calorimeter

[MichaelMendozaD1F]

A constant‑volume calorimeter was calibrated by carrying out a reaction known to release 2.22 kJ
of heat in 0.300 L
of solution in the calorimeter (q=−2.22 kJ)
, resulting in a temperature rise of 3.52 ∘C
. In a subsequent experiment, 150.0 mL
of 0.30 M HClO2(aq)
and 150.0 mL
of 0.30 M NaOH(aq)
were mixed in the same calorimeter and the temperature rose by 5.60 ∘C
. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?


could anyone help with this problem?

[Chem_Mod]

From what we are given, the reaction in the calorimeter releases -2.22 kJ of heat, which resulting in a temperature rise of 3.52 C in the calorimeter. The first step would to solve for heat capacity, this is the heat capacity of this one specific calorimeter that is used in this experiment. So it is the energy to raise the temperature by 1 C in the calorimeter.

After this first step, we can reword the question as if the energy just solved for is required to raise the temperature of the calorimeter by 1 C, how much heat is released if there's the rise in temperature by 5.60 C in the same calorimeter?

Note: heat capacity is the amount of energy required to raise the temperature of an object(in this case the calorimeter) one degree.

What we're looking at in this question is more about heat capacity of the calorimeter itself rather than specific heat capacity or molar heat capacity of single substances in the reaction.

[madison_bang_1L]

The first set of experimental details are going to be used to calibrate the calorimeter (that is, find the heat capacity). With the equation q = -Ccal*delta T, both q and delta T are known; solve for Ccal. You will then use this heat capacity value and plug it into the equation q = -Ccal*delta T with the new experiment values given. Since there is no work done, q = deltaU (change in internal energy). Ensure to designate when q is negative or positive. It makes sense for q to be negative because whatever heat lost by reaction is gained by the calorimeter.

[Matthew Li 1B]

you would use the equation q=-Ccal*(change in temperature). You would then find the heat capacity of the calorimeter by diving 2.22kj by3.52C. Then you would use that value of Ccal times 5.60 C to find the change in internal energy.