Ethanol (C2H5OH) boils at a temperature of 78.3oC.
What amount of energy, in joules, is necessary to heat to boiling and then completely vaporize a 22.8 g sample of ethanol initially at a temperature of 16.5oC.
The specific heat of ethanol is approximately constant at 2.44 JK-1g-1
The heat of vaporization of ethanol is 38.56 kJ mol-1.
See your other post above.
lmk
To find the amount of energy required to heat and vaporize the ethanol, we need to calculate the energy required to heat the ethanol from the initial temperature to its boiling point, and then the energy required to vaporize it.
First, let's calculate the energy required to heat the ethanol from 16.5oC to its boiling point at 78.3oC. We'll use the formula:
Q = m * c * ΔT
Where:
Q is the energy required (in joules),
m is the mass of ethanol (in grams),
c is the specific heat of ethanol (in J K-1 g-1), and
ΔT is the change in temperature (in Kelvin).
The change in temperature, ΔT, is calculated by subtracting the initial temperature from the final temperature:
ΔT = 78.3oC - 16.5oC = 61.8oC
Since the specific heat of ethanol is given as 2.44 J K-1 g-1, we can substitute the values into the formula:
Q1 = 22.8 g * 2.44 J K-1 g-1 * 61.8 K
Calculating this gives us Q1 = 35,569.856 J (or approximately 35.6 kJ).
Next, let's calculate the energy required to vaporize the ethanol. The energy required for vaporization can be calculated using the formula:
Q2 = moles * ΔHvap
Where:
Q2 is the energy required for vaporization (in joules),
moles is the number of moles of ethanol,
ΔHvap is the molar heat of vaporization of ethanol (in J mol-1).
To find the number of moles of ethanol, we can use the formula:
moles = mass / molar mass
The molar mass of ethanol (C2H5OH) can be calculated by summing up the atomic masses of its elements: 2 * atomic mass of carbon + 6 * atomic mass of hydrogen + atomic mass of oxygen.
The atomic masses of carbon, hydrogen, and oxygen can be found from the periodic table.
Atomic mass of carbon = 12.01 g mol-1
Atomic mass of hydrogen = 1.008 g mol-1
Atomic mass of oxygen = 16.00 g mol-1
Calculating the molar mass of ethanol:
molar mass of ethanol = (2 * 12.01 g mol-1) + (6 * 1.008 g mol-1) + (1 * 16.00 g mol-1)
= 46.07 g mol-1
Now we can calculate the number of moles:
moles = 22.8 g / 46.07 g mol-1
≈ 0.495 mol
The molar heat of vaporization of ethanol is given as 38.56 kJ mol-1, so substituting the values into the formula:
Q2 = 0.495 mol * 38.56 kJ mol-1 * 1000 J kJ-1
= 19,078 J (or approximately 19.1 kJ)
Finally, we can find the total energy required by adding Q1 and Q2:
Total energy = Q1 + Q2
= 35,569.856 J + 19,078 J
= 54,647.856 J (or approximately 54.6 kJ)
Therefore, it would take approximately 54.6 kJ of energy to heat and vaporize the 22.8 g sample of ethanol from 16.5oC to its boiling point.