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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
$q= m \times H_\text{fusion}$
$-23\,023\ \mathrm J = m \times 333\ \mathrm{J/g} = 69.12\ \mathrm g$

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  • $\begingroup$ The negative q is the amount of heat you need to add to the water to get its temperature from 55 to 0. The negative value means you are removing it from the water and adding the same amount of heat (with a plus sign) to the ice. $\endgroup$ – Chet Miller Nov 1 '15 at 23:10
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If you go by thermodynamic conventions,

$$\begin{align}q > 0 &\implies \text{ heat is gained by system}\\ q < 0 &\implies \text{ heat is lost by system} \end{align}$$

  • 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 $q$

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