# Question #35b3b

Sep 21, 2017

$6.0 \cdot {10}^{2}$ $\text{J}$

#### Explanation:

The key here is the enthalpy of fusion for ice, which is given to you as

$\Delta {H}_{\text{fus" = color(blue)("333.55 J") color(white)(.)"g}}^{- 1}$

For a given substance, the enthalpy of fusion tells you the amount of heat needed in order to convert $\text{1 g}$ of the substance from solid at its melting point to liquid at its melting point.

In your case, you know that in order to turn $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$, the normal melting point of ice, to liquid water at ${0}^{\circ} \text{C}$, you need $\textcolor{b l u e}{\text{333.55 J}}$ of heat.

This means that your sample will require

$1.8 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * color(blue)("333.55 J")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(6.0 * 10^2color(white)(.)"J}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of the sample.