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------------------------------------------------ What quantity of heat, in kJ, is required to convert 29.0 g of ethanol ([tex]C_2H_5OH[/tex]) at 23.0 [tex]^\circ C[/tex] to a vapor at 78.3 [tex]^\circ C[/tex] (its boiling point)?

Given:
- Specific heat capacity of ethanol = 2.46 J/g⋅C
- [tex]\Delta H_{\text{vap}} = 39.3 \text{ kJ/mol}[/tex]

To calculate the quantity of heat required to convert ethanol from a liquid to a vapor, we need to consider two processes: heating the liquid ethanol to its boiling point and vaporizing it at its boiling point. Firstly, we calculate the heat required to raise the temperature of the ethanol:

q1 = mass × specific heat capacity × temperature change

q1 = 29.0 g × 2.46 J/g·C × (78.3 - 23.0) C

q1 = 7455.3 J = 7.5 kJ

Next, we calculate the heat required for vaporization:

q2 = moles × molar heat of vaporization

First, we need to convert grams of ethanol to moles:

moles = mass / molar mass

moles = 29.0 g / 46 g/mol

moles = 0.63 mol

Now, we can calculate q2:

q2 = 0.63 mol × 39.3 kJ/mol

q2 = 24.72 kJ

The total quantity of heat required to convert 29.0 g of ethanol at 23.0°C to a vapor at 78.3°C is the sum of q1 and q2:

q = q1 + q2 = 7.5 kJ + 24.72 kJ = 32.22 kJ

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