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 28.6 g sample of ethanol initially at a temperature of 10.8oC.

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.

John, I think I worked this for you yesterday. Look at yesterday's posts.

To find the amount of energy required to heat and vaporize ethanol, we need to calculate two separate components: the energy required to heat the ethanol from 10.8°C to its boiling point at 78.3°C, and the energy required to completely vaporize the sample.

1. Calculate the energy required to heat the ethanol:
The formula to calculate the energy (Q) is given by Q = mcΔT, where:
- Q is the energy required
- m is the mass of the sample (28.6 g)
- c is the specific heat capacity of ethanol (2.44 JK^(-1)g^(-1))
- ΔT is the change in temperature (from 10.8°C to 78.3°C)

Q = (28.6 g) * (2.44 JK^(-1)g^(-1)) * (78.3°C - 10.8°C)

2. Calculate the energy required to vaporize the ethanol:
The heat of vaporization (ΔHvap) of ethanol is given as 38.56 kJ mol^(-1). However, we need to find the energy required to vaporize 28.6 g of ethanol, not 1 mole. The molar mass of ethanol is 46.07 g/mol.

Number of moles = mass / molar mass
Number of moles = 28.6 g / 46.07 g/mol

The energy required to vaporize the sample can be calculated using the formula:
Q = ΔHvap * moles

Q = (38.56 kJ mol^(-1)) * (28.6 g / 46.07 g/mol)

3. Add the two energy components to find the total energy required:
Total energy = energy required to heat + energy required to vaporize

Now, convert the units to joules:
1 kJ = 1000 J

Make sure all the units are consistent before performing the calculations.

Now, you can plug in the values and calculate the total energy required to heat and vaporize the ethanol.