The following information is given for bismuth at 1atm: boiling point = 1.627 ti
ID: 482614 • Letter: T
Question
The following information is given for bismuth at 1atm: boiling point = 1.627 times 10^3 degree C Delta H_vap(1.627 times 10^3 degree C) 822.9 J/g melting point = 271.0 degree C Delta H_fus(271.0 degree C) - 52.60 J/g specific heat solid = 0.1260 J/g degree C specific heat liquid 0.1510 J/g degree C A 34.30 g sample of liquid bismuth at 468.0 degree C is poured into a mold and allowed to cool to 20.0 degree C. How many kJ of energy arc released in this process. Report the answer as a positive number. _ kJExplanation / Answer
in this proccess there are first temperature (change 468.0 0C to 271.0 0C) and phase change (liquid to solid) then a temperature change (below 271.0 0C).
q = (34.30 g) (468.0 °C - 271 oC) (0.1510 J g¯1 °C¯1) = 1020.3221 J = 1.02 kJ
phase change
q = (52.60 J/g) (34.30 g ) = 1804.18 J = 1.804 KJ
temperature change in solid
q = (34.30 g) (271.0 °C -20.0 oC) (0.1260 J g¯1 °C¯1) = 1084.77 J = 1.084 kJ
total energy released = 1.084 KJ +1.804 KJ + 1.02 KJ
q = 3.908 kJ (rounded to the appropriate number of significant figures)
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