Answer :
Sure! Let's calculate the quantity of heat required to convert 50.0 grams of ethanol from 23.0°C to vapor at its boiling point, 78.3°C. We'll do this in two steps: heating the ethanol to its boiling point, and then converting it to vapor.
### Step 1: Calculate the Heat Required to Raise the Temperature
1. Identify the variables:
- Mass of ethanol ([tex]\(m\)[/tex]): 50.0 grams
- Initial temperature ([tex]\(T_i\)[/tex]): 23.0°C
- Boiling point ([tex]\(T_b\)[/tex]): 78.3°C
- Specific heat capacity ([tex]\(c\)[/tex]): 2.46 J/g°C
2. Calculate heat required to raise the temperature to the boiling point:
The formula to calculate the heat ([tex]\(q\)[/tex]) needed to change the temperature is:
[tex]\[
q = m \times c \times \Delta T
\][/tex]
where [tex]\(\Delta T\)[/tex] is the temperature change:
[tex]\[
\Delta T = T_b - T_i = 78.3°C - 23.0°C = 55.3°C
\][/tex]
So,
[tex]\[
q = 50.0 \text{ g} \times 2.46 \text{ J/g°C} \times 55.3°C
\][/tex]
Calculate [tex]\(q\)[/tex] and convert it to kilojoules by dividing by 1000:
[tex]\[
q \approx 6.802 \text{ kJ}
\][/tex]
### Step 2: Calculate the Heat Required for Vaporization
1. Determine the molar mass of ethanol ([tex]\(C_2H_5OH\)[/tex]):
- Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol
- Hydrogen (H): 6 atoms × 1.008 g/mol = 6.048 g/mol
- Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol
Total molar mass of ethanol = 24.02 + 6.048 + 16.00 = 46.068 g/mol
2. Calculate the moles of ethanol:
[tex]\[
\text{Moles of ethanol} = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \text{ g}}{46.068 \text{ g/mol}} \approx 1.085 \text{ moles}
\][/tex]
3. Use the heat of vaporization ([tex]\(\Delta H_{vap}\)[/tex]):
- Given [tex]\(\Delta H_{vap} = 39.3 \text{ kJ/mol}\)[/tex]
[tex]\[
\text{Heat for vaporization} = \text{moles} \times \Delta H_{vap} = 1.085 \text{ moles} \times 39.3 \text{ kJ/mol}
\][/tex]
[tex]\[
\approx 42.654 \text{ kJ}
\][/tex]
### Step 3: Calculate Total Heat Required
Finally, add the heat required to raise the temperature and the heat required for vaporization:
[tex]\[
\text{Total heat} = 6.802 \text{ kJ} + 42.654 \text{ kJ} \approx 49.456 \text{ kJ}
\][/tex]
Therefore, the total quantity of heat required to convert 50.0 g of ethanol from 23.0°C to vapor at its boiling point of 78.3°C is approximately 49.46 kJ.
### Step 1: Calculate the Heat Required to Raise the Temperature
1. Identify the variables:
- Mass of ethanol ([tex]\(m\)[/tex]): 50.0 grams
- Initial temperature ([tex]\(T_i\)[/tex]): 23.0°C
- Boiling point ([tex]\(T_b\)[/tex]): 78.3°C
- Specific heat capacity ([tex]\(c\)[/tex]): 2.46 J/g°C
2. Calculate heat required to raise the temperature to the boiling point:
The formula to calculate the heat ([tex]\(q\)[/tex]) needed to change the temperature is:
[tex]\[
q = m \times c \times \Delta T
\][/tex]
where [tex]\(\Delta T\)[/tex] is the temperature change:
[tex]\[
\Delta T = T_b - T_i = 78.3°C - 23.0°C = 55.3°C
\][/tex]
So,
[tex]\[
q = 50.0 \text{ g} \times 2.46 \text{ J/g°C} \times 55.3°C
\][/tex]
Calculate [tex]\(q\)[/tex] and convert it to kilojoules by dividing by 1000:
[tex]\[
q \approx 6.802 \text{ kJ}
\][/tex]
### Step 2: Calculate the Heat Required for Vaporization
1. Determine the molar mass of ethanol ([tex]\(C_2H_5OH\)[/tex]):
- Carbon (C): 2 atoms × 12.01 g/mol = 24.02 g/mol
- Hydrogen (H): 6 atoms × 1.008 g/mol = 6.048 g/mol
- Oxygen (O): 1 atom × 16.00 g/mol = 16.00 g/mol
Total molar mass of ethanol = 24.02 + 6.048 + 16.00 = 46.068 g/mol
2. Calculate the moles of ethanol:
[tex]\[
\text{Moles of ethanol} = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \text{ g}}{46.068 \text{ g/mol}} \approx 1.085 \text{ moles}
\][/tex]
3. Use the heat of vaporization ([tex]\(\Delta H_{vap}\)[/tex]):
- Given [tex]\(\Delta H_{vap} = 39.3 \text{ kJ/mol}\)[/tex]
[tex]\[
\text{Heat for vaporization} = \text{moles} \times \Delta H_{vap} = 1.085 \text{ moles} \times 39.3 \text{ kJ/mol}
\][/tex]
[tex]\[
\approx 42.654 \text{ kJ}
\][/tex]
### Step 3: Calculate Total Heat Required
Finally, add the heat required to raise the temperature and the heat required for vaporization:
[tex]\[
\text{Total heat} = 6.802 \text{ kJ} + 42.654 \text{ kJ} \approx 49.456 \text{ kJ}
\][/tex]
Therefore, the total quantity of heat required to convert 50.0 g of ethanol from 23.0°C to vapor at its boiling point of 78.3°C is approximately 49.46 kJ.
