How much heat (kJ) must be removed in order to change 1.40 moles of water at 56.0 oC to ice at -22.0 oC ?
The specific heat capacity of solid water is 2.03 J g-1 oC -1
2007-01-26 15:53:54 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Lets find the mass of the H2O
1.40 mol * 18.02gH2O/1mol = 25.228g H2O
Lets find the difference in temperature:
-22 - 56 = -78 C
25.228g * 2.03 = 51.21284J required to move the temperature of your 25.228g of solid H2O by 1 degree C.
Now multiply that by the amount of degrees (-78 C)
51.21284J * -78 C = 3994.60152 J = 3.99 kJ
Sorry, I have somewhat forgotten the real explanation and equations behind this problem since I took AP chem, and I just did this through common sense and experience on how values and data are handled in calculations in chem.
Hope it helps!
2007-01-26 16:06:48 · answer #1 · answered by haxxormaster 2 · 0⤊ 0⤋
Dang now I know why I flunked chemistry. I put my moles in dirt not water. :)
2007-01-27 00:02:21 · answer #2 · answered by southforty1961 3 · 0⤊ 0⤋