# In a calorimeter, 10.0 g of ice melts at 0°C. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?

Mar 26, 2016

$\text{3.34 kJ}$

#### Explanation:

The enthalpy of fusion, sometimes referred to as latent heat of fusion, tells you how much heat is required in order to convert $\text{1 g}$ of a given substance from solid at its melting point to liquid at its melting point.

In your case, you know that water has an enthalpy of fusion, $\Delta {H}_{\text{fus}}$, equal to ${\text{334 J g}}^{- 1}$.

This tells you that in order to melt $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to $\text{1 g}$ of liquid water at ${0}^{\circ} \text{C}$, you need to supply it with $\text{334 J}$ of heat.

So, if every gram of ice at ${0}^{\circ} \text{C}$ requires $\text{334 J}$ of heat in order to become liquid water at ${0}^{\circ} \text{C}$, it follows that $\text{10.0 g}$ of ice will require

$10.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * overbrace("334 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)(DeltaH_"fus")) = color(green)(|bar(ul(color(white)(a/a)"3,340 J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

If you want, you can express the answer in kilojoules by using the fact that $\text{1 kJ" = 10^3"J}$

"heat needed" = color(green)(|bar(ul(color(white)(a/a)"3.34 kJ"color(white)(a/a)|)))

The answer is rounded to three sig figs.