Question

How much heat is required to heat ethanol from `25^@ "C"` to a gas at `78.24^@ "C"`? `C_P = "2.460 J/g"^@ "C"`, `DeltaH_(vap)^@ = "38.56 kJ/mol"`, `T_b = 78.24^@ "C"`

Answer 1

I got `"44.04 kJ"` total.

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Well, ethanol starts as a liquid, with the following:

• `DeltaH_"vap"^@ = "38.56 kJ/mol"`
• `T_b = 78.24^@ "C"`

• `C_P = "2.460 J/g"^@ "C"`

• Heat it from room temperature to boiling point.

• Keep temperature constant at boiling point and put heat in to vaporize.

If we start at room temperature, `T = 25^@ "C"`, and we assume `DeltaH_"vap"^@` and `C_P` vary negligibly within the temperature range of `25^@ "C" -> 78.24^@ "C"`, then under natural conditions (constant pressure and varied temperature):

`color(green)(q_"heating") = mC_PDeltaT`

`= ("45.5 g")("2.460 J/g"^@ "C")(78.24^@ "C" - 25^@ "C")`

`=` `"5959.15 J"`

`=` `color(green)("5.959 kJ")`

That accounts for the temperature change up to the boiling point. Then, while boiling, the temperature is held constant, so that at the boiling point:

`DeltaH_"vap"^@ = q_"vap"/(n_"EtOH")`

`=> color(green)(q_"vap") = n_"EtOH"DeltaH_"vap"^@`

`= (("45.5 g EtOH")/("46.0684 g/mol"))("38.56 kJ/mol")`

`=` `color(green)("38.08 kJ")`

Thus, to vaporize `"45.5 g"` of ethanol starting at `25^@ "C"`, heating it up through to the boiling point of `78.24^@ "C"`, and turning it into the vapor phase, is:

`color(blue)(q_"tot") = q_"heating" + q_"vap"`

`= "5.959 kJ" + "38.08 kJ" =` `color(blue)("44.04 kJ")`