# What is the heat flow involved for "250 mL" of water that freezes? DeltaH_f = "6.02 kJ/mol". Assume the density of water is "1.00 g/mL" at this temperature.

Jul 31, 2017

Well, you have restricted yourself to two sig figs, so I get $- \text{84 kJ}$ for the heat flow with respect to the system, but a more exact answer would be $- \text{83.54 kJ}$ with respect to the system.

(What is the system?)

Freezing is a phase equilibrium where the temperature and pressure are constant. With constant pressure, we have

$q = \Delta H$, provided the units are the same.

Thus, we have the relationship

$\overline{\underline{| \stackrel{\text{ ")(" "q = n_wDeltaH_f" }}{|}}}$

where $q$ is the heat flow out of ${n}_{w}$ mols of ice that corresponds to a molar enthalpy of fusion $\Delta {H}_{f}$ at ${0}^{\circ} \text{C}$.

The density of water at ${0}^{\circ} \text{C}$ is near $\text{1.00 g/mL}$. That means you have $\text{250 g}$ of water. The units, however, must match the denominator of $\Delta {H}_{f}$ in order to cancel out.

Therefore, the heat flow involved is

$\textcolor{b l u e}{q} = {\overbrace{\left(250 \cancel{\text{mL" xx cancel"1.00 g"/cancel"mL" xx (cancel"1 mol")/(18.015 cancel"g")))^(n_w) xx overbrace((-"6.02 kJ"/cancel"mol}}\right)}}^{\Delta {H}_{f}}$

$=$ color(blue)(-83._(54) " kJ"

to two sig figs, you'd get $\underline{- \text{84 kJ}}$, as you have allowed yourself only two. Thus, $\text{84 kJ}$ of thermal energy were released.