Hi! So for this question, I understand how to solve it; I just am getting a different answer than is in the solution manual.
"A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solu- tion in the calorimeter (q 5 23.50 kJ), resulting in a temperature rise of 7.32 degrees C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 degrees C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?"
So I used the equation C = q/∆t to get Cv, but I put it in terms of Kelvin, whereas the Solutions Manual says C = (3.50 kJ)/(7.32K), even though the 7.32 given is in celsius... So I ended up using 280.47K, and I got a Cv of 0.012979 kJ/K. The Solutions Manual does it again, saying qcalorimeter = C* 2.49K instead of 2.49C. For these C values, should they be in terms of celsius or in terms of Kelvin? I also think this problem is solved a little strangely in the solutions manual because of this discrepancy, so I'd love it if someone could clarify this :) !
I got an end answer of ∆U = q = -3.44 kJ.
Textbook 4A.13 Clarification
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Re: Textbook 4A.13 Clarification
So for this problem I think that the solution manual is purposely exchanging the units of delta T from C to K to show us that the degree of change in temperature is the same in C and K. This is true because K=C+273.15, so if the temperature is changed by 1 C, then it is also changed by 1 K as well. So, even though the problem gives us delta T in terms of C, the delta T is the same in terms of K as well. Therefore, you could solve it as C = (3.50 kJ)/(7.32K), as the solution manual did, or as C = (3.50 kJ)/(7.32C) and you will get the same answer either way. Hopefully this helps.
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Re: Textbook 4A.13 Clarification
Hello! You should be getting the same value as given in the book regardless of if you use celsius or kelvin as Jagveer mentioned. However, you should still have a value of 7.32 because it is the change in T, so adding the 273.15 to 7.32 for the denominator is actually incorrect. Hope this helps.
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