Homework Answers
Ti = 41.9 oC
Tf = 114.4 oC
here
Cl = 4.184 J/g.oC
Heat required to convert liquid from 41.9 oC to 100.0 oC
Q1 = m*Cl*(Tf-Ti)
= 48.6 g * 4.184 J/g.oC *(100-41.9) oC
= 11814.1934 J
Lv = 2259.0 J/g
Heat required to convert liquid to gas at 100.0 oC
Q2 = m*Lv
= 48.6g *2259.0 J/g
= 109787.4 J
Cg = 2.01 J/g.oC
Heat required to convert vapour from 100.0 oC to 114.4 oC
Q3 = m*Cg*(Tf-Ti)
= 48.6 g * 2.01 J/g.oC *(114.4-100) oC
= 1406.6784 J
Total heat required = Q1 + Q2 + Q3
= 11814.1934 J + 109787.4 J + 1406.6784 J
= 123008 J
= 123 KJ
Answer: 123 KJ
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Use the References to access important valnes if needed for this question. The following information is given for benzene at 1 atm: boiling point 80.10 °C AHap(80.10 °C)-393.3 J/g melting point 5.500 °C AHf(5.500 °C)- 127.4 J/g specific heat gas 1.040 J/gC specific heat liquid= 1.740 J/g°C A 35.70 g sample of liquid benzene is initially at 40.90 °C. If the sample is heated at constant pressure (P 1 atm), kJ of energy are needed to raise the temperature of...
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