how much energy in calories, will be required to evaporate 125.0ml of ethanol at its boiling point?
Density=0.818g/ml
boiling point=78.5C
heat of vaporization=210 cal/g
Heat= mass*Hv= volume*density*Hv
21472.5 cal
To calculate the amount of energy required to evaporate a given volume of ethanol, you need to follow these steps:
Step 1: Convert the volume of ethanol from milliliters to grams.
Given: Density of ethanol = 0.818 g/ml, Volume = 125.0 ml.
By using the formula:
Mass = Density × Volume
Mass = 0.818 g/ml × 125.0 ml
Mass = 102.25 g (rounded to two decimal places)
Step 2: Calculate the energy required using the mass of ethanol.
Given: Heat of Vaporization = 210 cal/g, Mass of ethanol = 102.25 g.
Energy = Heat of Vaporization × Mass
Energy = 210 cal/g × 102.25 g
Energy ≈ 21473.5 cal (rounded to one decimal place)
Therefore, approximately 21,473.5 calories of energy will be required to evaporate 125.0 ml of ethanol at its boiling point.