# 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"

##### 1 Answer
Dec 16, 2016

I got $\text{44.04 kJ}$ total.

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

• $\Delta {H}_{\text{vap"^@ = "38.56 kJ/mol}}$
• ${T}_{b} = {78.24}^{\circ} \text{C}$
• ${C}_{P} = \text{2.460 J/g"^@ "C}$

• 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, $T = {25}^{\circ} \text{C}$, and we assume $\Delta {H}_{\text{vap}}^{\circ}$ and ${C}_{P}$ vary negligibly within the temperature range of ${25}^{\circ} \text{C" -> 78.24^@ "C}$, then under natural conditions (constant pressure and varied temperature):

$\textcolor{g r e e n}{{q}_{\text{heating}}} = m {C}_{P} \Delta T$

$= \left(\text{45.5 g")("2.460 J/g"^@ "C")(78.24^@ "C" - 25^@ "C}\right)$

$=$ $\text{5959.15 J}$

$=$ $\textcolor{g r e e n}{\text{5.959 kJ}}$

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")

$=$ $\textcolor{g r e e n}{\text{38.08 kJ}}$

Thus, to vaporize $\text{45.5 g}$ of ethanol starting at ${25}^{\circ} \text{C}$, heating it up through to the boiling point of ${78.24}^{\circ} \text{C}$, and turning it into the vapor phase, is:

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

$= \text{5.959 kJ" + "38.08 kJ} =$ $\textcolor{b l u e}{\text{44.04 kJ}}$