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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 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.
100 and 150 kg of nails is what is needed for the 100 square metre.
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65.807 kg
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31 gm to kg
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.
1650kj
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.
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.
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.