Using the following data for water, determine the energy required to melt 1.00 mole of ice (solid water) at its melting point. $$\begin{array} { | l | l | }
\hline \text { Boiling point } & 373 \mathrm {~K} \
\hline \text { Melting point } & 273 \mathrm {~K} \
\hline \text { Enthalpy of vaporization } & 2,260 \mathrm {~J} / \mathrm { g } \
\hline \text { Enthalpy of fusion } & 334 \mathrm {~J} / \mathrm { g } \
\hline \text { Specific heat capacity (solid) } & 2.11 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline \text { Specific heat capacity (liquid) } & 4.18 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline \text { Specific heat capacity (gas) } & 2.08 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline
\end{array}$$

Question

Using the following data for water, determine the energy required to melt 1.00 mole of ice (solid water) at its melting point. $$\begin{array} { | l | l | } 
\hline \text { Boiling point } & 373 \mathrm {~K} \
\hline \text { Melting point } & 273 \mathrm {~K} \
\hline \text { Enthalpy of vaporization } & 2,260 \mathrm {~J} / \mathrm { g } \
\hline \text { Enthalpy of fusion } & 334 \mathrm {~J} / \mathrm { g } \
\hline \text { Specific heat capacity (solid) } & 2.11 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline \text { Specific heat capacity (liquid) } & 4.18 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline \text { Specific heat capacity (gas) } & 2.08 \mathrm {~J} / ( \mathrm { g } \cdot \mathrm { K } ) \
\hline
\end{array}$$

Quizplus · Accepted Answer

The enthalpy of fusion for water is 334 J/g. To find the energy required to melt 1 mole of ice, convert moles to grams (1 mole of water = 18.02 g) and multiply by the enthalpy of fusion: 334 J/g * 18.02 g = 6016.68 J, which is approximately 6.02 kJ.

Using the Following Data for Water, Determine the Energy Required