posted by Dana on .
boiling point- 78.5 degrees C
heat of vaporization- 210cal/g
a)How much energy, in calories, will be required to evaporate 125.0 mL of ethanol at its boiling point?
b)How long will it take to add this energy, if the rate of heating is 1.24kcal/minute?
Please explain/show work. My lab manual gives no formula or instructions on how to do this. Thanks!
Convert 125.0 mL ethanol to grams using density and mass = volume x density. Then
grams ethanol x heat vaporization (in calories/gram) = heat required.
b)time (minutes)= cal heat required x (1 min/1.24 kcal) = ??