How much energy must be supplied to change $36 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ to water at…
How much energy must be supplied to change $36 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ to water at room temperature, $25^{\circ} \mathrm{C} \quad \Delta \mathrm{H}^{\circ}$ fusion $=6.01 \mathrm{~kJ} \mathrm{~mol}^{-1} \quad \mathrm{C}_{\mathrm{p}_{2} \text { liquid }}=4.18 \mathrm{JK}^{-1} \mathrm{~g}^{-1}$
$12 \mathbf{k J}$
$16 \mathrm{~kJ}$
$19 \mathbf{k J}$
$22 \mathrm{~kJ}$
Solution
$\Delta H=\Delta H\left(36 \mathrm{~g}ight.$ ice at $0^{\circ} \mathrm{C}$ to $36 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ at $\left.0^{\circ} \mathrm{C}ight)+\triangle \mathrm{H}\left(36 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}ight.$ at $0^{\circ} \mathrm{C}$ to $36 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}$ at $\left.25^{\circ} \mathrm{C}ight)$
$=\Delta \mathbf{H}_{\text {furion }} \times \mathbf{m o l}+\mathbf{C}_{\mathbf{p}} \mathbf{x}$ mass $\mathbf{x} \Delta \mathbf{T}$
$=\left(6.01 \times \frac{36}{18}ight)+\frac{4.18}{1000} \times 36 \times 25$
$\begin{array}{ll}\mathbf{k J} & \mathbf{k}\end{array}$
$=15.782 \approx 16 \mathrm{~kJ}$
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