# What is the enthalpy change when "5 g" of Y are dissolved in "100 g" of water to decrease its temperature from 22^@ "C" to 17^@ "C"? C_P = "4.184 J/g"^@ "C" for pure water near 25^@ "C".

Jul 7, 2017

$\Delta H = {q}_{P} = - \text{2196.6 J}$.

It doesn't really matter what $Y$ is, since we are only finding $\Delta H$ in $\text{J}$, not $\text{J/mol}$. We thus don't need the molar mass of $Y$.

At constant pressure, by definition, the heat flow ${q}_{P}$ is given by

${q}_{P} = \Delta H = m {C}_{P} \Delta T$,

where

• $\Delta H$ is the change in enthalpy of the solution.
• $m$ is the mass of the solution.
• ${C}_{P}$ is the specific heat capacity of the solution in $\text{J/g"^@ "C}$.
• $\Delta T$ is the change in temperature of the solution (in guess what units? $\text{K}$? Sure. But we'll use $\text{^@ "C}$).

As a result, there is not much of an extra step here. It's just regular heat flow calculations, with a different label for what $q$ is at constant pressure.

(If it were constant volume, we would have given you an empirically-obtained calorimeter heat capacity in $\text{kJ/"^@ "C}$, and asked for $\Delta E$ instead.)

We assume:

• the same specific heat capacity of water as usual, despite the solution not being pure water
• that ${C}_{P}$ does not change within the temperature range.

$\textcolor{b l u e}{{q}_{P}} = \left(\text{5 g + 100 g")("4.184 J/g"^@ "C")(17.0^@ "C" - 22.0^@ "C}\right)$

$=$ $\textcolor{b l u e}{- \text{2196.6 J}}$