Ethanol, C2H5OH, melts at -114°C and boils at 78°C.?

Question

The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/gK and 2.3 J/gK, respectively. How much heat is required to convert 62.0 g of ethanol at -135°C to the vapor phase at 78°C? Please, if anyone can figure this out, thank you so very much.

Answer 1

This problem must be done is several steps:

1) solid ethanol is raised from -135°C to -114°C.
2) solid ethanol at melting point is melted
3) liquid ethanol at -114 is raised to 78°C
4) liquid ethanol is vaporized

add the heat requirements for each step up:

specific heat * mass * dT = heat (in J)

1) 62g*.97*21° = 1263 J = 1.26 kJ
3) 62g *2.3*192° = 27400 J = 27.4 kJ

moles * heat of vaporization/fusion = heat (in kJ)

molecular weight of ethanol = 45 amu
moles = 62/45 = 1.37moles

2) 1.37 * 5.02 = 6.87 kJ
4) 1.37 * 38.56 = 52.8 kJ

total = 6.87 + 52.8 + 1.26 + 27.4 = 88.3 kJ

Answer 2

Step 1: solid at -135 to solid at -114. Use specific heat of solid.

Step 2: solid at -114 to liquid at -114. Use enthalpy of fusion

Step 3: liquid at -114 to liquid at 78. Use specific heat of liquid

Step 4: liquid at 78 to vapor at 78. Use enthalpy of vaporization

Done. Do your own math.

Answer 3

Break the problem up.

Compute the heat to bring the ice to melting point +
The heat to transform the ice to liquid +
The heat to raise the liquid to boiling point +
The heat to transform the liquid to gas

Each of the 4 computations is relatively simple, add them up and get your final answer.