6th edition 8.25 Calorimeter problem

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

6th edition 8.25 Calorimeter problem

Postby Kassidy Tran 1E » Fri Feb 08, 2019 6:23 pm

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.

Adrian C 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: 6th edition 8.25 Calorimeter problem

Postby Adrian C 1D » Fri Feb 08, 2019 7:13 pm

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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