# How much energy is needed to change 1 g of ice at 0°C to water at 0°C?

Feb 18, 2016

If you think about it for a moment---and it should be intuitive, you should recognize that ice changing into water is a melting event.

Melting is the opposite of fusion, or freezing.

Hence, we should consider "latent heat of fusion", or enthalpy of fusion, $\Delta {H}_{\text{fus}}$. Or, the molar enthalpy of fusion, $\Delta {\overline{H}}_{\text{fus}}$. For that we have the reference value:

color(green)(DeltabarH_("fus","H"_2"O") = "6.02 kJ/mol")

A natural melting event occurs at a constant temperature and constant pressure.

At a constant pressure, $\Delta H = {q}_{p}$, where $q$ is the heat flow through the system, and ${q}_{p}$ is that at constant pressure.

The "energy" you speak of is ${q}_{p}$. Therefore, we should solve for ${q}_{p}$:

$\setminus m a t h b f \left({n}_{\text{H"_2"O}} \Delta \overline{H} = {q}_{p}\right)$

where $n$ is the number of $\text{mol}$s and ${q}_{p}$ is in $\text{kJ}$.

The $\text{H"_2"O}$ in the system technically does not change in quantity while it melts, only in form, so we have:

n_("H"_2"O") = 1 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O"))

$= \text{0.0555 mol H"_2"O}$

Thus:

$\textcolor{b l u e}{{q}_{p}} = \text{0.0555 mol H"_2"O" xx "6.02 kJ/mol}$

= "0.3342 kJ" = color(blue)("334.2 J")

Since we have to supply heat energy into the system to melt the ice, the system absorbs energy. Thus, the sign of this answer should be positive.