Calorimetry

[@rinsim]

When we mix MgSO4 with water we end up with a system which is not just water. I think the assumption is that the mixture "behaves like water" so we can use the formula q=n*c_p*(T2-T1) where both n and c_p are for water only as if the entire system was made of water. Is this correct? Or this is always the case when we look at the water system only? Water doesn't react so 100g of water is always 100g of water with dissolved ions which is the other system (1.4g), so the formula for q_water is always applicable even if the general final system is a solution of 101.4g of H2O, Mg^2+ and So4^2-? And because the cup is adiabatic q_water = -q_reaction where q_reaction is the real thing we are interested in. One more question: when we derived that dH = dq @ const p, we assumed that the process was reversible, so dw cancels with p*dV, as p_ext = p. Can we assume that for this case? I guess not: the reaction is "spontaneous" and we need work to get back to the original system (get back to 1.4g of solid MgSo4(s)).

[@PhysicalChemistry]

Excellent questions. (1) Your first instinct was right: we should be using the heat capacity of this particular aqueous MgSO₄ solution. (The magnesium and sulfate ions heat up as well. ) The solution heat capacity won't be the same as Cₚ for pure water. But, because the solution is dilute, it will not be too different. So this is an approximation. (2) You're also right that we assumed that all of the heat produced by the reaction goes into the solution. In reality, some of it will escape into the surroundings (the air and the coffee cup). This is another approximation. (3) Again you're right that ΔH = qₚ only for a reversible process. In this case, though, there's not much PV work being done. The solution doesn't change volume by too much when the MgSO₄ dissolves. So the PV work is tiny. And any adjustments to the PV work due to the volume expanding irreversibly are even more tiny. So this is another approximation, but a good one.