## 4A.13

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 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 solution in the calorimeter
(q = -3.50 kJ), resulting in a temperature rise of 7.32 degrees C. 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 degrees C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?

So the order in which you use the equations is to:
Find the heat capacity from C= qcal / $\Delta$T.
Substitute the C into q = C$\Delta$T.

How do you know when to differentiate using the first equation, which you get from q = C$\Delta$T and when to use it again in the second step? Or in other words, how are you know the different from the heat (q) in the first step to the heat (q) in the second step? Also, how would you assign the signs (=/–) in these cases?

Shimei_2F
Posts: 100
Joined: Fri Aug 09, 2019 12:17 am

### Re: 4A.13

In the first equation, you're calculating for the heat capacity of the Calorimeter using the heat put into the Calorimeter which is why it's qcal. In the second equation, you use it to calculate the heat of the overall change in internal energy from the 2 rises in temperature.

KaleenaJezycki_1I
Posts: 127
Joined: Sat Aug 17, 2019 12:18 am
Been upvoted: 2 times

### Re: 4A.13

So like said above the first part of info (the first q stated) gives us qcal and allows us to solve for the specific heat capacity of the calorimeter. You would do this by using the same equation you stated as q=CdeltaT, but rearrange it as C= cal/delta T. Then you can use this value of C to solve the problem using the second half of the info as q=CdeltaT and use your C value given to find q. Finally, because volume is constant DeltaU=qv so DeltaU (internal energy) is equivalent to your q value you just found.