How much energy would be needed to vaporize 150 grams of ethanol when it is at its boiling point?
Hear of Vaporization (kj/kg)...841.00
To calculate the amount of energy needed to vaporize a substance, you can use the formula:
Energy = mass × specific heat of vaporization
In this case, the specific heat of vaporization is given as 841.00 kJ/kg. We need to convert the given mass from grams to kilograms before plugging the values into the formula.
1 kilogram (kg) = 1000 grams (g)
Therefore, 150 grams of ethanol is equal to 150/1000 = 0.15 kilograms (kg).
Now, let's substitute the values into the formula:
Energy = 0.15 kg × 841.00 kJ/kg
Energy = 126.15 kJ
Therefore, approximately 126.15 kJ of energy would be needed to vaporize 150 grams of ethanol when it is at its boiling point.