6th edition 8.25

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Riya Shah 4H
Posts: 65
Joined: Wed May 02, 2018 3:00 am

6th edition 8.25

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 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 C. What is the change in the internal energy of the neutralization reaction?

For this how do we know that q(calorimeter) = - q(reaction) ?

Chloe Likwong 2K
Posts: 70
Joined: Fri Sep 28, 2018 12:23 am

Re: 6th edition 8.25

I believe it is because the reaction is explicitly stated to release energy so -q(reaction)=-[-q(calorimeter)].
It would then be -q(reaction)=q(calorimeter).

Janice Park 1E
Posts: 36
Joined: Fri Sep 28, 2018 12:22 am

Re: 6th edition 8.25

Hey! Just an elaboration, we know that q(calorimeter) = - q(reaction) because any heat released from the reaction is going to be absorbed by the calorimeter.
Because of this, heat from the reaction should just be negative of the heat absorbed by the calorimeter.

Helen Zhao 1F
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: 6th edition 8.25

In a calorimeter, no heat is lost to the surroundings so the heat released from the reaction is absorbed by the calorimeter. q reaction is negative because heat is lost and q calorimeter is positive because heat is absorbed.