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?
