The following information is given for aluminum at 1atm: boiling point = 2.467 t
ID: 489158 • Letter: T
Question
The following information is given for aluminum at 1atm: boiling point = 2.467 times 10^3 degree C delta H_vap(2.467 times 10^3 degree C) = 1.053 times 10^4 J/g melting point = 660.0 degree C delta H_fus(660.0 degree C) = 398.4 J/g specific heat solid = 0.9000 J/g degree C specific heat liquid = 1.088 J/g degree C A 30.10 g sample of solid aluminum is initially at 648.0 degree C. If the sample is heated at constant pressure (P = 1 atm). kJ of heat are needed to raise the temperature of the sample to 1.190 times 10^3 degree C.Explanation / Answer
There are three processes involved:
1) heating solid Al from 648.0C to its melting point, 660C.
2) melting solid Al to liquid Al at 660C.
3) heating liquid Al from 660C to 1.190*103 C = 1190C
Find out the heat involved in these processes.
Q1 = (mass of solid Al)*(specific heat of solid Al)*(temperature change) = (30.10 g)*(0.9000 J/g.C)*(660 – 648)C = 325.08 J (ans).
Q2 = (mass of solid Al)*Hfus = (30.10 g)*(398.4 J/g) = 11991.84 J (ans).
Q3= (mass of liquid Al)*(specific heat of liquid Al)*(change in temperature) = (30.10 g)*(1.088 J/g.C)*(1190 – 660)C = 17356.864 J.
The total heat involved in the process is Q = Q1 + Q2 + Q3 = (325.08 + 11991.84 + 17356.864) J = 29673.784 J = (29673.784 J)*(1 kJ/1000 J) = 29.673784 kJ 29.7 kJ (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.