9460 kJ
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.
1650kj
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.
Oh, dude, to vaporize 2kg of water at 100°C, you'd need about 2260 kilojoules of energy. It's like making a really intense cup of tea, but instead of sipping it, you're just turning it into steam. So, yeah, it's a pretty hot process, literally.
414 kJ
9460 kJ
1650kj
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.
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.
2200 kj
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.
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.
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.
1650kj
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.
7095 kJ
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.