How much heat is required to convert 32.5 grams of ethanol at 28 C to the vapor phase at 78 C?

Ethanol (C2H5OH) boils at 75 C. Its density is 0.789 g/mL. The enthalpy of vaporization is 38.56 kJ/mol. The specific heat of liquid ethanol is 2.3 J/g-K.

First convert 32.5 g to moles. The molecular weight of C2H5OH is 24 + 16 + 6 = 46

You have 32.5/46 = 0.707 moles of ethanol.

The heat required (in kJ) is
0.707*38.56 + 32.5*50*2.3*10^-3

31

Well, let's heat things up with some calculations! To convert ethanol from a liquid to a vapor phase, we need to consider two steps: heating the liquid ethanol to its boiling point, and then vaporizing it.

First, let's calculate the heat required to raise the temperature of the ethanol from 28°C to its boiling point of 75°C. We'll use the specific heat of liquid ethanol, which is given as 2.3 J/g-K.

The temperature change (∆T) is 75°C - 28°C = 47°C.
The mass of ethanol is 32.5 grams.

Q1 = mass × specific heat × ∆T
Q1 = 32.5 g × 2.3 J/g-K × 47°C
Q1 = 3587.75 J

Next, we need to calculate the heat required to vaporize the ethanol at its boiling point. The enthalpy of vaporization of ethanol is given as 38.56 kJ/mol.

First, we need to calculate the number of moles of ethanol in 32.5 grams.
Molar mass of ethanol (C2H5OH) = 12.01 g/mol (C) + 1.01 g/mol (H) × 6 + 16.00 g/mol (O) + 1.01 g/mol (H)
Molar mass of ethanol = 46.07 g/mol

Number of moles = mass / molar mass
Number of moles = 32.5 g / 46.07 g/mol
Number of moles ≈ 0.706 mol

Q2 = number of moles × enthalpy of vaporization
Q2 = 0.706 mol × 38.56 kJ/mol
Q2 = 27.20 kJ = 27,200 J

Now, let's sum up the two heat values to find the total heat required.

Total heat = Q1 + Q2
Total heat = 3587.75 J + 27,200 J
Total heat ≈ 30,787.75 J

So, approximately 30,787.75 Joules of heat is required to convert 32.5 grams of ethanol at 28°C to the vapor phase at 78°C. That's enough heat to turn any party into a steamy affair!

To calculate the amount of heat required to convert 32.5 grams of ethanol from a liquid phase at 28°C to a vapor phase at 78°C, we need to consider the following:

1. Calculate the mass of ethanol using its density:
Density of ethanol = 0.789 g/mL
Volume of ethanol = mass / density = 32.5 g / 0.789 g/mL ≈ 41.19 mL (rounded to 2 decimal places)
Mass of ethanol = 41.19 mL * 0.789 g/mL ≈ 32.51 g (rounded to 2 decimal places)

2. Calculate the energy required to raise the temperature of ethanol from 28°C to its boiling point of 75°C:
Specific heat of liquid ethanol = 2.3 J/g-K
ΔT (change in temperature) = 75°C - 28°C = 47°C

Energy required = mass * specific heat * ΔT
= 32.51 g * 2.3 J/g-K * 47°C

3. Calculate the energy required for the phase change from liquid to vapor:
Enthalpy of vaporization of ethanol = 38.56 kJ/mol
Convert mass of ethanol to moles using its molar mass (46.07 g/mol):
Moles of ethanol = mass / molar mass
= 32.51 g / 46.07 g/mol ≈ 0.705 mol (rounded to 3 decimal places)

Energy required for phase change = moles of ethanol * enthalpy of vaporization
= 0.705 mol * 38.56 kJ/mol

4. Calculate the energy required to raise the temperature of the vapor from 75°C to 78°C:
Specific heat of vapor = 2.3 J/g-K (assumed same as liquid ethanol)

Energy required = mass * specific heat * ΔT
= 32.51 g * 2.3 J/g-K * 3°C

5. Sum up all the energies calculated in steps 2-4 to get the total heat required:
Total heat required = Energy for temperature change + Energy for phase change + Energy for temperature change
= (32.51 g * 2.3 J/g-K * 47°C) + (0.705 mol * 38.56 kJ/mol) + (32.51 g * 2.3 J/g-K * 3°C)

Note: Make sure to convert units properly and round the final answer to the appropriate number of significant figures.

To determine the amount of heat required to convert 32.5 grams of ethanol from a liquid phase at 28 °C to a vapor phase at 78 °C, we need to calculate the heat required for two separate steps:

1. Heating the liquid ethanol from 28 °C to its boiling point of 75 °C.
2. Vaporizing the liquid ethanol at its boiling point.

Let's calculate each step separately and then add them together to get the total heat required.

Step 1: Heating the liquid ethanol to its boiling point

To calculate the heat required to raise the temperature of the liquid ethanol, we can use the formula:

q = m * C * ΔT

where:
q is the heat in joules (J)
m is the mass of ethanol in grams (32.5 g)
C is the specific heat of liquid ethanol (2.3 J/g-K)
ΔT is the change in temperature (75 °C - 28 °C = 47 °C)

Plugging in the values:

q1 = 32.5 g * 2.3 J/g-K * 47 °C = 3772.75 J

So, the heat required to heat the liquid ethanol to its boiling point is 3772.75 J.

Step 2: Vaporizing the liquid ethanol at its boiling point

To calculate the heat required to vaporize the liquid ethanol, we can use the formula:

q = n * ΔHvap

where:
q is the heat in joules (J)
n is the number of moles of ethanol
ΔHvap is the enthalpy of vaporization of ethanol (38.56 kJ/mol = 38560 J/mol)

First, we need to calculate the number of moles using the given mass of ethanol and its molar mass.

The molar mass of ethanol (C2H5OH) =
(2 * atomic mass of C) + (6 * atomic mass of H) + (1 * atomic mass of O)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 46.08 g/mol

molar mass of ethanol

n = mass / molar mass
= 32.5 g / 46.08 g/mol
≈ 0.705 mol

Now we can calculate q2:

q2 = 0.705 mol * 38560 J/mol = 27164.8 J

So, the heat required to vaporize the liquid ethanol at its boiling point is 27164.8 J.

Total heat required:

The total heat required is the sum of q1 and q2:

Total heat = q1 + q2
= 3772.75 J + 27164.8 J
≈ 30937.55 J

Therefore, approximately 30937.55 J (or 30.9 kJ) of heat is required to convert 32.5 grams of ethanol from a liquid phase at 28 °C to a vapor phase at 78 °C.