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I was wondering if I could please get some help with this:

In a coffee-cup calorimeter $\pu{100.0 mL}$ of $\pu{1.0 M}\ \ce{NaOH}$ and $\pu{100.0 mL}$ of $\pu{1.0 M}\ \ce{HCl}$ are mixed. Both solutions are originally at $\pu{24.6 ^\circ C}$. After the reaction, the temperature is $\pu{31.3 ^\circ C}$. Assuming all solutions have a density of $\pu{1.0 g/cm^3}$ and a specific heat capacity of $\pu{4.181 J/(g ^\circ C)}$. What is the enthalpy change for the neutralization of $\ce{HCl}$ by $\ce{NaOH}$?"

This is what I did, but I'm not sure if it's correct:

$$ \begin{align} q(\text{surroundings}) &= m \cdot c \cdot \Delta T\\ &= \pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})\\ &= \pu{5.61 kJ}\\[2em] q(\text{system}) - \pu{5.61 kJ} &= n \cdot \Delta H - \pu{5.61 kJ}\\ &= \pu{0.1 mol} \cdot \Delta H \end{align} $$

Thus: $\Delta H = \pu{-56.1 kJ/mol}$

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In general, homework is not welcome here in the forum, which is probably why your question was voted down. However, since you have already given a solution here and only ask for a confirmation, I want to be so nice and reply to your question.

The short answer is: Yes!

In more detail, the following can be said: The heat from the neutralization added to the heat to warm the solution added to the heat to warm the calorimeter has to be equal to zero.

Thus: $Q(\text{neutralization}) + Q(\text{warm the solution}) + Q(\text{warm the calorimeter}) = 0$

In your example, it is assumed for simplicity that the calorimeter does not consume heat to heat up. Therefore $Q(\text{warm the calorimeter}) = 0$.

Thus:

$$ \begin{align} Q(\text{neutralization}) + Q(\text{warm the solution}) &= 0\\ n \Delta H + m c \Delta T &= 0\\ n \Delta H &= - m c \Delta T\\ \Delta H &= -\frac{m c \Delta T}{n}\\ &= \frac{\pu{200 g} \cdot \pu{4.184 J/(g ^\circ C)} \cdot (\pu{31.3 ^\circ C} - \pu{24.6 ^\circ C})}{\pu{0.1 mol}}\\ &= \pu{-56.1 kJ/mol} \end{align} $$

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    $\begingroup$ "In general, homework is not welcome here in the forum, which is probably why your question was voted down." This is absolutely not true. The reasons for the downvotes could be many, sloppy formatting, a title without information, not explaining the thought process. The particular cases like this one, i.e. a please check my homework request, however, are not encouraged (and disliked) here, because they hardy ever lead to a good answer, and they are generally to specific to be of future use. $\endgroup$ – Martin - マーチン Sep 25 '19 at 18:12

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