13.6 grams of ethanol are heated from 18.1°C to 105.2°C how many joules of energy were required? Ethanol heat of vaporization is ΔH(vap) 38.

Question

13.6 grams of ethanol are heated from 18.1°C to 105.2°C how many joules of energy were required? Ethanol heat of vaporization is ΔH(vap) 38.6 kJ/mol and heat of fusion is ΔH(fus) 4.60 kj/mol. Specific heats for the three phases of ethanol are: cp (liquid) ethanol 2.46 J/g°C, cp (gas) ethanol 1.43 J/g°C, cp (solid) ethanol 2.42 J/g°C. Ethanol is 46.0 g/mol. The melting point of ethanol is -114.1 °C and the boiling point is 78.4 °C. Answer value

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Edana Edana 4 years 2021-08-11T11:32:29+00:00 1 Answers 16 views 0

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  1. Ben
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    2021-08-11T11:33:32+00:00
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    Answer:

    13.9kJ of energy

    Explanation:

    This problem must be solved by stages: The heat of the liquid ethanol until boiling point. The change from liquid to gas and the heating of ethanol gas:

    Increasing liquid ethanol temperature from 18.1°C – 78.4°C

    Q = S*ΔT*m

    Where Q is heat, S is specific heat of liquid ethanol: 2.46J/g°C, ΔT is change in temperature: 78.4°C – 18.1°C = 60.3°C, m is mass of ethanol = 13.6g

    Q = 2.46J/g°C*60.3°C*13.6g

    Q = 2017J

    Change from liquid to gas:

    The heat of vaporization is defined as the heat required to convert 1 mole of a substance from liquid to gas. The energy required is:

    Q = ΔH(vap)*m / MW

    Where ΔH is 38.6kJ/mol, m is mass of ethanol = 13.6g, MW is molar weight of ethanol: 46.07g/mol

    Q = 11.4kJ

    Increasing gas ethanol temperature from 78.4°C – 105.2°C

    Q = S*ΔT*m

    Where Q is heat, S is specific heat of gas ethanol: 1.43J/g°C, ΔT is change in temperature: 26.8°C, m is mass of ethanol = 13.6g

    Q = 1.43J/g°C*26.8°C*13.6g

    Q = 521J

    Total energy in kJ:

    2.0kJ + 11.4kJ + 0.5kJ =

    13.9kJ of energy

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