The following information is given for aluminum at 1 atm: boiling point = 2.467E3 °C Hvap(2.467E3°C) = 1.053E4 J/g melting point = 660.0 °C Hfus(660.0°C) = 398.4 J/g specific heat solid= 0.9000 J/g ⋅ °C specific heat liquid = 1.088 J/g ⋅ °C A 46.10 g sample of liquid aluminum at 951.0 °C is poured into a mold and allowed to cool to 28.0 °C. How many kJ of energy are released in this process. Report the answer as a positive number. __kJ
Ti = 951.0 oC
Tf = 128.0 oC
Cl = 1.088 J/g.oC
Heat released to convert liquid from 951.0 oC to 660.0 oC
Q1 = m*Cl*(Ti-Tf)
= 46.1 g * 1.088 J/g.oC *(951-660) oC
= 14595.6288 J
Lf = 398.4 J/g
Heat released to convert liquid to solid at 660.0 oC
Q2 = m*Lf
= 46.1g *398.4 J/g
= 18366.24 J
Cs = 0.9 J/g.oC
Heat released to convert solid from 660.0 oC to 128.0 oC
Q3 = m*Cs*(Ti-Tf)
= 46.1 g * 0.9 J/g.oC *(660-128) oC
= 22072.68 J
Total heat released = Q1 + Q2 + Q3
= 14595.6288 J + 18366.24 J + 22072.68 J
= 55035 J
= 55.0 KJ
Answer: 55.0 KJ
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