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### 6th edition 8.25

Posted: Tue Feb 12, 2019 2:44 pm
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) ?

### Re: 6th edition 8.25

Posted: Tue Feb 12, 2019 2:59 pm
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).

### Re: 6th edition 8.25

Posted: Fri Feb 15, 2019 3:14 am
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.

### Re: 6th edition 8.25

Posted: Fri Feb 15, 2019 12:58 pm
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.

### Re: 6th edition 8.25

Posted: Fri Feb 15, 2019 2:21 pm
A calorimeter is an isolated system so no heat is lost to the surroundings. Any heat released from the reaction will be absorbed by the calorimeter. Therefore, qcalorimeter = -qreaction