Some properties of aluminum are summarized in the following list.
normal melting point 658°C
heat of fusion 0.395 kJ/g
normal boiling point 2467°C
heat of vaporization 10.52 kJ/g
specific heat of the solid 0.902 J/g°C
(a) Calculate the quantity of energy required to heat 1.75 mol of aluminum from 17°C to its normal melting point.
(b) Calculate the quantity of energy required to melt 1.37 mol of aluminum at 658°C.
(c) Calculate the amount of energy required to vaporize 1.36 mol of aluminum at 2467°C.
chemistry - Jake, Saturday, May 1, 2010 at 5:58pm
I'm the one who posted the question... this is how I'm doing part A. I'm told it's the wrong answer, but I can't figure out what I'm doing wrong. I'm hesitant on going to parts B and C until I understand part A.
Step one: I changed the mols of Al to grams.
13.9mol Al * (26.9815g Al / 1mol Al) = 47.2176g Al
Step two: I plugged the numbers into the Q = s * m * change in T
Q = (0.902 J/g*C)(47.2176g Al) ( 641C)
Q = 27300KJ = 2.73 * 10^4KJ
Again, I'm told I'm wrong here. Any help would be appreciated thanks!
chemistry - DrBob222, Saturday, May 1, 2010 at 7:27pm
I answered this above but generically.
I don't know where you came up with 13.9 mol Al; however, it must just be a typo because the 47.2 is correct for grams.
The Q step is ok, also; however, the answer is not in kJ.
mass (g) x specific heat (J/g*C) x (T in C) gives Joules.
I think that is the trouble you are having.