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I want to calculate the energy change when a solute is dissolved in water. I know that I can achieve this by using the equation $q=mc \Delta T$. My question is does the mass ($m$) change with the amount of solute added to the solvent?

For example, if I had 25g of water ($c=4.18$) as my solvent and 5g of sodium chloride as my solute. Would the energy change equal:

$$q=25 \cdot 4.18 \cdot \Delta T$$ or $$q=(25+5) \cdot 4.18 \cdot \Delta T$$

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    $\begingroup$ Use 30 grams as your mass since the heat capacity of a sodium chloride solution is about the same as the heat capacity of water. You could use physical chem as a tag. $\endgroup$ – Brinn Belyea Jan 3 '15 at 18:57
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    $\begingroup$ For a rough estimate you can ignore the mass of the solute if it is relatively small compared to the mass of the solvent (probably not in this case but oh well). $\endgroup$ – bon Jan 3 '15 at 19:06
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Your energy conservation equation is $$q_{rxn} + q_{soln} = 0$$ so $$q_{rxn} = -q_{soln} = -m_{soln}c_{soln}\Delta T_{soln}$$ Note the minus sign; if $\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too.

If you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c_{soln} \approx c_{\rm H_2O}$.

If you need more precision, though, you'll need to measure or look up $c_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.

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You have to take the mass of the mixture. So, the second expression is true.

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    $\begingroup$ but you also have to take into account the different specific heat capacity of the solution to water and its temperature dependence $\endgroup$ – bon Jan 3 '15 at 21:10
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    $\begingroup$ You are right. But, generally, we assume that the specific heat capacity of the solution is the specific heat capacity of water. $\endgroup$ – Yomen Atassi Jan 3 '15 at 21:16

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