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

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 =

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