How much energy (in kilojoules) is released when 21.0 of ethanol vapor at 94.5 is cooled to -19.0?
You need to look up the boiling point temperature and the freezing point T, then use these general formulas.
q1 = mass ethanol vapor x specific heat ethanol vapor x (Tfinal-Tinitial) for all temperatures above the boiling point.
q2 = mass ethanol x heat vporization at the boiling point.
q3 = mass ethanol liquid x specific heat ethanol liquid x (Tfinal-Tinitial) for temperatures from the boiling point to its freezing point.
q4 = mass ethanol liquid x heat fusiion at the freezing point.
Total Q = q1 + q2 + q3 + q4.
To find the amount of energy released when ethanol vapor is cooled from 94.5 °C to -19.0 °C, we can use the equation:
Q = mcΔT
Where:
Q is the energy released or absorbed (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)
First, we need to find the mass of ethanol. The given quantity is 21.0 g.
Next, we need to determine the specific heat capacity of ethanol. The specific heat capacity of ethanol is usually around 2.44 J/g°C.
Finally, we calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = (-19.0 °C) - (94.5 °C)
ΔT = -113.5 °C
Now, we can substitute the values into the equation:
Q = (21.0 g) * (2.44 J/g°C) * (-113.5 °C)
Since the unit of energy is given in kilojoules, we need to convert our answer from joules to kilojoules:
Q = (21.0 g) * (2.44 J/g°C) * (-113.5 °C) / 1000
Calculating the value gives us the amount of energy released in joules. To convert it to kilojoules, we divide by 1000:
Q = -59.853 J
Therefore, approximately 59.853 kilojoules of energy is released when 21.0 g of ethanol vapor at 94.5 °C is cooled to -19.0 °C.