## Textbook question 4A.13

Paige Lee 1A
Posts: 136
Joined: Sat Sep 07, 2019 12:16 am

### Textbook question 4A.13

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 8C. 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 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

I keep getting +1.19kJ instead of -1.19kJ. Could someone please explain how to get -1.19kJ?

Christine Honda 2I
Posts: 116
Joined: Sat Sep 14, 2019 12:17 am

### Re: Textbook question 4A.13

Qreaction=-Qcalorimeter and so that is why the answer is negative 1.19kJ!

Posts: 97
Joined: Tue Feb 12, 2019 12:15 am

### Re: Textbook question 4A.13

The reaction is mixing with the products so heat will transfer to the insulated chamber of the calorimeter, causing the internal energy to decrease.

aishwarya_atmakuri
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am

### Re: Textbook question 4A.13

The reaction is transferring heat to its surroundings, so the system itself is losing heat. Therefore, it will have a negative value.

Shrayes Raman
Posts: 129
Joined: Sat Jul 20, 2019 12:15 am

### Re: Textbook question 4A.13

If a reaction is exothermic it loses heat to surroundings so the change must be negative. I think you are getting value for calorimeter not system.

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Textbook question 4A.13

try to think of the - and + signs as how we express the energy change from one perspective. Like it was said before, the reaction is exothermic so it'll leave to it's surroundings, lessening the amount of heat available. From this perspective, heat is being taken away and moved somewhere else.