaporize 2 kg of copper?
9460 kJ
How much energy is required to vaporized 2 kg of copper?
9460 kJ
How much energy is required to vaporize 1.5 kg of copper?
1650kj
How much energy is required to vaporize 2 kg of gold Use the table below and this equation?
The heat of vaporization of gold is 158 kJ/kg. To find the total energy required to vaporize 2 kg of gold, you can use the equation: Energy = mass * heat of vaporization. Substitute the values to get: Energy = 2 kg * 158 kJ/kg = 316 kJ. Therefore, 316 kJ of energy is required to vaporize 2 kg of gold.
How much energy is required to vaporize 2 kg of copper Use the table below and this equation Q Melva?
2200 kj
How much energy is required to vaporize 1 oz of copper?
To vaporize 1 ounce (approximately 28.35 grams) of copper, you would need about 1,800 joules of energy per gram, as the heat of vaporization of copper is around 1,800 kJ/kg. Therefore, the total energy required for 1 ounce of copper would be approximately 51 kJ (or 51,000 joules). This calculation assumes that the copper is initially at its melting point before vaporization occurs.
How much energy is required to melt 2 kg of copper Use the table below and this equation Q mLfusion.?
414 kJ
How much energy is required to vaporize 1.5 of aluminum?
1650kj
How much energy of is required to vaporize 1.5 kg of aluminum?
1650kj
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 is the Hfusion used to calculate the energy required to vaporize a volume of liquid?
Liters liquid 1000ml/1L g/ml mol/g Hfusion
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.
