[s3a]

Here is a question that I get partially:

When 1.045 g sample of solid CaO is added to 49.0 mL of water at 25.0 C in a calorimeter,
o
Ca(OH)2(aq) is formed and the temperature of the water rises to 32.3 C. Assuming that the final
o
solution weighs 50.0 g and has a specific heat of 4.184 J/g C. a) Calculate the enthalpy change q, in
joules, accompanying this reaction. Is the process exothermic or endothermic? b) Write the
chemical equation and evaluate ∆H, for the reaction, in kJ/mol.
[ a) q = – 1,527.16 = –1.5 x 103 J b) CaO(s) + H2O(l) = Ca(OH)2(aq), ∆H = – 82 kJ/mol ] (Answers)

I got q = (s)(W)(ΔT) aka (heat capacity)(mass)(Change in Temperature)
q = (1.00cal/gC)(50.0g)(32.3C-25C)
q = 365 calories
q = (365cal)(4.184J/cal)
q = 1527.16J

I got the numerical answer right but given that there are no negative signs, how must I know that this is an exothermic process? Also, to my knowledge q and ΔH are the same thing...what is going on here?

Any help would be greatly appreciated!
Thanks in advance!

[UG]

You know that it is exothermic because the temperature rose from 25 to 32.3. If it was endothermic, the final temperature would be lower than the initial

[s3a]

Well, if it ROSE, it GAINED energy/temperature, didn't it? I would classify that as endothermic. Also, if the final temperature is larger than the initial that means large - small = positive number which still indicates to me that this will be endothermic.

[renge ishyo]

s3A, what UG said was completely correct.

I think you are getting confused between the system (the chemical reaction) and the surroundings (the water). What rose in temperature was the water surrounding the chemical reaction. The only way that can happen is for the reaction to release energy into the surrounding water. Since the reaction is releasing energy to the water it is exothermic. In an endothermic reaction, the reaction would absorb energy from the water, so the temperature of the water would go down.

[s3a]

Oh, now I get it! I was assuming it was all one big thing and I didn't think about the surroundings which in this case would be the water. Thanks to both of you!

Edit: Actually, I get the theory part but the mathematical part, do I just do:

q = (s)(W)(ΔT) aka (heat capacity)(mass)(Change in Temperature)
q = (1.00cal/gC)(50.0g)(32.3C-25C)
q = 365 calories
q = (365cal)(4.184J/cal)
q = 1527.16J

and then add a negative in the end? Or is there a number that I have incorrectly inserted into the formula?