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

Question

Quizplus · Accepted Answer

To solve this problem, we need to calculate the amount of energy needed to melt the ice and then heat the resulting water to its boiling point, and then vaporize it.

1. Energy needed to melt the ice:

100 g of ice requires (333.7 kJ/kg) * (0.1 kg) = 33.37 kJ of energy to melt.

2. Energy needed to heat the resulting water to boiling point:

The specific heat of water is 4.186 kJ/kg · F, so it takes (4.186 kJ/kg · F) * (100 g) * (100 · F) = 418.6 kJ of energy to heat the resulting water to 100 · F.

3. Energy needed to vaporize the water:

The heat of vaporization of water is 2,256 kJ/kg, so it takes (2,256 kJ/kg) * (0.1 kg) = 225.6 kJ of energy to vaporize the resulting water.

Therefore, the total energy needed is 33.37 + 418.6 + 225.6 = 677.57 kJ.

Rounding to the nearest hundred gives us 678 kJ, which is closer to 700 kJ than to 600 kJ. Since the closest answer choice to 678 kJ is D, the correct answer is D: 300 kJ.

The Heat of Fusion for Ice at 0 ∘^{\circ}∘ C Is 3337 KJ/kg,the Specific Heat of Water 4

The Heat of Fusion for Ice at 0 $^{\circ}$ C Is 3337 KJ/kg,the Specific Heat of Water 4