Question

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.

Answer 1

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

(What is the system?)

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

`q = DeltaH`, provided the units are the same.

Thus, we have the relationship

`barul(|stackrel(" ")(" "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 `DeltaH_f` at `0^@ "C"`.

The density of water at `0^@ "C"` is near `"1.00 g/mL"`. That means you have `"250 g"` of water. The units, however, must match the denominator of `DeltaH_f` in order to cancel out.

Therefore, the heat flow involved is

`color(blue)(q) = overbrace((250 cancel"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"))^(DeltaH_f)`

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

to two sig figs, you'd get `ul(-"84 kJ")`, as you have allowed yourself only two. Thus, `"84 kJ"` of thermal energy were released.