The following information is given for bismuth at 1 atm:
boiling point = 1.627E3 °C | Hvap(1.627E3°C) = 822.9 J/g |
melting point = 271.0 °C | Hfus(271.0°C) = 52.60 J/g |
specific heat solid= 0.1260 J/g ⋅ °C | |
specific heat liquid = 0.1510 J/g ⋅ °C |
A 48.90 g sample of liquid bismuth at 504.0 °C is poured into a mold and allowed to cool to 24.0 °C. How many kJ of energy are released in this process. Report the answer as a positive number.
Ti = 504.0 oC
Tf = 24.0 oC
Cl = 0.151 J/g.oC
Heat released to convert liquid from 504.0 oC to 271.0 oC
Q1 = m*Cl*(Ti-Tf)
= 48.9 g * 0.151 J/g.oC *(504-271) oC
= 1720.4487 J
Lf = 52.6 J/g
Heat released to convert liquid to solid at 271.0 oC
Q2 = m*Lf
= 48.9g *52.6 J/g
= 2572.14 J
Cs = 0.126 J/g.oC
Heat released to convert solid from 271.0 oC to 24.0 oC
Q3 = m*Cs*(Ti-Tf)
= 48.9 g * 0.126 J/g.oC *(271-24) oC
= 1521.8658 J
Total heat released = Q1 + Q2 + Q3
= 1720.4487 J + 2572.14 J + 1521.8658 J
= 5814.4545 J
= 5.814 KJ
Answer: 5.814 KJ
Get Answers For Free
Most questions answered within 1 hours.