The following information is given for lead at 1atm: boiling point = 1.740 times
ID: 496007 • Letter: T
Question
The following information is given for lead at 1atm: boiling point = 1.740 times 10^3 degree C melting point = 328.0 degree C specific heat solid = 0.1300 J/g degree C specific heat liquid = 0.1380 J/g degree C delta H_vap (1.740 times 10^3 degree C) = 858.2 J/g delta H_ (328.0 degree C) = 23.00 J/g A 42.10 g sample of liquid lead at 533.0 degree C poured into a mold and allowed to cool to 20.0 degree C. How many kJ of energy are released in this process. Report the answer as a positive number. kJExplanation / Answer
Melting point --> 328 °C...
total Q:
+ Qliquid from 533°C to 328 °C
+ Qliquid to solid ( freezing )
+ Qsolid from 328°C to 20°C
so...
Qliquid from 533°C to 328 °C = m*Cliquid*(Tfreezing - Tinitial)
Qliquid to solid ( freezing ) = latent Heat of freezing * mass
Qsolid from 328°C to 20°C= m*Csolid *(Tfinal- Tfreezing )
substitute values
Qliquid from 533°C to 328 °C = m*Cliquid*(Tfreezing - Tinitial) = (42.10*0.138)(328-533) = -1191.009 J
Qliquid to solid ( freezing ) = latent Heat of freezing * mass = 42.10*(-23) = -968.3 J
Qsolid from 328°C to 20°C= m*Csolid *(Tfinal- Tfreezing ) = 42.10*0.13*(20-328) = -1685.684 J
total Q = -(1191.009 +968.3 +1685.684 ) = -3844.993 J
Q in kJ = (3844.993)/1000 = 3.8449 kJ of heat is released
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