Calculating amount of ice required for heat loss Question:


  Given $400.0\ \mathrm{g}$ of hot tea at $80.0 °\mathrm{C}$, what mass of ice at $0 °\mathrm{C}$ must
  be added to obtain iced tea at $10.0 °\mathrm{C}$? The specific heat of the
  tea is $4.18\ \mathrm{J}/(\mathrm{g}\ \cdot°\mathrm{C})$ , and $\Delta H_{\mathrm{fusion}}$ for ice is $+6.01\ \mathrm{kJ\ mol}^{-1}$ .


$q$ is the amount of heat transferred. Since the solution is decreasing heat, then $q_{\mathrm{ice\ tea}}=-q_{\mathrm{ice}}$ We can then solve q using the information given.

$q_{\mathrm{ice}\,\mathrm{tea}}=\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\cdot ^{\circ}\mathrm{C}}\right)(400\,\mathrm{g})(-70\,^{\circ}\mathrm{C})=-1.17\times 10^5\,\mathrm{J}$

The next part cofuses me, this is what the solution manual does.

$q_{\mathrm{ice}}=1.17\times 10^{5}\,\mathrm{J}=(6.01\,\mathrm{kJ\ mol}^
{-1})\left(\dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\right)\biggl(m_{\mathrm{ice}}\times\dfrac{1\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}{18.02\,\mathrm{g}\,\mathrm{H}_{2}\mathrm{O}}\biggr)$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\ \cdot\ ^{\circ}\mathrm{C}}\right)(m_{\mathrm{ice}})(10.0\ ^{\circ}\mathrm{C})$

I'm not too sure where this equation is derived from or of its purpose. The top part is determining the the enthalpy of ice but why is it being added to the specific heat of ice?
HINT: 

$\pu{4.18 J/g K}$ is the specific heat of water. Also question assumes that due to addition of tea leaves the specific heat doesn't change much and remains equal to that of water.

So if you mix ice and tea and want the final temperature at $\pu{10^\circ C}$ then what do you expect of ice. Shouldn't it must melt and reach to $\pu{10^\circ C}$ and that will require a further energy to raise it temperature from $\pu{0^\circ C}$ to $\pu{10^\circ C}$. Also the mass of ice will remain same even if it get converted to water