Molar enthelpy heat of vaporization

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To calculate the energy required to heat and vaporize a 23 g sample of ethanol from 12 degrees C to its boiling point of 78.29 degrees C, first use the equation q = mass x specific heat x change in temperature, with the specific heat of ethanol at 2.44 J/gK. This will provide the thermal energy needed to reach the boiling point. Next, apply the molar enthalpy of vaporization, which is 35.56 kJ/mol, to determine the energy required for the phase change from liquid to gas. Ensure unit consistency when performing calculations, as the mass is in grams and the heat of vaporization is in kJ/mol. This two-step approach will yield the total energy needed for the process.
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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!
 
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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.
 
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