Textbook 4A.13 Clarification

[Gwendolyn Hill 2F]

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.

[Jagveer 1I]

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.

[HannahRobinson3L]

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.