Answer :
The total quantity of heat required to convert 17 g of ethanol at 23°C to a vapor at 78.3°C is approximately 16.85 kJ, by calculating the heat needed to increase the temperature and the heat needed for phase transition.
The quantity of heat required to convert 17.0 g of ethanol from 23.0 ∘C to a vapor at 78.3∘C consists of two components: the heat needed to raise the temperature to the boiling point, and the heat needed to cause phase transition (from liquid to gas).
To calculate the first component, we use the formula:q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.
So, q = 17.0 g * 2.46 J/g°C * (78.3°C - 23.0°C) ≈ 2,349 J or 2.35 kJ.
For the second component, we should first convert the grams to moles using ethanol's molar mass (46.07g/mol). ΔHvap is given per mole,so we need moles of ethanol which is 17.0 g ÷ 46.07 g/mol = 0.369 mol of ethanol.
So, q for the phase transition = 0.369 mol * 39.3 kJ/mol ≈ 14.5 kJ
By adding these two quantities, we find that the total quantity of heat required is approximately 16.85 kJ.
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