Calculate the amount of heat energy, in units of kilojoules, required to convert 50.0 g of liquid ethanol, CH₃CH₂OH, at 25.5°C to gaseous ethanol at 92.0°C. (C_{ethanol liquid} = 2.44 J/g°C; C_{ethanol gas} = 1.42 J/g°C; molar heat of vaporization of liquid ethanol = 3.86 × 10⁴ J/mol. The boiling point of ethanol is 78.4°C and its molar mass is 46.07 g/mol.)

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To calculate the total heat energy required, we need to consider three steps: heating the liquid ethanol from 25.5°C to its boiling point (78.4°C), vaporizing the ethanol, and then heating the vapor from 78.4°C to 92.0°C.1. Heating the liquid ethanol to its boiling point:$$q_1 = m \cdot c_{liquid} \cdot \Delta T = 50.0 \, \text{g} \cdot 2.44 \, \text{J/g°C} \cdot (78.4°C - 25.5°C) = 50.0 \cdot 2.44 \cdot 52.9 = 6456.6 \, \text{J}$$2. Vaporizing the ethanol:The molar heat of vaporization is given per mole, so first, we find the moles of ethanol:$$n = \frac{50.0 \, \text{g}}{46.07 \, \text{g/mol}} = 1.085 \, \text{mol}$$Then, we calculate the heat required to vaporize this amount:$$q_2 = n \cdot \Delta H_{vap} = 1.085 \, \text{mol} \cdot 3.86 \times 10^4 \, \text{J/mol} = 41905.1 \, \text{J}$$3. Heating the vapor to the final temperature:$$q_3 = m \cdot c_{vapor} \cdot \Delta T = 50.0 \, \text{g} \cdot 1.42 \, \text{J/g°C} \cdot (92.0°C - 78.4°C) = 50.0 \cdot 1.42 \cdot 13.6 = 964.8 \, \text{J}$$Adding these together:$$q_{total} = q_1 + q_2 + q_3 = 6456.6 \, \text{J} + 41905.1 \, \text{J} + 964.8 \, \text{J} = 49326.5 \, \text{J}$$Converting to kilojoules:$$q_{total} = 49.3265 \, \text{kJ}$$Rounded to three significant figures, the total heat energy required is approximately 49.3 kJ, which is closest to option A, 49.2 kJ, considering possible rounding differences in calculation steps.

Calculate the Amount of Heat Energy, in Units of Kilojoules