Assuming the density of Aluminum is 2.7 g/cubic cm
Then 1.259 kg occupies (1.259 x 1000 /2.7) cm3 = 466.3 cm3
1 kg 250g is equal to 1250 grams.
just 1250 kg :))
800 kj
1250 g = 1.25 kgTo convert from g to kg, divide by 1000.
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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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.
1.25 kg = 1250 g
1250 kg
1.25 kg
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.
1.25 tonnes.