Molar enthelpy heat of vaporization

[GLprincess02]

Homework Statement

Ethanol boils at a temp. of 78.29 degrees C. What amount of energy, in joules, is necessary to heat to boiling and then completely vaporize a 23 g sample of ethanol initially at 12 degrees C?

Homework Equations

q=mass x constant x change in temp ?

The Attempt at a Solution

I know that the molar enthalpy heat of vaporization is 35.56 kJ/mol and the specific heat of ethanol (C2H5OH) is approximately constant at 2.44 J/gK. However, I am unsure of what to do with these numbers or even which equation to use. Any help at all will be appreciated!

 

[hage567]

You can consider this in two parts: the first part is the thermal energy needed to bring the ethanol from 12 degrees to its boiling point. You can use the equation you posted for this. Then you must consider the heat of vaporization when the ethanol undergoes the phase change to the gas state. You should be able to find what you need to do this in your textbook. Give that a try.

Caution: be careful of your units when doing the calculations, since you have quantities expressed in both grams and moles.

 

[Blueshift5]

I understand your confusion and will be happy to assist you with this problem. To solve this, we will need to use the equation q=mass x constant x change in temp, as you have correctly identified. Here is the step-by-step solution:

1. First, we need to calculate the change in temperature (ΔT) of the ethanol. This can be done by subtracting the initial temperature (12°C) from the boiling temperature (78.29°C). So, ΔT = 78.29°C - 12°C = 66.29°C.

2. Next, we need to convert the mass of ethanol from grams to kilograms, as the specific heat constant is given in J/gK. So, the mass of ethanol in kilograms is 23 g / 1000 = 0.023 kg.

3. Now, we can plug these values into the equation q=mass x constant x change in temp. So, q = (0.023 kg) x (2.44 J/gK) x (66.29°C) = 3.03 kJ.

4. However, this value only represents the energy needed to heat the ethanol to its boiling point. To completely vaporize the ethanol, we need to use the molar enthalpy heat of vaporization (ΔHvap). To do this, we need to first calculate the number of moles of ethanol in the sample. This can be done by dividing the mass (23 g) by the molar mass of ethanol (46.07 g/mol). So, the number of moles of ethanol is 23 g / 46.07 g/mol = 0.499 mol.

5. Now, we can calculate the energy needed to vaporize the ethanol by multiplying the number of moles by the molar enthalpy heat of vaporization. So, q = (0.499 mol) x (35.56 kJ/mol) = 17.74 kJ.

6. Finally, we can find the total energy needed by adding the two values we calculated in step 3 and 5. So, the total energy needed is 3.03 kJ + 17.74 kJ = 20.77 kJ.

Therefore, it would take approximately 20.77 kJ of energy to heat and vaporize a 23 g sample of ethanol from 12°C to its boiling point of