Answer :
To calculate the amount of heat needed to boil 81.1 g of octane (C8H18), starting from a temperature of 100.5 °C, we need to consider two main components: the heat required to raise the temperature of octane to its boiling point and the heat required to vaporize it.
Here's the step-by-step solution:
1. Calculate Moles of Octane:
- First, determine how many moles of octane you have. Use the formula:
[tex]\[
\text{moles of octane} = \frac{\text{mass of octane}}{\text{molar mass of octane}}
\][/tex]
- Given:
- Mass of octane = 81.1 g
- Molar mass of octane = 114.22 g/mol
- Calculation:
[tex]\[
\text{moles of octane} = \frac{81.1 \, \text{g}}{114.22 \, \text{g/mol}} \approx 0.710 \, \text{mol}
\][/tex]
2. Calculate Heat Required to Increase Temperature to Boiling Point:
- Use the formula for heat change due to temperature change:
[tex]\[
q_1 = \text{mass} \times \text{specific heat} \times \Delta T
\][/tex]
- Given:
- Specific heat of octane = 2.25 J/g°C
- Initial temperature = 100.5 °C
- Boiling point of octane = 125.7 °C
- [tex]\(\Delta T = \text{boiling point} - \text{initial temperature} = 125.7 - 100.5 = 25.2\)[/tex] °C
- Calculation:
[tex]\[
q_1 = 81.1 \, \text{g} \times 2.25 \, \text{J/g°C} \times 25.2 \, \text{°C} \approx 4570 \, \text{J} = 4.57 \, \text{kJ}
\][/tex]
3. Calculate Heat Required to Vaporize Octane:
- Use the formula for heat required to vaporize:
[tex]\[
q_2 = \text{moles of octane} \times \text{heat of vaporization}
\][/tex]
- Given:
- Heat of vaporization for octane = 26.49 kJ/mol
- Calculation:
[tex]\[
q_2 = 0.710 \, \text{mol} \times 26.49 \, \text{kJ/mol} \approx 18.81 \, \text{kJ}
\][/tex]
4. Calculate Total Heat Required:
- Add the heats calculated in steps 2 and 3:
[tex]\[
\text{Total heat} = q_1 + q_2 = 4.57 \, \text{kJ} + 18.81 \, \text{kJ} \approx 23.38 \, \text{kJ}
\][/tex]
Therefore, the total amount of heat needed to boil 81.1 g of octane, starting from 100.5 °C, is approximately [tex]\(23.4 \, \text{kJ}\)[/tex].
Here's the step-by-step solution:
1. Calculate Moles of Octane:
- First, determine how many moles of octane you have. Use the formula:
[tex]\[
\text{moles of octane} = \frac{\text{mass of octane}}{\text{molar mass of octane}}
\][/tex]
- Given:
- Mass of octane = 81.1 g
- Molar mass of octane = 114.22 g/mol
- Calculation:
[tex]\[
\text{moles of octane} = \frac{81.1 \, \text{g}}{114.22 \, \text{g/mol}} \approx 0.710 \, \text{mol}
\][/tex]
2. Calculate Heat Required to Increase Temperature to Boiling Point:
- Use the formula for heat change due to temperature change:
[tex]\[
q_1 = \text{mass} \times \text{specific heat} \times \Delta T
\][/tex]
- Given:
- Specific heat of octane = 2.25 J/g°C
- Initial temperature = 100.5 °C
- Boiling point of octane = 125.7 °C
- [tex]\(\Delta T = \text{boiling point} - \text{initial temperature} = 125.7 - 100.5 = 25.2\)[/tex] °C
- Calculation:
[tex]\[
q_1 = 81.1 \, \text{g} \times 2.25 \, \text{J/g°C} \times 25.2 \, \text{°C} \approx 4570 \, \text{J} = 4.57 \, \text{kJ}
\][/tex]
3. Calculate Heat Required to Vaporize Octane:
- Use the formula for heat required to vaporize:
[tex]\[
q_2 = \text{moles of octane} \times \text{heat of vaporization}
\][/tex]
- Given:
- Heat of vaporization for octane = 26.49 kJ/mol
- Calculation:
[tex]\[
q_2 = 0.710 \, \text{mol} \times 26.49 \, \text{kJ/mol} \approx 18.81 \, \text{kJ}
\][/tex]
4. Calculate Total Heat Required:
- Add the heats calculated in steps 2 and 3:
[tex]\[
\text{Total heat} = q_1 + q_2 = 4.57 \, \text{kJ} + 18.81 \, \text{kJ} \approx 23.38 \, \text{kJ}
\][/tex]
Therefore, the total amount of heat needed to boil 81.1 g of octane, starting from 100.5 °C, is approximately [tex]\(23.4 \, \text{kJ}\)[/tex].
