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Question

::-owLv2] Online teaching an × tvity ass C Search Use the Refereaces to access importaat values if meeded for this question The following information is given for aluminum at latm 2467.00°C dHrap(2467.000-1.053 × 104 Lig (660.00%) = 398.41g Tm = 660.00°C Specific heat solid = 0.9000 J, g Specific heat liquid .oss ig °C A 24.10 g sample of liquid aluminum at 923.00°C is poured into a mold and allowed to cool to 25.00°C. How many kJ of energy are released in this proces (Report the answer as a positive number) ) Energy kJ Submit Answer 8 more group attempts remaining

Explanation / Answer

Ti = 923.0 oC

Tf = -25.0 oC

Cl = 0.9 J/g.oC

Heat released to convert liquid from 923.0 oC to 660.0 oC

Q1 = m*Cl*(Ti-Tf)

= 24.1 g * 0.9 J/g.oC *(923-660) oC

= 5704.47 J

Lf = 398.4 J/g

Heat released to convert liquid to solid at 660.0 oC

Q2 = m*Lf

= 24.1g *398.4 J/g

= 9601.44 J

Cs = 1.088 J/g.oC

Heat released to convert solid from 660.0 oC to -25.0 oC

Q3 = m*Cs*(Ti-Tf)

= 24.1 g * 1.088 J/g.oC *(660--25) oC

= 17961.248 J

Total heat released = Q1 + Q2 + Q3

= 5704.47 J + 9601.44 J + 17961.248 J

= 33267 J

= 33.3 KJ

Answer: 33.3 KJ

Feel free to comment below if you have any doubts or if this answer do not work

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