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Answer :
We begin by calculating the amount of heat needed to raise the temperature of 18.0 g of ethanol from
[tex]$$23.0\,^\circ\text{C}$$[/tex]
to its boiling point at
[tex]$$78.3\,^\circ\text{C}.$$[/tex]
1. First, determine the temperature change:
[tex]$$
\Delta T = 78.3\,^\circ\text{C} - 23.0\,^\circ\text{C} = 55.3\,^\circ\text{C}.
$$[/tex]
2. Next, calculate the heat required for heating using the formula
[tex]$$
q_{\text{heating}} = m \cdot c \cdot \Delta T,
$$[/tex]
where
- [tex]$m = 18.0\,\text{g}$[/tex] is the mass of ethanol,
- [tex]$c = 2.46\,\text{J/(g}\cdot^\circ\text{C)}$[/tex] is the specific heat capacity of ethanol.
Substituting the values:
[tex]$$
q_{\text{heating}} = 18.0\,\text{g} \cdot 2.46\,\text{J/(g}\cdot^\circ\text{C)} \cdot 55.3\,^\circ\text{C} \approx 2448.68\,\text{J}.
$$[/tex]
Converting joules to kilojoules:
[tex]$$
q_{\text{heating}} \approx \frac{2448.68\,\text{J}}{1000} \approx 2.45\,\text{kJ}.
$$[/tex]
3. Now, determine the number of moles of ethanol. The molar mass of ethanol is approximately
[tex]$$46.07\,\text{g/mol}.$$[/tex]
Thus,
[tex]$$
n = \frac{18.0\,\text{g}}{46.07\,\text{g/mol}} \approx 0.3907\,\text{mol}.
$$[/tex]
4. Calculate the heat required for vaporization using the enthalpy of vaporization:
[tex]$$
q_{\text{vap}} = n \cdot \Delta H_{\text{vap}},
$$[/tex]
where
- [tex]$\Delta H_{\text{vap}} = 39.3\,\text{kJ/mol}$[/tex].
Hence,
[tex]$$
q_{\text{vap}} = 0.3907\,\text{mol} \cdot 39.3\,\text{kJ/mol} \approx 15.35\,\text{kJ}.
$$[/tex]
5. Finally, add the heat required for heating and vaporization to obtain the total heat:
[tex]$$
q_{\text{total}} = q_{\text{heating}} + q_{\text{vap}} \approx 2.45\,\text{kJ} + 15.35\,\text{kJ} \approx 17.80\,\text{kJ}.
$$[/tex]
Thus, the total quantity of heat required to convert 18.0 g of ethanol at [tex]$23.0\,^\circ\text{C}$[/tex] to vapor at [tex]$78.3\,^\circ\text{C}$[/tex] is approximately
[tex]$$\boxed{17.80\,\text{kJ}}.$$[/tex]
[tex]$$23.0\,^\circ\text{C}$$[/tex]
to its boiling point at
[tex]$$78.3\,^\circ\text{C}.$$[/tex]
1. First, determine the temperature change:
[tex]$$
\Delta T = 78.3\,^\circ\text{C} - 23.0\,^\circ\text{C} = 55.3\,^\circ\text{C}.
$$[/tex]
2. Next, calculate the heat required for heating using the formula
[tex]$$
q_{\text{heating}} = m \cdot c \cdot \Delta T,
$$[/tex]
where
- [tex]$m = 18.0\,\text{g}$[/tex] is the mass of ethanol,
- [tex]$c = 2.46\,\text{J/(g}\cdot^\circ\text{C)}$[/tex] is the specific heat capacity of ethanol.
Substituting the values:
[tex]$$
q_{\text{heating}} = 18.0\,\text{g} \cdot 2.46\,\text{J/(g}\cdot^\circ\text{C)} \cdot 55.3\,^\circ\text{C} \approx 2448.68\,\text{J}.
$$[/tex]
Converting joules to kilojoules:
[tex]$$
q_{\text{heating}} \approx \frac{2448.68\,\text{J}}{1000} \approx 2.45\,\text{kJ}.
$$[/tex]
3. Now, determine the number of moles of ethanol. The molar mass of ethanol is approximately
[tex]$$46.07\,\text{g/mol}.$$[/tex]
Thus,
[tex]$$
n = \frac{18.0\,\text{g}}{46.07\,\text{g/mol}} \approx 0.3907\,\text{mol}.
$$[/tex]
4. Calculate the heat required for vaporization using the enthalpy of vaporization:
[tex]$$
q_{\text{vap}} = n \cdot \Delta H_{\text{vap}},
$$[/tex]
where
- [tex]$\Delta H_{\text{vap}} = 39.3\,\text{kJ/mol}$[/tex].
Hence,
[tex]$$
q_{\text{vap}} = 0.3907\,\text{mol} \cdot 39.3\,\text{kJ/mol} \approx 15.35\,\text{kJ}.
$$[/tex]
5. Finally, add the heat required for heating and vaporization to obtain the total heat:
[tex]$$
q_{\text{total}} = q_{\text{heating}} + q_{\text{vap}} \approx 2.45\,\text{kJ} + 15.35\,\text{kJ} \approx 17.80\,\text{kJ}.
$$[/tex]
Thus, the total quantity of heat required to convert 18.0 g of ethanol at [tex]$23.0\,^\circ\text{C}$[/tex] to vapor at [tex]$78.3\,^\circ\text{C}$[/tex] is approximately
[tex]$$\boxed{17.80\,\text{kJ}}.$$[/tex]
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