density- 0.818g/mL
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!
a)
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) = ??
To solve this problem, we'll need to use a combination of basic formulas and conversion factors. Here's how you can find the answers to both parts (a) and (b) of the question:
a) To calculate the amount of energy required to evaporate the ethanol, use the formula:
Energy = Mass x Heat of Vaporization
First, we need to convert the volume of ethanol to mass using the given density.
1. Convert the volume from milliliters (mL) to grams (g) using the formula:
Mass = Volume x Density
Mass = 125.0 mL x 0.818 g/mL
Next, multiply the mass by the heat of vaporization to find the energy required:
Energy = Mass x Heat of Vaporization
Substituting the values:
Energy = (125.0 mL x 0.818 g/mL) x 210 cal/g
Calculate the result to find the total energy required in calories.
b) To determine the time taken to add this energy, use the formula:
Time = Energy / Rate of Heating
Substitute the values:
Time = Energy / Rate of Heating
Time = Energy / (1.24 kcal/minute)
Now, calculate the result to find the time in minutes.
By following these steps, you can find the answers to both parts (a) and (b) of the question.