## 6th edition 8.25 Calorimeter problem

Kassidy Tran 1E
Posts: 77
Joined: Fri Sep 28, 2018 12:15 am

### 6th edition 8.25 Calorimeter problem

The question is
“A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in .200L of solution in the calorimeter (q = -3.50 kJ), resulting in a temperature rise of 7.32 C. In a subsequent experiment, 100.0 mL of .200 M HBr(aq) and 100.0 mL of KOH(aq) were mixed in the same calorimeter and the temperature rose by 2.49 C. What is the change in the internal energy of the neutralization reaction?”

I attempted this problem by starting off with finding the specific heat capacity of the calorimeter with q/dT, but in the solutions manual, the units of the temperature value 7.32 C was changed to 7.32 K. I was wondering why the units were changed from Celsius to Kelvin, and if this was of any significance to the final answer or not.

Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

### Re: 6th edition 8.25 Calorimeter problem

Normally, it wouldn't matter because celsius and kelvin use the same scale. The difference in temperatures between celsius and kelvin would the same. However, in this case it is probably a typo because 2.49 is given in celsius. Both temperatures have to be using the same temperature scale for the calculation to work.

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