heat of fusion in an equation You add $100.0\ \mathrm g$ of water at $55.0\ \mathrm{^\circ C}$ to $100.0\ \mathrm g$ of ice at $0.00\ \mathrm{^\circ C}$.  Some of the ice melts and cools the water to $0.00\ \mathrm{^\circ C}$.  When thermal equilibrium is established at $0.00\ \mathrm{^\circ C}$, what mass of ice has melted?
  $\Delta H$ of fusion for water is $333\ \mathrm{J/g}$. 


My question is I got a negative for $q$.  Would this change to a positive because you need heat to melt the ice?  I no you can't have a negative. So I am not sure if it should be a neg or a positive

$q = m \times \text{specific heat} \times (T_\text{final} - T_\text{initial})$
specific heat of water is $4.186$ (should memorize or on a chart)
$q= 100\ \mathrm g \times 4.186\ \mathrm J \times (0 -55) = -23\,023\ \mathrm J$

mass ice × heat of fusion
$q= m \times H_\text{fusion}$
$-23\,023\ \mathrm J = m \times 333\ \mathrm{J/g} = 69.12\ \mathrm g$
If you go by thermodynamic conventions,

$$\begin{align}q &gt; 0 &amp;\implies \text{ heat is gained by system}\\
q &lt; 0 &amp;\implies \text{ heat is lost by system} \end{align}$$


You initially considered water as your system, and it lost heat, therefore  $q$ was negative.
Now when you do calculation for ice, it becomes your system, and the heat is gained by ice, $\implies$ positive $q$