## 4A.13

$\Delta U=q+w$

804994652
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

### 4A.13

In the 7th Ed. Section 4A Q13, the problem states "a constant volume calorimeter was calibrated by carrying out a rxn known to release 3.5 kJ if heat in 0.2 L of sol. in the calorimeter (q=-3.50kJ). When calculating C, the solutions manual changed the q from -3.50 kj to + 3.50 kj...why?

aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

### Re: 4A.13

the q=-Ccal*deltaT when you are calibrating the calorimeter; q=-q(cal), another way is to note that the specific heat of a calorimeter should not be negative so if you get a negative value you should make it positive. hope this helps!

804994652
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

### Re: 4A.13

Thank you!
Do you think that this explanation makes sense: Since the reaction released the 3.5 kj, the calorimeter is getting the 3.50kj of work done on it.. So when work is done on the system (calorimeter), it is (+)?