CaCl2(s) -> Ca(aq) + 2Cl(aq) ΔH_rxn = +198.0 KJ

Question

"If that g sample of calcium chloride was dissolved in 1000.0g of water, with both substances originally at 25.0C, what would be the final temperature of the solution be? Assume no loss of heat to the surroundings and that the entire solution has a specific heat capacity of 4.184J/gC."

I'm not quite sure if there are three q's (one for CaCl2, one for H20 and one for the heat absorbed), or just 2 q's which is juse about CaCl2 and H20.). Also getting an answer between 0 to 25C is impossible.

For additional information, the amount of heat that would be absorbed for the same amount of CaCl2 is +40.2KJ.

Answer 1

Ok, if the dissolving gave off 176kj, that will heat the solution

176kj=1000*4.18*deltaTemp

deltTemp=44 C about
final temp, add intial temp 25

Answer 2

Oh god, I messed up my question. Apologizes. It was 22.33g calcium chloride dissolved in 1000.0g of water.

Answer 3

So, from what you're saying:

(Heat Absorbed is 40.2KJ, need to convert that to J or g/J?) = 1000g * 4.184 (T_f - 25)

Answer 4

1) 40200J = (1000g + 22.33g) * 4.184J/gC * (T_f - 25)

2) 40200J = (1022.33) * 4.184J/gC * (T_f - 25)

3) 40200J = 4277.42872 * (T_f - 25)

3) 40200J = 4277.42872T_f - 106935.718

4) 40200J + 106935.718 = 4277.42872T_f - 106935.718 + 106935.718

5) 147135.718 = 4277.42872T_f

6) 147135.718/4277.42872 = 4277.42872T_f/4277.42872

7)34.3981694685 = T_f

=> T_f ~= 34.40C

If I am wrong, please tell me.

Answer 5

Thank you bobpursley, I was unable to get that question right on my homework. ;)