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 =

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )

catya that's not even a b or c