a)
q = mass x specific heat x (Tfinal-Tinitial).
b)
q = mass x heat fusion.
c)
q = mass x heat vap
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.
4 answers
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.
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
So the answer to part A is 2.73*10^4kJ
Again, I'm told I'm wrong here. Any help would be appreciated thanks!
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
So the answer to part A is 2.73*10^4kJ
Again, I'm told I'm wrong here. Any help would be appreciated thanks!
Forget it... I got it thanks!!!!!
I answered this below where you showed your work. I think the problem is that you have kJ and it is actually J.