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Question 10 0.5 pts Consider the heating curve of water below. What is the energy change (in kJ) when 38.5 g of water vapor (
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Answer #1

Ti = 135.0 oC

Tf = 95.0 oC

Cg = 2.0 J/g.oC

Heat released to convert vapour from 135.0 oC to 100.0 oC

Q1 = m*Cg*(Ti-Tf)

= 38.5 g * 2 J/g.oC *(135-100) oC

= 2695 J

Lv = 40.7KJ/mol =

40700J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 38.5/18.016

= 2.137 mol

Heat released to convert gas to liquid at 100.0 oC

Q2 = n*Lv

= 2.137 mol *40700 J/mol

= 86975.4663 J

Cl = 4.18 J/g.oC

Heat released to convert liquid from 100.0 oC to 95.0 oC

Q3 = m*Cl*(Ti-Tf)

= 38.5 g * 4.18 J/g.oC *(100-95) oC

= 804.65 J

Total heat released = Q1 + Q2 + Q3

= 2695 J + 86975.4663 J + 804.65 J

= 90475 J

= 90.5 KJ

Since it is heat released, please enter your answer with negative sign

Answer: -90.6 KJ

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