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The following information is given for cadmium at latm: T = 765.00°C T. - 321.00°C Specific heat solid = 0.2300 J/g °C Specif
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Answer #1

Ti = 386.0 oC

Tf = 25.0 oC

Cl = 0.264 J/g.oC

Heat released to convert liquid from 386.0 oC to 321.0 oC

Q1 = m*Cl*(Ti-Tf)

= 32.1 g * 0.264 J/g.oC *(386-321) oC

= 550.836 J

Lf = 54.4 J/g

Heat released to convert liquid to solid at 321.0 oC

Q2 = m*Lf

= 32.1g *54.4 J/g

= 1746.24 J

Cs = 0.23 J/g.oC

Heat released to convert solid from 321.0 oC to 25.0 oC

Q3 = m*Cs*(Ti-Tf)

= 32.1 g * 0.23 J/g.oC *(321-25) oC

= 2185.368 J

Total heat released = Q1 + Q2 + Q3

= 550.836 J + 1746.24 J + 2185.368 J

= 4482.4 J

= 4.4824 KJ

Answer: 4.4824 KJ

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