Difficult (for me) thermochem question

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Swenis

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Hey guys, I've been practicing a lot recently for an upcoming general chem test I have, and I've come across a problem that I can't solve for the life of me. If any of you guys can share any insight into the question, I would be so thankful! 😍

When 21.45 g of potassium nitrate were dissolved in 100.0g of water in a calorimeter, the temperature fell from 25.00oC to 14.14oC. What is the dH for the solution process? The heat capacity for the calorimeter
is 263.6J/oC.

KNO3(s) ----> K+(aq) + NO3-(aq)
 
Swenis said:
Hey guys, I've been practicing a lot recently for an upcoming general chem test I have, and I've come across a problem that I can't solve for the life of me. If any of you guys can share any insight into the question, I would be so thankful! 😍

When 21.45 g of potassium nitrate were dissolved in 100.0g of water in a calorimeter, the temperature fell from 25.00oC to 14.14oC. What is the dH for the solution process? The heat capacity for the calorimeter
is 263.6J/oC.

KNO3(s) ----> K+(aq) + NO3-(aq)


Your question seems kinda odd, I'll assume your calorimeter is a constant pressure calorimeter, in which case the dH is equal to the Q that escapes and warms up the water. Usually though, the heat capacity of the calorimeter is not given, as it merely acts as a form of insulation. Thus your answer can be either the (mass of water)X(heat capacity of water)(dT), or it can be that, plus the change in temperature of the calorimeter i.e. and additional 263.6X(dT).
Or, as calorimeters usually go, it will just be the heatCapacity of the calorimeter times the change in temperature.

Hope that helps. Check to see if any of those three gives you the right answer
 
Well, since q=mc(dT) and c is the specific heat capacity, I first took the 263.6 J/oC and divided it by the amount of solution I have, which is 121.45g (21.45g + 100.0g). This gave me 2.17J/g-oC as the specific heat capacity.

I then plugged everything into the equation: q=(121.45g)(2.17J/g-oC)(-10.86oC)
q= -2862.11 J
q= -2.862 kJ = dH

Did I do this right?
 
Any input, please?
 
don't think so, C would be the specific heat capacity of water °C = 4.186 joule/gram
 
young skywalker said:
don't think so, C would be the specific heat capacity of water °C = 4.186 joule/gram

Yeah, I think you're right. I just don't know why the question provided the heat capacity if it isn't needed. 😕
 
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