1. Calculate the heat required to melt 9.43 g of benzene at its normal melting p
ID: 1002224 • Letter: 1
Question
1. Calculate the heat required to melt 9.43 g of benzene at its normal melting point. Heat of fusion (benzene) = 9.92 kJ/mol Heat = _______ kJ 2.Calculate the heat required to vaporize 9.43 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol Heat = _______ kJ 3.What quantity of energy does it take to convert 0.800 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g·°C; liquid, 4.2 J/g·°C; steam, 2.0 J/g·°C; delta H vap = 40.7 kJ/mol; delta H fus = 6.02 kJ/mol. Energy = _______ kJ 4. A 23.2-g sample of ice at -12.1°C is mixed with 118.2 g of water at 80.0°C. Calculate the final temperature of the mixture, assuming no heat loss to the surroundings. The heat capacities of H2O(s) and H2O(l) are 2.03 and 4.18 J/g°C, respectively, and the enthalpy of fusion for ice is 6.02 kJ/mol. Final temperature = ______ CExplanation / Answer
1. heat required to melt benzene
q = mdHf
with,
m = 9.43 g
dHf = 9.92 kJ/mol
we get heat required to melt,
q = 9.43 x 9.92/78 = 1.20 kJ
2. Heat required to vaporise benzene
q = mdHv
with,
m = 9.43 g
dHf = 30.7 kJ/mol
we get heat required to melt,
q = 9.43 x 30.7/78 = 3.71 kJ
3. Energy required to convert ice to steam
q1 = ice at -20 oC to ice at 0 oC
= mCpdT = 800 g x 2.03 x (0 - (-20) = 32.48 kJ
q2 = ice at -20 oC to water at 0 oC
= mdHf = 800 g x 6.02/18 = 267.55 kJ
q3 = water at 0 oC to water at 100 oC
= mCpdT = 800 g x 4.2 x (100 - 0) = 336 kJ
q4 = water at 100 oC to steam at 100 oC
= mdHv = 800 g x 40.7/18 = 1809 kJ
q5 = steam at 100 oC to steam at 250 oC
= mCpdT = 800 g x 2.0 x (250 - 100) = 240 kJ
Total energy required = q1 + q2 + q3 + q4 + q5 = 2685.03 kJ
4. Let Tf be the final temperature
heat lost = heat gained
118.2 x 4.18 x (80 - Tf) = 23.2 x 2.03 x (Tf - (-12.1)) + 23.2 x 6.02/18
39526.08 - 494.076Tf = 47.096Tf + 569.8616 + 7.76
Tf = 38948.458/541.172 = 71.97 oC
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