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To calculate the energy required to vaporize aluminum, we need to use its heat of vaporization, which is approximately 10.4 MJ/kg. For 2 kg of aluminum, the energy required would be 2 kg × 10.4 MJ/kg = 20.8 MJ. Thus, 20.8 megajoules of energy is needed to vaporize 2 kg of 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.
32 weighing operations are required to measure 31 kg of rice with a 1 kg stone. Start by weighing 1 kg, then 2 kg, 3 kg, and so on up to 31 kg. Each weighing operation will add 1 kg to the total.
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 gold, you need to consider its heat of vaporization, which is approximately 1,500 kJ/kg. For 2 kg of gold, the energy required would be 2 kg × 1,500 kJ/kg = 3,000 kJ. Thus, it takes about 3,000 kJ of energy to vaporize 2 kg of gold.
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65.807 kg
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31 gm to kg
To calculate the energy required to vaporize aluminum, we need to use its heat of vaporization, which is approximately 10.4 MJ/kg. For 2 kg of aluminum, the energy required would be 2 kg × 10.4 MJ/kg = 20.8 MJ. Thus, 20.8 megajoules of energy is needed to vaporize 2 kg of 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.
32 weighing operations are required to measure 31 kg of rice with a 1 kg stone. Start by weighing 1 kg, then 2 kg, 3 kg, and so on up to 31 kg. Each weighing operation will add 1 kg to the total.
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.
1650kj
lf = 3.35 x 105 J kg-1 This much amount of heat required to convert 1 kg of ice to liquid Mani.Ra
The heat of fusion for gold is 64.4 kJ/mol. To convert this to energy required to melt 1.5 kg of gold, we need to calculate the number of moles in 1.5 kg of gold (1.5 kg of gold is approximately 0.047 moles). Then, the energy required would be approximately 3.03 kJ.