Question

The following information is given for aluminum at 1 atm: boiling point = 2.467E3 °C Hvap(2.467E3°C)...

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

Homework Answers

Answer #1

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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