The heat of fusion for ice at $0{ }^{\circ} \mathrm{C}$ is $333.7 \mathrm{~kJ} / \mathrm{kg}$, the specific heat of water $4.186 \mathrm{~kJ} / \mathrm{kg}{ }^{\circ} \mathrm{C}$, and the heat of vaporization of water at $100{ }^{\circ} \mathrm{C}$ is $2,256 \mathrm{~kJ} / \mathrm{kg}$. What is the energy needed to vaporize 100 grams of ice at starting at $0{ }^{\circ} \mathrm{C}$ ?

Question

Quizplus · Accepted Answer

To vaporize 100 grams of ice starting at $0{ }^{\circ} \mathrm{C}$, first, the ice must melt into water, which requires the heat of fusion. Then, the water must be heated from $0{ }^{\circ} \mathrm{C}$ to $100{ }^{\circ} \mathrm{C}$, requiring energy calculated using the specific heat of water. Finally, the water at $100{ }^{\circ} \mathrm{C}$ must be vaporized, which requires the heat of vaporization. The total energy is the sum of these three components:1. Melting ice: $0.1 \, \text{kg} \times 333.7 \, \text{kJ/kg} = 33.37 \, \text{kJ}$2. Heating water from $0{ }^{\circ} \mathrm{C}$ to $100{ }^{\circ} \mathrm{C}$: $0.1 \, \text{kg} \times 4.186 \, \text{kJ/kg}^{\circ} \mathrm{C} \times 100{ }^{\circ} \mathrm{C} = 41.86 \, \text{kJ}$3. Vaporizing water: $0.1 \, \text{kg} \times 2256 \, \text{kJ/kg} = 225.6 \, \text{kJ}$Adding these together: $33.37 \, \text{kJ} + 41.86 \, \text{kJ} + 225.6 \, \text{kJ} = 300.83 \, \text{kJ}$, which rounds to $300.8 \, \text{kJ}$.

The Heat of Fusion for Ice At 0∘C0{ }^{\circ} \mathrm{C}0∘C Is 333.7 kJ/kg333.7 \mathrm{~kJ} / \mathrm{kg}333.7 kJ/kg

The Heat of Fusion for Ice At $0{ }^{\circ} \mathrm{C}$ Is $333.7 \mathrm{~kJ} / \mathrm{kg}$