## Homework Problem 8.25

elizabethrojas1G
Posts: 44
Joined: Fri Sep 25, 2015 3:00 am

### Homework Problem 8.25

#25: A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200L of solution in the calorimeter, (q=-3.50kJ) resulting in a temperature rise of 7.32C. In a subsequent experiment, 100.0 mL of 0.200 M HBr(aq) and 100.0 mL of 0.200M KOH (aq) were mixed in the same calorimeter and the temperature rose by 2.49C. What is the change in the internal energy of the neutralization reaction?

I don't really know how to go about solving the second part of this problem. I used the equation C= (change in internal energy)/(change in temperature) and got C=0.478 kJ/K. How can I use that to find the change in the internal energy of the neutralization reaction?

Thanks!

Divya Prajapati 1E
Posts: 40
Joined: Fri Sep 25, 2015 3:00 am

### Re: Homework Problem 8.25

Hi! Based on your calculations, you found C, the heat capacity of the entire calorimeter (rather than its specific heat capacity) is 0.478 kJ/K.

We can use the equation q = - C x delta T to find how much heat the reaction released into the calorimeter. We know delta T is 2.49 K using the given change in temperature for the calorimeter. When you plug in both C (0.478 kJ/K) and delta T (2.49 K), you find that q is -1.19 kJ. This means the reaction released 1.19 kJ into the calorimeter.

Because the reaction occurs in constant volume, the system cannot perform any work (w = 0). So, change in internal energy is equal to q. Therefore, delta U = - 1.19 kJ.

elizabethrojas1G
Posts: 44
Joined: Fri Sep 25, 2015 3:00 am

### Re: Homework Problem 8.25

Divya Prajapati 1E wrote:Hi! Based on your calculations, you found C, the heat capacity of the entire calorimeter (rather than its specific heat capacity) is 0.478 kJ/K.

We can use the equation q = - C x delta T to find how much heat the reaction released into the calorimeter. We know delta T is 2.49 K using the given change in temperature for the calorimeter. When you plug in both C (0.478 kJ/K) and delta T (2.49 K), you find that q is -1.19 kJ. This means the reaction released 1.19 kJ into the calorimeter.

Because the reaction occurs in constant volume, the system cannot perform any work (w = 0). So, change in internal energy is equal to q. Therefore, delta U = - 1.19 kJ.

Why wouldn't the delta U be positive 1.19kJ if the calorimeter absorbed the heat that the reaction lost? Or is it that the delta U is equal to the heat of the reaction? If so, is it always this case?