Chapter 10, Problem 40QP

Calculated the heat released when 84.6 g of gaseous ethanol at $90.0°\text{C}$ is converted to liquid ethanol at  $54.2°\text{C}\text{.}$ The boiling point of ethanol is $78.4°\text{C}\text{.}$ (See Question 10.39 for heat constant that may be useful in answering this question).

Interpretation Introduction

Interpretation:

The heat required for the conversion of $\text{84}\text{.6}\text{g}$ gaseous ethanol at $90°\text{C}$ to liquid ethanol at $54.2°\text{C}$ is to be determined.

Concept Introduction:

Molar heat of evaporation is the heat required for a mole of liquid to convert it into gas without changing the temperature.

Specific heat of a substance is the heat required for the substance to increase the temperature of $1.0\text{g}$ of a substance by $1.0°\text{C}$ .

Answer to Problem 40QP

Solution:

The heat required for the conversion of $\text{84}\text{.6}\text{g}$ gaseous ethanol at $90°\text{C}$ to liquid alcohol at $54.2°\text{C}$ is $-77.4\text{kJ}$ .

Explanation of Solution

The heat required is calculated by a three-step process. The first step involves the cooling of gaseous ethanol from $90°\text{C}$ till its boiling point $78.4°\text{C}$ , after which all of the ethanol will be converted into liquid form, which requires heat of evaporation. The final step involves the cooling the liquid ethanol till $54.2°\text{C}$ .

The heat change associated with the given process is given by the expression as shown below.

$q=mc\Delta T$ …… (1)

Here, $q$ is heat in joules, $m$ is the mass of the substance, $c$ is the specific heat of the substance, and $\Delta T$ is the temperature change.

Now, heat released in the cool of gaseous ethanol is calculated by considering the temperature change from $90°\text{C}$ to $78.4°\text{C}$ in equation (1).

So, the temperature change is calculated as shown below.

$\begin{array}{c}\Delta T=T_2-T_1\\ =78.4°\text{C}-90°\text{C}\\ =-\text{11}\text{.6}°\text{C}\end{array}$

So, theheat released is calculated as,

$\begin{array}{c}q=-84.6\text{<menclose notation="updiagonalstrike">g </menclose>}\times 1.42\text{J/(<menclose notation="updiagonalstrike">g </menclose>}°\text{<menclose notation="updiagonalstrike">C </menclose>)}\times 11.6°\text{<menclose notation="updiagonalstrike">C </menclose>}\\ =-\text{1}\text{.393}\times {\text{10}}^3\text{J}\end{array}$ $$

So, the number moles of ethanol in $\text{84}\text{.6}\text{g}$ is calculated as shown below.

$\begin{array}{c}n=84.6\cancel{\text{g}\text{ethanol}}\times \frac{1\text{ mol}\text{ethanol}}{46.07\cancel{\text{g}\text{ethanol}}}\\ =1.84\text{mol}\end{array}$

Now, the heat of vaporization for ethanol is $\text{3}{\text{.86×10}}^{\text{4}}\text{J/mol}$ , which can be used to calculate the heat required to condense all the gaseous ethanol.

$\begin{array}{c}{q}_{\text{condensation}}=1.84\cancel{\text{mol}}\text{×}\left(-\text{3}{\text{.86×10}}^4\text{J}/\cancel{\text{mol}}\right)\\ =-7.10{\text{×10}}^4\text{J}\end{array}$

Now, the heat released in the cool of liquid ethanol is calculated by considering the temperature change from $78.4°\text{C}$ to $54.2°\text{C}$ in equation (1).

The temperature change is calculated as shown below.

$\begin{array}{c}\Delta T=T_2-T_1\\ =54.2°\text{C}-78.4°\text{C}\\ =-\text{24}\text{.2}°\text{C}\end{array}$

So, the heat released is calculated as shown below.

$\begin{array}{c}q=-84.6\text{<menclose notation="updiagonalstrike">g </menclose>}\times 2.44\text{J/(<menclose notation="updiagonalstrike">g </menclose>°<menclose notation="updiagonalstrike">C </menclose>)}\times 24.2°\text{<menclose notation="updiagonalstrike">C </menclose>}\\ =-5.00\times {10}^3\text{J}\end{array}$

Total heat released is the sum of the above three steps.

$\begin{array}{c}q=\left(-1.393\times {10}^3\text{J}\right)+\left(-7.10\times {10}^4\text{J}\right)+\left(-5.00\times {10}^3\text{J}\right)\\ =-7.74\times {10}^4\text{J}\\ =-77.4\text{kJ}\end{array}$

Conclusion

The total heat change associated with the given process is $-77.4\text{kJ}$ .