Is mass (m) ever negligible when solving for heat (q)?

[AndresElizarraras 2A]

I was doing textbook problem 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 5 23.50 kJ), resulting in a temperature rise of 7.32 8C. 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 8C. What is the change in the internal energy of the reaction mixture as a result of the neutralization reaction?"

I got stuck and looked at the solution manual. In their calculations, I noticed that in Q = mcT, they did not account for any sort of mass. Why wasn't mass accounted for if our equation for heat (q) requires it to be solved?

I also noticed that the solution for this problem is written in kelvin (K) despite the solution manual not showing any conversions to this unit. Is the answer not in celsius according to the solution manual's work?

[Alyssa Cheung 3I]

In this scenario, you're looking at the calorimeter as a whole. Therefore, you don't need the mass because you're trying to solve for the heat capacity (aka the energy needed to raise the temperature of the object/entire calorimeter). If you were looking at specific heat capacity, then that would be a different story, and you would have to account for the mass since specific heat capacity is the energy needed to raise the temperature of specifically one gram of the object.

[Sarah Wu 2B]

When solving for just the heat capacity, C = q/T. However, if you are solving for the specific heat capacity, Cs= q/(m*T) which accounts for mass and is taken from the equation q= mCsT. In regards to the solution to the problem, I believe the solution manual should be in degrees C, not K like it is written. However, the final answer, -1.19kJ should still be correct.

[Ben Hsieh 3I]

For this question in particular, you are solving for just the heat capacity not specific heat capacity in regards to a particular situation therefore C can be seen as equal to q/T. Generally, I would not worry about the units for which your answer is as long as they equal the same value post conversion unless specified in the problem. K is the SI unit for temperature though.